Chapter 5: Quadratic Equations in One Variable

Exercise 5.1

Question 1

In each of the following, determine whether the given numbers are roots of the given equations or not:

(i) $x^2 - 5x + 6 = 0; 2, -3$

(ii) $3x^2 - 13x - 10 = 0 ; 5, -\dfrac{2}{3}$

Answer

(i) Given,

$x^2 - 5x + 6 = 0$

Substituting x = 2 in the LHS of the given equation, we get:

$\text{LHS} = 2^2 - 5 \times 2 + 6 \\[0.5em] = 4 - 10 + 6 \\[0.5em] = 10 - 10 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ 2 is a root of the given equation.

Substituting x = -3 in the LHS of the given equation, we get:

$\text{LHS} = (-3)^2 - 5 \times (-3) + 6 \\[0.5em] = 9 + 15 + 6 \\[0.5em] = 30$

≠ $\text{RHS}$

∴ 3 is not a root of the given equation.

Hence, 2 is a root but -3 is not a root of the given equation.

(ii) Given,

$3x^2 - 13x - 10 = 0$

Substituting x = 5 in the LHS of the given equation, we get:

$\text{LHS} = 3 \times (5)^2 - 13 \times 5 - 10 \\[0.5em] = 3 \times 25 - 65 -10 \\[0.5em] = 75 - 75 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ 5 is a root of the equation

Substituting x = $-\dfrac{2}{3}$ in the LHS of the given equation, we get:

$\text{LHS} = 3 \times \Big(-\dfrac{2}{3}\Big)^2 - 13 \times \Big( -\dfrac{2}{3} \Big) - 10 \\[1em] = 3 \times \dfrac{4}{9} + \dfrac{26}{3} - 10 \\[1em] = \dfrac{4}{3} + \dfrac{26}{3} - 10 \\[1em] = \dfrac{30}{3} - 10 \\[1em] = 10 - 10 \\[1em] = 0 \\[1em] = \text{RHS}$

∴ $-\dfrac{2}{3}$ is a root of the equation

Hence, both 5 and $-\dfrac{2}{3}$ are roots of the given equation.

Question 2

In each of the following, determine whether the given numbers are solutions of the given equations or not:

(i) $x^2 - 3\sqrt{3}x + 6 = 0; \sqrt{3}, \space -2\sqrt{3}$

(ii) $x^2 - \sqrt{2}x - 4 = 0; -\sqrt{2}, \space 2\sqrt{2}$

Answer

(i) Given,

$x^2 - 3\sqrt{3}x + 6 = 0$

Substituting x = $\sqrt{3}$ in the LHS of the given equation, we get:

$\text{LHS} = (\sqrt{3})^2 - 3\sqrt{3} \times \sqrt{3} + 6 \\[0.5em] = 3 - 9 + 6 \\[0.5em] = 9 - 9 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $\sqrt{3}$ is a solution of the given equation

Substituting x = $-2\sqrt{3}$ in the LHS of the given equation, we get:

$\text{LHS} = (-2\sqrt{3})^2 - 3\sqrt{3} \times -2\sqrt{3} + 6 \\[0.5em] = 12 + 18 + 6 \\[0.5em] = 36$

≠ $\text{RHS}$

∴ $-2\sqrt{3}$ is not a solution of the given equation

Hence, $\sqrt{3}$ is a solution of the given equation but $-2\sqrt{3}$ is not.

(ii) Given,

$x^2 - \sqrt{2}x - 4 = 0$

Substituting x = $-\sqrt{2}$ in the LHS of the given equation, we get:

$\text{LHS} = (-\sqrt{2})^2 - \sqrt{2} \times (-\sqrt{2}) - 4 \\[0.5em] = 2 + 2 - 4 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $-\sqrt{2}$ is a solution of the given equation

Substituting x = $2\sqrt{2}$ in the LHS of the given equation, we get:

$\text{LHS} = (2\sqrt{2})^2 - \sqrt{2} \times 2\sqrt{2} - 4 \\[0.5em] = 8 - 4 - 4 \\[0.5em] = 8 - 8 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $2\sqrt{2}$ is a solution of the given equation

Hence, $-\sqrt{2}$ and $2\sqrt{2}$ both are solutions of the given equation.

Question 3(i)

If $-\dfrac{1}{2}$ is the solution of the equation 3x² + 2kx - 3 = 0, find the value of k.

Answer

Since, $-\dfrac{1}{2}$ is a solution of the equation 3x² + 2kx - 3 = 0, x = $-\dfrac{1}{2}$ satisfies the given equation.

Substituting x = $-\dfrac{1}{2}$ and solving for k we get:

$3\Big(-\dfrac{1}{2}\Big)^2 + 2k \Big(-\dfrac{1}{2}\Big) -3 = 0 \\[1em] \Rightarrow 3 \times \dfrac{1}{4} - k - 3 = 0 \\[1em] \Rightarrow -k = 3 - \dfrac{3}{4} \\[1em] \Rightarrow k = \dfrac{3}{4} - 3 \\[1em] \Rightarrow k = \dfrac{3 - 12}{4} \\[1em] \Rightarrow k = -\dfrac{9}{4}$

Hence, the value of k is $-\dfrac{9}{4}$

Question 3(ii)

If $\dfrac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, find the value of k.

Answer

Since, $\dfrac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, x = $\dfrac{2}{3}$ satisfies the given equation.

Substituting x = $-\dfrac{2}{3}$ and solving for k we get:

$\Rightarrow 7\Big(\dfrac{2}{3}\Big)^2 + k \times \dfrac{2}{3} -3 = 0 \\[1em] \Rightarrow 7 \times \dfrac{4}{9} + \dfrac{2k}{3} -3 = 0 \\[1em] \Rightarrow \dfrac{28}{9} - 3 + \dfrac{2k}{3} = 0 \\[1em] \Rightarrow \dfrac{28}{9} - \dfrac{27}{9} + \dfrac{6k}{9} = 0 \text{ (Taking L.C.M.) } \\[1em] \Rightarrow \dfrac{28 - 27 + 6k}{9} = 0 \\[1em] \Rightarrow 1 + 6k = 0 \\[1em] \Rightarrow 6k = -1 \\[1em] \Rightarrow k = -\dfrac{1}{6}$

Hence, the value of k is $-\dfrac{1}{6}.$

Question 4(i)

If $\sqrt{2}$ is a root of the equation $kx^2 + \sqrt{2}x - 4 = 0$ , find the value of k.

Answer

Since $\sqrt{2}$ is a root of the equation $kx^2 + \sqrt{2}x - 4 = 0$ , $x = \sqrt{2}$ satisfies the given equation.

Substituting $x = \sqrt{2}$ in the given equation:

$k ( \sqrt{2} )^2 + \sqrt{2} \times \sqrt{2} - 4 = 0 \\[0.5em] \Rightarrow 2k + 2 - 4 = 0 \\[0.5em] \Rightarrow 2k - 2 = 0 \\[0.5em] \Rightarrow 2k = 2 \\[0.5em] \Rightarrow k = 1$

Hence, the value of k is 1

Question 4(ii)

If a is the root of the equation x² - (a + b)x + k = 0, find the value of k.

Answer

Since, a is the root of the equation x² - (a + b)x + k = 0 , x = a satisfies the given equation.

Substituting x = a in the given equation:

$\Rightarrow a^2 - (a+b)a + k = 0 \\[0.5em] \Rightarrow a^2 - (a^2 + ab) + k = 0 \\[0.5em] \Rightarrow a^2 - a^2 - ab + k = 0 \\[0.5em] \Rightarrow k - ab = 0 \\[0.5em] \Rightarrow k = ab$

Hence, the value of k is ab.

Question 5

If $\dfrac{2}{3}$ and -3 are roots of the equation px² + 7x + q = 0, find the values of p and q .

Answer

The given equation is px² + 7x + q = 0.

As $\dfrac{2}{3}$ is a root of the equation, so x = $\dfrac{2}{3}$ satisfies the given equation.

Substituting x = $\dfrac{2}{3}$ in the given equation:

$p\big(\dfrac{2}{3} \big)^2 + 7 \times \dfrac{2}{3} + q = 0 \\[1em] \Rightarrow \dfrac{4p}{9} + \dfrac{14}{3} + q = 0 \\[1em] \Rightarrow q = -\dfrac{4p}{9} - \dfrac{14}{3} \\[1em] \Rightarrow q = -\big(\dfrac{4p + 42}{9}\big) \space \text{...(i)} \\[1em]$

Also -3 is a root of the given equation, so x = -3 satisfies the given equation.

Substituting x = -3 in the given equation:

$p(-3)^2 + 7 \times (-3) + q = 0 \\[0.5em] \Rightarrow 9p - 21 + q = 0 \space \text{...(ii)}$

Putting value of q from eqn (i) into eqn (ii)

$9p - 21 - \Big(\dfrac{4p + 42}{9}\Big) = 0$

Taking 9 as the LCM

$\Rightarrow \dfrac{81p - 189 - 4p - 42}{9} = 0 \\[1em] \Rightarrow 77p - 231 = 0 \\[1em] \Rightarrow 77p = 231 \\[1em] \Rightarrow p = \dfrac{231}{77} \\[1em] \Rightarrow p = 3$

Substituting p = 3 in (i), we get

$q = -\Big( \dfrac{4\times 3 + 42}{9} \Big) \\[1em] \Rightarrow q= -\Big(\dfrac{12 + 42}{9}\Big) \\[1em] \Rightarrow q= -\Big(\dfrac{54}{9}\Big) \\[1em] \Rightarrow q= -6$

Hence, p = 3 and q = -6

Exercise 5.2

Question 1(i)

Solve the following equation by factorisation:

x² - 3x - 10 = 0

Answer

Given,

$x^2 - 3x - 10 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 2x - 10 = 0 \\[0.5em] \Rightarrow x(x - 5) + 2(x - 5) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 5) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 5 = 0 \text{ ( Zero - product rule) }\\[0.5em] \Rightarrow x = -2 \text{ or } x = 5$

Hence, the roots of given equation are -2, 5.

Question 1(ii)

Solve the following equation by factorisation:

x(2x + 5) = 3

Answer

Given,

$x(2x + 5) = 3 \\[0.5em] \Rightarrow 2x^2 + 5x = 3 \\[0.5em] \Rightarrow 2x^2 + 5x - 3 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[0.5em] \Rightarrow 2x^2 + 6x - x - 3 = 0 \\[0.5em] \Rightarrow 2x(x + 3) - 1(x + 3) = 0 \\[0.5em] \Rightarrow (2x - 1)(x + 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x - 1 = 0 \text{ or } x + 3 = 0 \text{ (Zero product rule) } \\[0.5em] \Rightarrow 2x = 1 \text{ or } x = -3 \\[0.5em] \Rightarrow x = \dfrac{1}{2} \text{ or } x = -3 \\[0.5em]$

Hence, the roots of given equation are $\dfrac{1}{2}$ , -3.

Question 2(i)

Solve the following equation by factorisation:

3x² - 5x - 12 = 0

Answer

Given,

$3x^2 - 5x - 12 = 0 \\[0.5em] \Rightarrow 3x^2 - 9x + 4x - 12 = 0 \\[0.5em] \Rightarrow 3x(x - 3) + 4(x - 3) = 0 \\[0.5em] \Rightarrow (x - 3)(3x + 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ or } 3x + 4 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = 3 \text{ or } x = -\dfrac{4}{3}$

Hence, the roots of given equation are 3, $-\dfrac{4}{3}$ .

Question 2(ii)

Solve the following equation by factorisation:

21x² - 8x - 4 = 0

Answer

Given,

$21x^2 - 8x - 4 = 0 \\[0.5em] \Rightarrow 21x^2 - 14x + 6x - 4 = 0 \\[0.5em] \Rightarrow 7x(3x - 2) + 2(3x - 2) = 0 \\[0.5em] \Rightarrow (7x + 2)(3x - 2) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 7x + 2 = 0 \text{ or } 3x - 2 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = -\dfrac{2}{7} \text{ or } x = \dfrac{2}{3}$

Hence, the roots of given equation are $-\dfrac{2}{7}$ , $\dfrac{2}{3}$ .

Question 3(i)

Solve the following equation by factorisation:

3x² = x + 4

Answer

Given,

$3x^2 = x + 4 \\[0.5em] \Rightarrow 3x^2 - x - 4 = 0 \text{ (Writing as } ax^2 + bx + c = 0)\\[0.5em] \Rightarrow 3x^2 - 4x + 3x - 4 = 0 \\[0.5em] \Rightarrow x(3x - 4) + 1(3x - 4) = 0 \\[0.5em] \Rightarrow (x + 1)(3x - 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 1 = 0 \text{ or } 3x - 4 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = -1 \text{ or } x = \dfrac{4}{3}$

Hence, the roots of given equation are -1, $\dfrac{4}{3}$ .

Question 3(ii)

Solve the following equation by factorisation:

x(6x - 1) = 35

Answer

Given,

$x(6x - 1) = 35 \\[0.5em] \Rightarrow 6x^2 - x - 35 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[0.5em] \Rightarrow 6x^2 - 15x + 14x - 35 = 0 \\[0.5em] \Rightarrow 3x(2x - 5) + 7(2x - 5) = 0 \\[0.5em] \Rightarrow (3x + 7)(2x - 5) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 3x + 7 = 0 \text{ or } 2x - 5 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow x = -\dfrac{7}{3} \text{ or } x = \dfrac{5}{2}$

Hence, the roots of given equation are $-\dfrac{7}{3}$ , $\dfrac{5}{2}$ .

Question 4(i)

Solve the following equation by factorisation:

6p² + 11p - 10 = 0

Answer

Given,

$6p^2 + 11p - 10 = 0 \\[0.5em] \Rightarrow 6p^2 + 15p - 4p - 10 = 0 \\[0.5em] \Rightarrow 3p(2p + 5) - 2(2p + 5) = 0 \\[0.5em] \Rightarrow (2p + 5)(3p - 2) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2p + 5 = 0 \text{ or } 3p - 2 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow 2p = -5 \text{ or } 3p = 2 \\[0.5em] \Rightarrow p = -\dfrac{5}{2} \text{ or } p = \dfrac{2}{3}$

Hence, the roots of given equation are $-\dfrac{5}{2}$ , $\dfrac{2}{3}$ .

Question 4(ii)

Solve the following equation by factorisation:

$\dfrac{2}{3}x^2 - \dfrac{1}{3}x = 1$

Answer

Given,

$\dfrac{2}{3}x^2 - \dfrac{1}{3}x = 1 \\[1em] \Rightarrow \dfrac{2}{3}x^2 - \dfrac{1}{3}x - 1 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow \dfrac{2}{3}x^2 \times 3 - \dfrac{1}{3}x \times 3 - 1 \times 3 = 0 \times 3 \text{ (Multiplying the equation by 3) } \\[1em] \Rightarrow 2x^2 - x - 3 = 0 \\[1em] \Rightarrow 2x^2 - 3x + 2x - 3 = 0 \\[1em] \Rightarrow x(2x - 3) + 1(2x - 3) = 0 \\[1em] \Rightarrow (x + 1)(2x - 3) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 1 = 0 \text{ or } 2x - 3 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -1 \text{ or } x = \dfrac{3}{2}$

Hence, the roots of given equation are -1, $\dfrac{3}{2}$ .

Question 5(i)

Solve the following equation by factorisation:

3(x - 2)² = 147

Answer

Given,

$3(x-2)^2 = 147 \\[1em] \Rightarrow 3( x^2 - 4x + 4) = 147 \\[1em] \Rightarrow 3x^2 - 12x + 12 = 147 \\[1em] \Rightarrow 3x^2 - 12x + 12 - 147 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 3x^2 - 12x - 135 = 0 \\[1em] \Rightarrow \dfrac{3x^2 - 12x - 135 }{3} = \dfrac{0}{3} \\[1em] \text{ (Dividing the complete equation by 3) } \\[1em] \Rightarrow x^2 - 4x - 45 = 0 \\[1em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[1em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[1em] \Rightarrow (x - 9)(x + 5) = 0 \text{ (Factorising left side) }\\[1em] \Rightarrow x - 9 = 0 \text{ or } x + 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 9 \text { or } x = -5.$

Hence, the roots of given equation are 9, -5.

Question 5(ii)

Solve the following equation by factorisation:

$\dfrac{1}{7}(3x-5)^2 = 28$

Answer

Given,

$\dfrac{1}{7}(3x-5)^2 = 28 \\[1em] \Rightarrow \dfrac{1}{7}( 9x^2 - 30x + 25) = 28 \\[1em] \Rightarrow \dfrac{9}{7}x^2 - \dfrac{30}{7}x + \dfrac{25}{7} = 28 \\[1em] \Rightarrow \dfrac{9}{7}x^2 \times 7 - \dfrac{30}{7}x \times 7 + \dfrac{25}{7} \times 7 = 28 \times 7 \text{ (Multiplying the complete equation by 7) } \\[1em] \Rightarrow 9x^2 - 30x + 25 = 196 \\[1em] \Rightarrow 9x^2 - 30x + 25 - 196 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 9x^2 - 30x - 171 = 0 \\[1em] \Rightarrow 9x^2 - 57x + 27x - 171 = 0 \\[1em] \Rightarrow 3x(3x - 19) + 9(3x - 19) = 0 \\[1em] \Rightarrow (3x + 9)(3x - 19) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 3x + 9 = 0 \text{ or } 3x - 19 = 0 \text{ (Zero - product rule) }\\[1em] \Rightarrow 3x = -9 \text { or } 3x = 19. \\[1em] \Rightarrow x = -\dfrac{9}{3} \text{ or } x = \dfrac{19}{3} \\[1em] \Rightarrow x = -3 \text{ or } x = \dfrac{19}{3} \\[1em]$

Hence, the roots of given equation are -3, $\dfrac{19}{3}$ .

Question 6

Solve the following equation by factorisation:

x² - 4x - 12 = 0 when x ∈ N

Answer

Given,

$x^2 - 4x - 12 = 0 \\[0.5em] \Rightarrow x^2 - 6x + 2x - 12 = 0 \\[0.5em] \Rightarrow x(x - 6) + 2(x - 6) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 6) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 6 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = -2 \text{ or } x = 6$

Since x ∈ N hence x = -2 is not the root.
Hence, the root of given equation is 6.

Question 7

Solve the following equation by factorisation:

2x² - 9x + 10 = 0 , when

(i) x ∈ N
(ii) x ∈ Q

Answer

Given,

$2x^2 - 9x + 10 = 0 \\[0.5em] \Rightarrow 2x^2 - 5x - 4x + 10 = 0 \\[0.5em] \Rightarrow x(2x - 5) - 2(2x - 5) = 0 \\[0.5em] \Rightarrow (x - 2)(2x - 5) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 2 = 0 \text{ or } 2x - 5 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = 2 \text{ or } x =\dfrac{5}{2}$

(i) Hence, the root of given equation is 2 , when x ∈ N

(ii) Hence, the root of given equation is 2, $\dfrac{5}{2}$ , when x ∈ Q

Question 8(i)

Solve the following equation by factorisation:

a²x² + 2ax + 1 = 0 , a ≠ 0.

Answer

Given,

$a^2x^2 + 2ax + 1 = 0 \\[0.5em] \Rightarrow a^2x^2 + ax + ax + 1 = 0 \\[0.5em] \Rightarrow ax(ax + 1) + 1(ax + 1) = 0 \\[0.5em] \Rightarrow (ax + 1)(ax + 1) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow ax + 1 = 0 \text{ (Zero-product rule) }\\[0.5em] \Rightarrow ax = -1 \\[0.5em] \Rightarrow x = -\dfrac{1}{a}$

Hence, the roots of given equation are $-\dfrac{1}{a} , -\dfrac{1}{a}$

Question 8(ii)

Solve the following equation by factorisation:

x² - (p + q)x + pq = 0

Answer

Given,

$x^2 - (p + q)x + pq = 0\\[0.5em] \Rightarrow x^2 - px - qx + pq = 0 \\[0.5em] \Rightarrow x(x - p) - q(x - p) = 0 \\[0.5em] \Rightarrow (x - q)(x - p) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x -q = 0 \text{ or } x - p = 0 \text{ (Zero-product rule)}\\[0.5em] \Rightarrow x = q \text{ or } x = p. \\[0.5em]$

Hence, the roots of given equation are p, q.

Question 9

Solve the following equation by factorisation:

a²x² + (a² + b²)x + b² = 0, a ≠ 0.

Answer

Given,

$a^2x^2 + (a^2 + b^2)x + b^2 = 0 \\[0.5em] \Rightarrow a^2x^2 + a^2x + b^2x + b^2 = 0 \\[0.5em] \Rightarrow a^2x(x + 1) + b^2(x + 1) = 0 \\[0.5em] \Rightarrow (a^2x + b^2)(x + 1) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow a^2x + b^2 = 0 \text{ or } x + 1 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow a^2x = -b^2 \text{ or } x = -1 \\[0.5em] \Rightarrow x = -\dfrac{b^2}{a^2} \text{ or } x =-1 \\[0.5em]$

Hence, the roots of given equation are $-\dfrac{b^2}{a^2}, -1.$

Question 10(i)

Solve the following equation by factorisation:

$\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0$

Answer

Given,

$\sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \\[1em] \Rightarrow x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + \sqrt{3} = 0 \text{ or } \sqrt{3}x + 7 = 0 \text{ (Zero- product rule) } \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } \sqrt{3}x + 7 = 0 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } \sqrt{3}x = -7 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\dfrac{7}{\sqrt{3}} \\[1em] x = -\sqrt{3} \text{ or } x = -\dfrac{7\sqrt{3}}{3}$

Hence, the roots of given equation are $-\sqrt{3}, -\dfrac{7\sqrt{3}}{3}$

Question 10(ii)

Solve the following equation by factorisation:

$4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0$

Answer

Given,

$4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \\[1em] \Rightarrow 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0 \\[1em] \Rightarrow 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \\[1em] \Rightarrow (\sqrt{3}x + 2)(4x - \sqrt{3}) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow \sqrt{3}x + 2 = 0 \text{ or } 4x - \sqrt{3} = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow \sqrt{3}x = -2 \text{ or } 4x = \sqrt{3} \\[1em] \Rightarrow x = -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4} \\[1em] \Rightarrow x = -\dfrac{2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4} \\[1em] x = -\dfrac{2\sqrt{3}}{3} \text{ or } x = \dfrac{\sqrt{3}}{4}$

Hence, the roots of given equation are $-\dfrac{2\sqrt{3}}{3}$ , $\dfrac{\sqrt{3}}{4}.$

Question 11(i)

Solve the following equation by factorisation:

$x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0$

Answer

Given,

$x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0 \\[1em] \Rightarrow x^2 - x - \sqrt{2}x + \sqrt{2} = 0 \\[1em] \Rightarrow x(x - 1) - \sqrt{2}(x - 1) = 0 \\[1em] \Rightarrow (x - \sqrt{2})(x - 1) = 0 \text{ (Factorising left side) } \\[1em] x - \sqrt{2} = 0 \text{ or } x - 1 = 0 \text{ (Zero-product rule) } \\[1em] x = \sqrt{2} \text{ or } x = 1$

Hence, the roots of given equation are $\sqrt{2}$ , 1.

Question 11(ii)

Solve the following equation by factorisation:

$x + \dfrac{1}{x} = 2\dfrac{1}{20}$

Answer

Given,

$x + \dfrac{1}{x} = 2\dfrac{1}{20} \\[1em] \Rightarrow x \times x + \dfrac{1}{x} \times x = \dfrac{41}{20} \times x \\[1em] \Rightarrow x^2 + 1 = \dfrac{41}{20}x \\[1em] \Rightarrow 20(x^2 + 1) = 41x \\[1em] \Rightarrow 20x^2 + 20 = 41x \\[1em] \Rightarrow 20x^2 - 41x + 20 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 20x^2 - 25x - 16x + 20 = 0 \\[1em] \Rightarrow 5x(4x - 5) - 4(4x - 5) = 0 \\[1em] \Rightarrow (5x - 4)(4x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 5x - 4 = 0 \text{ or } 4x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 5x = 4 \text{ or } 4x = 5 \\[1em] x = \dfrac{4}{5} \text{ or } x = \dfrac{5}{4} \\[1em]$

Hence, the roots of given equation are $\dfrac{4}{5}$ , $\dfrac{5}{4}$ .

Question 12(i)

Solve the following equation by factorisation:

$\dfrac{2}{x^2} - \dfrac{5}{x} + 2 = 0, x$ ≠ 0

Answer

Given,

$\dfrac{2}{x^2} - \dfrac{5}{x} + 2 = 0 \\[1em] \Rightarrow \dfrac{2 - 5x + 2x^2}{x^2} = 0 \\[1em] \Rightarrow 2 - 5x + 2x^2 = 0 \times x^2 \\[1em] \Rightarrow 2x^2 - 5x + 2 = 0 \\[1em] \Rightarrow 2x^2 - 4x - x + 2 = 0 \\[1em] \Rightarrow2x(x - 2) - 1(x - 2) = 0 \\[1em] \Rightarrow (2x - 1)(x - 2) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x - 1 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 2x = 1 \text{ or } x = 2 \\[1em] x = \dfrac{1}{2} \text{ or } x = 2 \\[1em]$

Hence, the roots of given equation are $\dfrac{1}{2}$ , 2.

Question 12(ii)

Solve the following equation by factorisation:

$\dfrac{x^2}{15} - \dfrac{x}{3} -10 = 0.$

Answer

Given,

$\dfrac{x^2}{15} - \dfrac{x}{3} -10 = 0 \\[1em] \Rightarrow \dfrac{x^2 - x \times 5 - 10 \times 15}{15} = 0 \\[1em] \Rightarrow x^2 - 5x - 150 = 0 \\[1em] \Rightarrow x^2 - 15x + 10x - 150 = 0 \\[1em] \Rightarrow x(x - 15) + 10(x - 15) = 0 \\[1em] \Rightarrow (x + 10)(x - 15) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 10 = 0 \text{ or } x - 15 = 0 \text{ (Zero-product rule) } \\[1em] x = -10 \text{ or } x = 15 \\[1em]$

Hence, the roots of given equation are -10 , 15.

Question 13(i)

Solve the following equation by factorisation:

$3x - \dfrac{8}{x} = 2$

Answer

Given,

$3x - \dfrac{8}{x} = 2 \\[1em] \Rightarrow \dfrac{3x^2 - 8}{x} = 2 \\[1em] \Rightarrow 3x^2 - 8 = 2x \\[1em] \Rightarrow 3x^2 - 2x - 8 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 3x^2 - 6x + 4x - 8 = 0 \\[1em] \Rightarrow 3x(x - 2) + 4(x - 2) = 0 \\[1em] \Rightarrow (3x + 4)(x - 2) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 3x + 4 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow 3x = -4 \text { or } x = 2 \\[1em] x = -\dfrac{4}{3} \text{ or } x = 2 \\[1em]$

Hence, the roots of given equation are $-\dfrac{4}{3}$ , 2.

Question 13(ii)

Solve the following equation by factorisation:

$\dfrac{x + 2}{x + 3} = \dfrac{2x - 3}{3x - 7}$

Answer

Given,

$\dfrac{x + 2}{x + 3} = \dfrac{2x - 3}{3x - 7} \\[1em] \Rightarrow (x + 2) \times (3x - 7) = (2x - 3) \times (x + 3) \\[1em] \Rightarrow 3x^2 - 7x + 6x - 14 = 2x^2 + 6x - 3x - 9 \\[1em] \Rightarrow 3x^2 - x - 14 = 2x^2 + 3x - 9 \\[1em] \Rightarrow 3x^2 - 2x^2 - x - 3x - 14 + 9 = 0 \\[1em] \Rightarrow x^2 - 4x - 5 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 - 5x + x - 5 = 0 \\[1em] \Rightarrow x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (x - 5)(x + 1) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 5 = 0 \text{ or } x + 1 = 0 \text{ (Zero-product rule) } \\[1em] x = 5 \text{ or } x = -1 \\[1em]$

Hence, the roots of given equation are -1, 5.

Question 14(i)

Solve the following equation by factorisation:

$\dfrac{8}{x + 3} - \dfrac{3}{2 - x} = 2$

Answer

Given,

$\dfrac{8}{x + 3} - \dfrac{3}{2 - x} = 2 \\[1em] \Rightarrow \dfrac{8(2 - x) - 3(x + 3)}{(x + 3)(2 - x)} = 2 \\[1em] \Rightarrow 8(2 - x) - 3(x + 3) = 2(x + 3)(2 - x) \\[1em] \Rightarrow 16 - 8x - 3x - 9 = 2(2x - x^2 + 6 - 3x) \\[1em] \Rightarrow 7 - 11x = 2(- x^2 - x + 6) \\[1em] \Rightarrow 7 - 11x = -2x^2 - 2x + 12 \\[1em] \Rightarrow 7 - 11x + 2x^2 + 2x - 12 = 0 \\[1em] \Rightarrow 2x^2 - 9x - 5 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow 2x^2 - 10x + x - 5 = 0 \\[1em] \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (2x + 1)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] x = -\dfrac{1}{2} \text{ or } x = 5$

Hence, the roots of given equation are $-\dfrac{1}{2}$ , 5.

Question 14(ii)

Solve the following equation by factorisation:

$\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 2\dfrac{1}{2}$

Answer

Given,

$\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 2\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{x \times x + (x - 1)(x - 1)}{x(x - 1)} = 2\dfrac{1}{2} \\[1em] \Rightarrow x^2 + x^2 - x - x + 1 = \dfrac{5x(x - 1)}{2} \\[1em] \Rightarrow 2x^2 - 2x + 1 = \dfrac{5x^2 - 5x}{2} \\[1em] \Rightarrow 2(2x^2 - 2x + 1) = 5x^2 - 5x \\[1em] \Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x \\[1em] \Rightarrow 4x^2 - 5x^2 - 4x + 5x + 2 = 0 \\[1em] \Rightarrow -x^2 + x + 2 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 - x - 2 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) \text{ (Factorising left side) } \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \text{ (Zero-product rule) } \\[1em] x = -1 \text{ or } x = 2 \\[1em]$

Hence, the roots of given equation are -1, 2.

Question 15(i)

Solve the following equation by factorisation:

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3$

Answer

Given,

$\dfrac{x + 1}{x - 1} + \dfrac{x - 2}{x + 2} = 3 \\[1em] \Rightarrow \dfrac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = 3 \\[1em] \Rightarrow \dfrac{x^2 + 2x + x + 2 + x^2 - x - 2x + 2}{(x - 1)(x + 2)} = 3 \\[1em] \Rightarrow x^2 + 2x + x + 2 + x^2 - x - 2x + 2 = 3(x - 1)(x + 2) \\[1em] \Rightarrow x^2 + x^2 + 3x - 3x + 2 + 2 = 3(x^2 + 2x - x - 2) \\[1em] \Rightarrow 2x^2 + 4 = 3(x^2 + x - 2) \\[1em] \Rightarrow 2x^2 + 4 = 3x^2 + 3x - 6 \\[1em] \Rightarrow 2x^2 - 3x^2 - 3x + 4 + 6 = 0 \\[1em] \Rightarrow -x^2 - 3x + 10 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 + 3x - 10 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow x^2 + 5x - 2x - 10 = 0 \\[1em] \Rightarrow x(x + 5) - 2(x + 5) = 0 \\[1em] \Rightarrow (x - 2)(x + 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 5 = 0 \text{ (Zero-product rule) } \\[1em] x = 2 \text{ or } x = -5 \\[1em]$

Hence, the roots of given equation are 2 , -5.

Question 15(ii)

Solve the following equation by factorisation:

$\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6}$

Answer

Given,

$\dfrac{1}{x - 3} - \dfrac{1}{x + 5} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \dfrac{1}{6} \\[1em] \Rightarrow x + 5 - x + 3 = \dfrac{(x - 3)(x + 5)}{6} \\[1em] \Rightarrow 8 = \dfrac{x^2 + 5x - 3x - 15 }{6} \\[1em] \Rightarrow 8 \times 6 = x^2 + 2x - 15 \\[1em] \Rightarrow 48 = x^2 + 2x - 15 \\[1em] \Rightarrow x^2 + 2x - 15 = 48 \\[1em] \Rightarrow x^2 + 2x - 15 - 48 = 0 \\[1em] \Rightarrow x^2 + 2x - 63 = 0 \text{ (Writing as } ax^2 + bx + c = 0) \\[1em] \Rightarrow x^2 + 9x - 7x - 63 = 0 \\[1em] \Rightarrow x(x + 9) - 7(x + 9) = 0 \\[1em] \Rightarrow (x - 7)(x + 9) = 0 \text{ (Factorising left side) } \\[1em] x - 7 = 0 \text{ or } x + 9 = 0 \text{ (Zero-product rule) } \\[1em] x = 7 \text{ or } x = -9 \\[1em]$

Hence, the roots of given equation are -9 , 7.

Question 16(i)

Solve the following equation by factorisation:

$\dfrac{a}{ax - 1} + \dfrac{b}{bx - 1} = a + b, a + b$ ≠ 0, $ab$ ≠ 0

Answer

Given,

$\dfrac{a}{ax - 1} + \dfrac{b}{bx - 1} = a + b \\[1em] \Rightarrow \dfrac{a}{ax - 1} + \dfrac{b}{bx - 1} - a - b = 0 \\[1em] \Rightarrow \big( \dfrac{a}{ax - 1} - b \big) + \big(\dfrac{b}{bx -1} - a \big) = 0 \\[1em] \Rightarrow \dfrac{a - b(ax - 1)}{ax - 1} + \dfrac{b - a(bx - 1)}{bx -1} = 0 \\[1em] \Rightarrow \dfrac{a - abx + b}{ax -1} + \dfrac{b - abx + a}{bx - 1} = 0 \\[1em] \Rightarrow (a - abx + b) \big( \dfrac{1}{ax - 1} + \dfrac{1}{bx - 1}\big) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow a - abx + b = 0 \text{ or } \dfrac{1}{ax - 1} + \dfrac{1}{bx - 1} = 0 \text{ (Zero-product rule) }\\[1em] \Rightarrow a + b = abx \text{ or } \dfrac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)} = 0 \\[1em] \Rightarrow x = \dfrac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \times (ax - 1)(bx - 1) \\[1em] \Rightarrow x = \dfrac{a + b}{ab} \text{ or } ax + bx - 2 = 0 \\[1em] \Rightarrow x = \dfrac{a + b}{ab} \text{ or } x(a + b) = 2 \\[1em] x = \dfrac{a + b}{ab} \text{ or } x = \dfrac{2}{(a + b)} \\[1em]$

Hence, the roots of given equation are $\dfrac{a + b}{ab} , \dfrac{2}{(a + b)}$ .

Question 16(ii)

Solve the following equation by factorisation:

$\dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x}$

Answer

Given,

$\dfrac{1}{2a + b + 2x} = \dfrac{1}{2a} + \dfrac{1}{b} + \dfrac{1}{2x} \\[1em] \Rightarrow \dfrac{1}{2a + b + 2x} - \dfrac{1}{2x} = \dfrac{1}{2a} + \dfrac{1}{b} \\[1em] \Rightarrow \dfrac{2x - (2a + b + 2x)}{(2a + b + 2x)(2x)} = \dfrac{b + 2a}{2ab} \\[1em] \Rightarrow \dfrac{ -(2a + b )}{(2a + b + 2x)(2x)} = \dfrac{b + 2a}{2ab}\\[1em] \Rightarrow \dfrac{ -1}{(2a + b + 2x)(2x)} = \dfrac{1}{2ab}\\[1em] \Rightarrow -2ab = (2a + b + 2x)(2x) \\[1em] \Rightarrow -2ab = 4ax + 2bx + 4x^2 \\[1em] \Rightarrow -ab = 2ax + bx + 2x^2 \text{ (Dividing the complete equation by 2) } \\[1em] \Rightarrow 2ax + bx + 2x^2 + ab = 0 \\[1em] \Rightarrow 2x^2 + 2ax + bx + ab = 0 \\[1em] \Rightarrow 2x(x + a) + b(x + a) = 0 \\[1em] \Rightarrow (2x + b)(x + a) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow 2x + b = 0 \text{ or } x + a = 0 \text{ (Zero-product rule) } \\[1em] x = -\dfrac{b}{2} \text{ or } x = -a \\[1em]$

Hence, the roots of given equation are $-\dfrac{b}{2}$ , -a.

Question 17

Solve the following equation by factorisation:

$\dfrac{1}{x + 6} + \dfrac{1}{x - 10} = \dfrac{3}{x - 4}$

Answer

Given,

$\dfrac{1}{x + 6} + \dfrac{1}{x - 10} = \dfrac{3}{x - 4} \\[1em] \Rightarrow \dfrac{x - 10 + x + 6}{(x + 6)(x - 10)} = \dfrac{3}{x - 4} \\[1em] \Rightarrow \dfrac{2x - 4}{(x + 6)(x - 10)} = \dfrac{3}{x - 4} \\[1em] \Rightarrow (2x - 4)(x - 4) = 3(x + 6)(x - 10) \\[1em] \Rightarrow 2x^2 - 8x - 4x + 16 = 3(x^2 - 10x + 6x - 60) \\[1em] \Rightarrow 2x^2 - 12x + 16 = 3(x^2 - 4x - 60) \\[1em] \Rightarrow 2x^2 - 12x + 16 = 3x^2 - 12x - 180 \\[1em] \Rightarrow 2x^2 - 3x^2 -12x + 12x + 16 + 180 = 0 \\[1em] \Rightarrow -x^2 + 196 = 0 \\[1em] \Rightarrow x^2 = 196 \\[1em] \Rightarrow x = \sqrt{196} \\[1em] x = 14 , -14 \\[1em]$

Hence, the roots of given equation are 14 , -14.

Question 18(i)

Solve the following equation by factorisation:

$\sqrt{3x + 4} = x$

Answer

Given,

$\sqrt{3x + 4} = x$

On squaring both sides, we get

$3x + 4 = x^2 \\[0.5em] \Rightarrow x^2 - 3x - 4 = 0 \\[0.5em] \Rightarrow x^2 - 4x + x - 4 = 0 \\[0.5em] \Rightarrow x(x - 4) + 1(x - 4) = 0 \\[0.5em] \Rightarrow (x + 1)(x - 4) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x + 1 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) }\\[0.5em] x = -1 \text{ or } x = 4$

As equation is squared so roots need to be checked so putting x = -1 and x = 4 in the equation $\sqrt{3x + 4} = x.$
Checking for x = -1

$\Rightarrow \sqrt{3 \times -1 + 4} = -1 \\[0.5em] \Rightarrow \sqrt{-3 + 4} = -1 \\[0.5em] \Rightarrow \sqrt{1} = -1 \text{ (This equation is false) }$

Checking for x = 4

$\Rightarrow \sqrt{3 \times 4 + 4} = 4 \\[0.5em] \Rightarrow \sqrt{12 + 4} = 4 \\[0.5em] \Rightarrow \sqrt{16} = 4 \text{ (This equation is true ) }\\[0.5em]$

Since for x = -1 equation is false hence x = -1 is not the root of the given equation.
Hence, the root of given equation is 4.

Question 18(ii)

Solve the following equation by factorisation:

$\sqrt{x(x - 7)} = 3\sqrt{2}$

Answer

Given,

$\sqrt{x(x - 7)} = 3\sqrt{2}$

On squaring both sides, we get

$x(x - 7) = 18 \\[0.5em] \Rightarrow x^2 - 7x = 18 \\[0.5em] \Rightarrow x^2 - 7x - 18 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 2x - 18 = 0 \\[0.5em] \Rightarrow x(x - 9) + 2(x - 9) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 9) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 9 = 0 \text{ (Zero-product rule) } \\[0.5em] x = -2 \text{ or } x = 9.$

As equation is squared so roots need to be checked so putting x = -2 and x = 9 in the equation $\sqrt{x(x - 7)} = 3\sqrt{2}$
Checking for x = -2

$\Rightarrow \sqrt{-2(-2 - 7)} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{-2(-9)} = 3\sqrt{(2)}\\[0.5em] \Rightarrow \sqrt{18} = 3\sqrt{2} \text{ (This equation is true) }\\[0.5em]$

Checking for x = 9

$\Rightarrow \sqrt{9(9 - 7)} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{9 \times 2} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{18} = 3\sqrt{2} \text{ (This equation is true) }$

As the above two equations are true,

∴ The roots of given equation are -2, 9.

Question 19

Use the substitution y = 3x + 1 to solve for x :

5(3x + 1)² + 6(3x + 1) - 8 = 0.

Answer

Given,

$5(3x + 1)^2 + 6(3x + 1) - 8 = 0 \\[1em] \Rightarrow 5y^2 + 6y - 8 = 0 \text{ (Putting 3x + 1 = y) } \\[1em] \Rightarrow 5y^2 + 10y - 4y - 8 = 0 \\[1em] \Rightarrow 5y(y + 2) - 4(y + 2) = 0 \\[1em] \Rightarrow (5y - 4)(y + 2) = 0 \text{ (Factorising left side) }\\[1em] \Rightarrow 5y - 4 = 0 \text{ or } y + 2 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow y = \dfrac{4}{5} \text{ or } y = -2 \\[1em] \text{When } y = \dfrac{4}{5}, 3x + 1 = \dfrac{4}{5} \\[1em] \Rightarrow 3x = \dfrac{4}{5} - 1 \\[1em] \Rightarrow 3x = -\dfrac{1}{5} \\[1em] \Rightarrow x = -\dfrac{1}{15} \\[1em] \text{When } y = -2 , 3x + 1 = -2 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1$

Hence, the roots of given equation are $-\dfrac{1}{15}$ , -1.

Question 20

Find the values of x if p + 1 = 0 and x² + px - 6 = 0.

Answer

Since, p + 1 = 0 it means p = -1.

Given,

$x^2 + px - 6 = 0 \\[0.5em] \Rightarrow x^2 + (-1)x - 6 = 0 \\[0.5em] \Rightarrow x^2 - x - 6 = 0 \\[0.5em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[0.5em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] x = -2 \text{ or } x = 3$

Hence, the values of x are -2 , 3.

Question 21

Find the values of x if p + 7 = 0, q - 12 = 0 and x² + px + q = 0.

Answer

Since , p + 7 = 0, q - 12 = 0 it means p = -7 and q = 12.

Given,

$x^2 + px + q = 0 \\[0.5em] \Rightarrow x^2 + (-7)x + 12 = 0 \\[0.5em] \Rightarrow x^2 - 7x + 12 = 0 \\[0.5em] \Rightarrow x^2 - 4x - 3x + 12 = 0 \\[0.5em] \Rightarrow x(x - 4) - 3(x - 4) = 0 \\[0.5em] \Rightarrow (x - 3)(x - 4) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) } \\[0.5em] x = 3 \text{ or } x = 4$

Hence, the values of x are 3 , 4.

Question 22

If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p.

Answer

If x = p is a solution of the equation x(2x + 5) = 3 , then x = p satisfies the equation.
Putting x = p in equation,

$p(2p + 5) = 3 \\[0.5em] \Rightarrow 2p^2 + 5p = 3 \\[0.5em] \Rightarrow 2p^2 + 5p - 3 = 0 \\[0.5em] \Rightarrow 2p^2 + 6p - p - 3 = 0 \\[0.5em] \Rightarrow 2p(p + 3) - 1(p + 3) = 0 \\[0.5em] \Rightarrow (2p - 1)(p + 3) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow 2p - 1 = 0 \text{ or } p + 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow 2p = 1 \text{ or } p = -3 \\[0.5em] p = \dfrac{1}{2} \text{ or } p = -3$

Hence, the values of p are -3 , $\dfrac{1}{2}$ .

Question 23

If x = 3 is a solution of the equation (k + 2)x² - kx + 6 = 0 , find the value of k. Hence, find the other root of the equation.

Answer

If x = 3 is a solution of the equation (k + 2)x² - kx + 6 = 0 , then x = 3 satisfies the equation.
Putting x = 3 in equation,

$(k + 2)3^2 - 3k + 6 = 0 \\[0.5em] \Rightarrow 9(k + 2) - 3k + 6 = 0 \\[0.5em] \Rightarrow 9k + 18 - 3k + 6 = 0 \\[0.5em] \Rightarrow 6k + 24 = 0 \\[0.5em] \Rightarrow 6k = -24 \\[0.5em] \Rightarrow k = -\dfrac{24}{6} \\[0.5em] k = -4 \\[0.5em]$

Putting value of k in equation in order to find other root

$\Rightarrow (-4 + 2)x^2 - (-4)x + 6 = 0 \\[0.5em] \Rightarrow -2x^2 + 4x + 6 = 0 \\[0.5em] \Rightarrow 2x^2 - 4x - 6 = 0 \text{ (Multiplying equation by -1) }\\[0.5em] \Rightarrow 2x^2 - 6x + 2x - 6 = 0 \\[0.5em] \Rightarrow 2x(x - 3) + 2(x - 3) = 0 \\[0.5em] \Rightarrow (2x + 2)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x + 2 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow 2x = -2 \text{ or } x = 3 \\[0.5em] x = -1 \text{ or } x = 3 .$

Hence, the values of k is -4 ,and the other root is -1.

Exercise 5.3

Question 1(i)

Solve the following equations by using formula:

2x² - 7x + 6 = 0

Answer

The given equation is 2x² - 7x + 6 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -7 , c = 6

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-7) ± \sqrt{(-7^2) - 4\times 2 \times 6}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{7 ± \sqrt{49 - 48}}{4} \\[1em] \Rightarrow x = \dfrac{7 ± \sqrt{1}}{4} \\[1em] \Rightarrow x = \dfrac{7 + 1}{4} \text{ or } \dfrac{7 - 1}{4} \\[1em] \Rightarrow x = \dfrac{8}{4} \text { or } \dfrac{6}{4} \\[1em] x = 2 \text{ or } \dfrac{3}{2}$

Hence roots of the given equation are 2 , $\dfrac{3}{2}$ .

Question 1(ii)

Solve the following equations by using formula:

2x² - 6x + 3 = 0

Answer

The given equation is 2x² - 6x + 3 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -6 , c = 3

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6^2) - 4\times 2 \times 3}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 24}}{4} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{12}}{4} \\[1em] \Rightarrow x = \dfrac{6 + \sqrt{12}}{4} \text{ or } \dfrac{6 - \sqrt{12}}{4} \\[1em] \Rightarrow x = \dfrac{6 + 2\sqrt{3}}{4} \text { or } \dfrac{6 - 2\sqrt{3}}{4} \\[1em] x = \dfrac{3 + \sqrt{3}}{2} \text{ or } \dfrac{3 - \sqrt{3}}{2}$

Hence roots of the given equation are $\dfrac{3 + \sqrt{3}}{2} , \dfrac{3 - \sqrt{3}}{2}$ .

Question 2(i)

Solve the following equations by using formula:

256x² - 32x + 1 = 0

Answer

The given equation is 256x² - 32x + 1 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 256 , b = -32 , c = 1

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-32) ± \sqrt{(-32)^2 - 4\times 256 \times 1}}{2 \times 256} \\[1em] \Rightarrow x = \dfrac{32 ± \sqrt{1024 - 1024}}{512} \\[1em] \Rightarrow x = \dfrac{32 ± \sqrt{0}}{512} \\[1em] \Rightarrow x = \dfrac{32 + 0}{512} \text{ or } \dfrac{32 - 0}{512} \\[1em] \Rightarrow x = \dfrac{32}{512} \text{ or } \dfrac{32 - 0}{512} \\[1em] x = \dfrac{1}{16} \text{ or } \dfrac{1}{16}$

Hence roots of the given equation are $\dfrac{1}{16} , \dfrac{1}{16}$ .

Question 2(ii)

Solve the following equations by using formula:

25x² + 30x + 7 = 0

Answer

The given equation is 25x² + 30x + 7 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 25 , b = 30 , c = 7

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(30) ± \sqrt{(30^2) - 4\times 25 \times 7}}{2 \times 25} \\[1em] \Rightarrow x = \dfrac{-30 ± \sqrt{900 - 700}}{50} \\[1em] \Rightarrow x = \dfrac{-30 ± \sqrt{200}}{50} \\[1em] \Rightarrow x = \dfrac{-30 + \sqrt{200}}{50} \text{ or } \dfrac{-30 - \sqrt{200}}{50} \\[1em] \Rightarrow x = \dfrac{-30 + 10\sqrt{2}}{50} \text { or } \dfrac{-30 - 10\sqrt{2}}{50} \\[1em] x = \dfrac{-3 + \sqrt{2}}{5} \text{ or } \dfrac{-3 - \sqrt{2}}{5}$

Hence roots of the given equation are $\dfrac{-3 + \sqrt{2}}{5} , \dfrac{-3 - \sqrt{2}}{5}$ .

Question 3(i)

Solve the following equations by using formula:

$2x^2 + \sqrt{5}x - 5 = 0$

Answer

The given equation is $2x^2 + \sqrt{5}x - 5 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = $\sqrt{5}$ , c = -5

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(\sqrt{5}) ± \sqrt{(\sqrt{5})^2 - 4\times 2 \times -5}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} ± \sqrt{5 + 40}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} ± \sqrt{45}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} + \sqrt{45}}{4} \text{ or } \dfrac{-\sqrt{5} - \sqrt{45}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} + 3\sqrt{5}}{4} \text { or } \dfrac{-\sqrt{5} - 3\sqrt{5}}{4} \\[1em] \Rightarrow x = \dfrac{\sqrt{5}(-1 + 3)}{4} \text{ or } \dfrac{\sqrt{5}(-1 - 3)}{4} \\[1em] x = \dfrac{\sqrt{5}}{2} \text{ or } -\sqrt{5}$

Hence roots of the given equations are $\dfrac{\sqrt{5}}{2} , -\sqrt{5}$ .

Question 3(ii)

Solve the following equations by using formula:

$\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0$

Answer

The given equation is $\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0$

Comparing it with ax² + bx + c = 0, we get
$a = \sqrt{3}, \space b =10, \space c = -8\sqrt{3}$

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(10) ± \sqrt{(10)^2 - 4\times \sqrt{3} \times -8\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow x = \dfrac{-10 ± \sqrt{100 + 96}}{2\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{-10 ± \sqrt{196}}{2\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{-10 + 14}{2\sqrt{3}} \text{ or } \dfrac{-10 - 14}{2\sqrt{3}}\\[1em] \Rightarrow x = \dfrac{4}{2\sqrt{3}} \text { or } \dfrac{-24}{2\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{2}{\sqrt{3}} \text{ or } \dfrac{-12}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \text{ or } \dfrac{-12}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \text{ (Multiplying both roots by } \dfrac{\sqrt{3}}{\sqrt{3}}) \\[1em] \Rightarrow x = \dfrac{2\sqrt{3}}{3} \text{ or } -4\sqrt{3}.$

Hence roots of the given equations are $\dfrac{2\sqrt{3}}{3} , -4\sqrt{3}$ .

Question 4(i)

Solve the following equations by using formula:

$\dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4$

Answer

Given,

$\dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4 \\[1em] \Rightarrow \dfrac{(x - 2)^2 + (x + 2)^2}{(x - 2)(x + 2)} = 4 \\[1em] \Rightarrow \dfrac{x^2 + 4 - 4x + x^2 + 4 + 4x}{x^2 - 2x + 2x - 4} = 4 \\[1em] \Rightarrow \dfrac{2x^2 + 8}{x^2 - 4} = 4 \\[1em] \Rightarrow 2x^2 + 8 = 4(x^2 - 4) \\[1em] \Rightarrow 2x^2 + 8 = 4x^2 - 16 \\[1em] \Rightarrow 2x^2 - 4x^2 + 8 + 16 = 0 \\[1em] \Rightarrow -2x^2 + 24 = 0 \\[1em] 2x^2 - 24 = 0 \text{ (Multiplying equation by -1) }$

The equation is $2x^2 - 24 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 0 , c = -24

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-0 ± \sqrt{0^2 - 4 \times 2 \times (-24)}}{2 \times 2} \\[1em] \Rightarrow \dfrac{ ±\sqrt{192}}{4} \\[1em] \Rightarrow \dfrac{±8\sqrt{3}}{4} \\[1em] +2\sqrt{3} \text { or } -2\sqrt{3} \\[1em]$

Hence roots of the given equations are $2\sqrt{3} , -2\sqrt{3}$ .

Question 4(ii)

Solve the following equations by using formula:

$\dfrac{x + 1}{x + 3} = \dfrac{3x + 2}{2x + 3}$

Answer

Given,

$\dfrac{x + 1}{x + 3} = \dfrac{3x + 2}{2x + 3} \\[0.5em] \Rightarrow (x + 1)(2x + 3) = (3x + 2)(x + 3) \text{ On cross multiplication}\\[0.5em] \Rightarrow 2x^2 + 3x + 2x + 3 = 3x^2 + 9x + 2x + 6 \\[0.5em] \Rightarrow 2x^2 + 5x + 3 = 3x^2 + 11x + 6 \\[0.5em] \Rightarrow 2x^2 - 3x^2 + 5x - 11x + 3 - 6 = 0 \\[0.5em] \Rightarrow -x^2 - 6x - 3 = 0 \\[0.5em] \Rightarrow x^2 + 6x + 3 = 0 \text{ (Multiplying equation by -1) } \\[0.5em]$

The equation is x² + 6x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 6 , c = 3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-6 ± \sqrt{6^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] \Rightarrow \dfrac{-6 ±\sqrt{24}}{2} \\[1em] \Rightarrow \dfrac{-6 + \sqrt{24}}{2} \text { or } \dfrac{-6 - \sqrt{24}}{2} \\[1em] \Rightarrow \dfrac{-6 + 2\sqrt{6}}{2} \text{ or } \dfrac{-6 - 2\sqrt{6}}{2} \\[1em] -3 + \sqrt{6} \text{ or } -3 - \sqrt{6}$

Hence roots of the given equations are $-3 + \sqrt{6} , -3 - \sqrt{6}$ .

Question 5(i)

Solve the following equations by using formula:

a(x² + 1) = (a² + 1)x , a ≠ 0

Answer

Given,

a(x² + 1) = (a² + 1)x

First converting the equation in the form ax² + bx + c = 0.

$\Rightarrow ax^2 + a = a^2x + x \\[0.5em] \Rightarrow ax^2 - a^2x - x + a = 0 \\[0.5em] \Rightarrow ax^2 -(a^2 + 1)x + a = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0, we get
a = a , b = -(a² + 1) , c = a

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-(a^2 + 1)) ± \sqrt{(-(a^2 + 1))^2 - 4 \times a \times a}}{2 \times a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{(a^2 + 1)^2 - 4a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{a^4 + 1 + 2a^2 - 4a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{a^4 + 1 - 2a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{(a^2 - 1)^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± (a^2 - 1) }{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 + a^2 - 1}{2a} \text{ or } \dfrac{a^2 + 1 - a^2 + 1}{2a} \\[1em] \Rightarrow \dfrac{2a^2}{2a} \text{ or } \dfrac{2}{2a} \\[1em] \Rightarrow a \text{ or } \dfrac{1}{a}$

Hence roots of the given equations are a , $\dfrac{1}{a}$ .

Question 5(ii)

Solve the following equations by using formula:

4x² - 4ax + (a² - b²) = 0

Answer

The given equation is 4x² - 4ax + (a² - b²) = 0.

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -4a , c = a² - b²

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-4a) ± \sqrt{(-4a)^2 - 4 \times 4 \times (a^2 - b^2)}}{2 \times 4} \\[1em] \Rightarrow \dfrac{4a ± \sqrt{16a^2 - 16(a^2 - b^2)}}{8} \\[1em] \Rightarrow \dfrac{4a ± \sqrt{16a^2 - 16a^2 + 16b^2}}{8} \\[1em] \Rightarrow \dfrac{4a ± \sqrt{16b^2}}{8} \\[1em] \Rightarrow \dfrac{4a ± 4b}{8} \\[1em] \Rightarrow \dfrac{4a + 4b}{8} \text{ or } \dfrac{4a - 4b}{8} \\[1em] \Rightarrow \dfrac{a + b}{2} \text{ or } \dfrac{a - b}{2} \\[1em]$

Hence roots of the given equations are $\dfrac{a + b}{2} , \dfrac{a - b}{2}$ .

Question 6(i)

Solve the following equations by using formula:

$x - \dfrac{1}{x} = 3 , x$ ≠ 0

Answer

Given,

$x - \dfrac{1}{x} = 3 , x$ ≠ 0

$\Rightarrow \dfrac{x^2 - 1}{x} = 3 \text{ (By taking L.C.M) } \\[0.5em] \Rightarrow x^2 - 1 = 3x \\[0.5em] \Rightarrow x^2 - 3x - 1 = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -3 , c = -1

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-3) ± \sqrt{(-3)^2 - 4 \times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{9 - (-4)}}{2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \dfrac{3 + \sqrt{13}}{2} \text{ or } \dfrac{3 - \sqrt{13}}{2} \\[1em]$

Hence roots of the given equations are $\dfrac{3 + \sqrt{13}}{2} , \dfrac{3 - \sqrt{13}}{2}$ .

Question 6(ii)

Solve the following equations by using formula:

$\dfrac{1}{x} + \dfrac{1}{x - 2} = 3 , x$ ≠ $0, 2.$

Answer

Given,

$\dfrac{1}{x} + \dfrac{1}{x - 2} = 3 \\[1em] \Rightarrow \dfrac{x - 2 + x }{x(x - 2)} = 3 \\[1em] \Rightarrow \dfrac{2x - 2}{x^2 - 2x} = 3 \\[1em] \Rightarrow 2x - 2 = 3(x^2 - 2x) \\[1em] \Rightarrow 2x - 2 = 3x^2 - 6x \\[1em] \Rightarrow 3x^2 - 6x - 2x + 2 = 0 \\[1em] \Rightarrow 3x^2 - 8x + 2 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -8 , c = 2

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-8) ± \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \\[1em] \Rightarrow \dfrac{-(-8) ± \sqrt{(-8)^2 - 24}}{6} \\[1em] \Rightarrow \dfrac{8 ± \sqrt{64 - 24}}{6} \\[1em] \Rightarrow \dfrac{8 ± \sqrt{40}}{6} \\[1em] \Rightarrow \dfrac{8 ± 2\sqrt{10}}{6} \\[1em] \Rightarrow \dfrac{4 ± \sqrt{10}}{3} \\[1em] \Rightarrow \dfrac{4 + \sqrt{10}}{3} \text{ or } \dfrac{4 - \sqrt{10}}{3}$

Hence roots of the given equations are $\dfrac{4 + \sqrt{10}}{3} , \dfrac{4 - \sqrt{10}}{3}$ .

Question 7

Solve for x :

$2\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5, x$ ≠ $-3, \dfrac{1}{2}$

Answer

Given,

$2\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5 \\[1em] \Rightarrow \text{ Taking } y = \dfrac{2x - 1}{x + 3} \text{ the equation becomes} \\[1em] \Rightarrow 2y - \dfrac{3}{y} = 5 \\[1em] \Rightarrow 2y^2 - 3 = 5y \\[1em] \Rightarrow 2y^2 - 5y - 3 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -5 , c = -3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-5) ± \sqrt{(-5)^2 - 4 \times 2 \times -3}}{2 \times 2} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{25 + 24}}{4} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{49}}{4} \\[1em] \Rightarrow \dfrac{5 ± 7}{4} \\[1em] \Rightarrow \dfrac{5 + 7}{4} \text{ or } \dfrac{5 - 7}{4}\\[1em] \Rightarrow \dfrac{12}{4} \text{ or } \dfrac{-2}{4} \\[1em] \Rightarrow 3 \text{ or } -\dfrac{1}{2} \\[1em]$

But,

$y = \dfrac{2x - 1}{x + 3} \\[1em] \therefore 3 = \dfrac{2x - 1}{x + 3} \text{ or } -\dfrac{1}{2} = \dfrac{2x - 1}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 2x - 1 \text{ or } -(x + 3) = 2(2x - 1) \\[1em] \Rightarrow 3x + 9 = 2x - 1 \text{ or } -x - 3 = 4x - 2 \\[1em] \Rightarrow 3x - 2x = -1 - 9 \text{ or } -x - 4x = -2 + 3 \\[1em] \Rightarrow x = -10 \text{ or } -5x = 1 \\[1em] \Rightarrow x = -10 \text{ or } x = -\dfrac{1}{5} \\[1em]$

Hence roots of the given equations are -10 , $-\dfrac{1}{5}$ .

Question 8

Solve the following quadratic equations for x and give your answer correct to 2 decimal places :

(i) x² - 5x - 10 = 0

(ii) x² + 7x = 7

Answer

(i) The given equation is x² - 5x - 10 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -5 , c = -10

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-5) ± \sqrt{(-5)^2 - 4\times 1 \times -10}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{25 + 40}}{2} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{65}}{2} \\[1em] \Rightarrow x = \dfrac{5 + \sqrt{65}}{2} \text{ or } \dfrac{5 - \sqrt{65}}{2} \\[1em] \text{ Also } \sqrt{65} = 8.062 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{5 + 8.062}{2} \text { or } \dfrac{5 - 8.062}{2} \\[1em] \Rightarrow x = \dfrac{13.062}{2} \text{ or } \dfrac{-3.062}{2} \\[1em] \Rightarrow x = 6.531 \text{ or } -1.531 \\[1em] x = 6.53 \text{ or } -1.53 \text{ (correct to two decimal places) }$

Hence roots of the given equations are 6.53 , -1.53.

(ii) Given, x² + 7x = 7

or , x² + 7x - 7 = 0

The given equation is x² + 7x - 7 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 7 , c = -7

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(7) ± \sqrt{(7)^2 - 4\times 1 \times -7}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{-7 ± \sqrt{49 + 28}}{2} \\[1em] \Rightarrow x = \dfrac{-7 ± \sqrt{77}}{2} \\[1em] \Rightarrow x = \dfrac{-7 + \sqrt{77}}{2} \text{ or } \dfrac{-7 - \sqrt{77}}{2} \\[1em] \text{ Also } \sqrt{77} = 8.775 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{-7 + 8.775}{2} \text { or } \dfrac{-7 - 8.775}{2} \\[1em] \Rightarrow x = \dfrac{1.775}{2} \text{ or } \dfrac{-15.775}{2} \\[1em] \Rightarrow x = 0.885 \text{ or } -7.885 \\[1em] \Rightarrow x = 0.89\text{ or } -7.89 \text{ (correct to two decimal places) }$

Hence roots of the given equations are 0.89 , -7.89.

Question 9

Solve the following equations by using quadratic formula and give answer in correct to 2 decimal places :

(i) 4x² - 5x - 3 = 0

(ii) x² - 7x + 3 = 0

Answer

(i) The given equation is 4x² - 5x - 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -5 , c = -3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-5) ± \sqrt{(-5)^2 - 4\times 4 \times -3}}{2 \times 4} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{25 + 48}}{8} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{73}}{8} \\[1em] \Rightarrow x = \dfrac{5 + \sqrt{73}}{8} \text{ or } \dfrac{5 - \sqrt{73}}{8} \\[1em] \text{ Also } \sqrt{73} = 8.54 \text{ (From tables)} \\[1em] \Rightarrow x = \dfrac{5 + 8.54}{8} \text { or } \dfrac{5 - 8.54}{8} \\[1em] \Rightarrow x = \dfrac{13.54}{8} \text{ or } \dfrac{-3.54}{8} \\[1em] \Rightarrow x = 1.69 \text{ or } -0.44$

Hence roots of the given equations are 1.69, -0.44.

(ii) For a quadratic equation in the form :

ax² + bx + c = 0

The solutions are :

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Comparing equation x² - 7x + 3 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -7 and c = 3

$\Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] = \dfrac{7 \pm \sqrt{49 - 12}}{2} \\[1em] = \dfrac{7 \pm \sqrt{37}}{2} \\[1em] = \dfrac{7 \pm 6.08}{2} \\[1em] = \dfrac{13.08}{2} \text{ or } \dfrac{0.92}{2}\\[1em] = 6.54 \text{ or } 0.46$

Hence, the value of x = 6.54 or 0.46

Question 10

Solve the following quadratic equations and give your answer correct to two significant figures :

(i) x² - 4x - 8 = 0

(ii) $x -\dfrac{18}{x} = 6$

Answer

(i) The given equation is x² - 4x - 8 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -4 , c = -8

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-4) ± \sqrt{(-4)^2 - 4\times 1 \times -8}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{4 ± \sqrt{16 + 32}}{2} \\[1em] \Rightarrow x = \dfrac{4 ± \sqrt{48}}{2} \\[1em] \Rightarrow x = \dfrac{4 + \sqrt{48}}{2} \text{ or } \dfrac{4 - \sqrt{48}}{2} \\[1em] \text{ Also } \sqrt{48} = 6.928 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{4 + 6.928}{2} \text { or } \dfrac{4 - 6.928}{2} \\[1em] \Rightarrow x = \dfrac{10.928}{2} \text{ or } \dfrac{-2.928}{2} \\[1em] \Rightarrow x = 5.464 \text{ or } -1.464 \\[1em] x = 5.5 \text{ or } -1.5 \text{ (correct to two significant figures) }$

Hence roots of the given equations are 5.5 , -1.5.

(ii) Given,

$\Rightarrow x - \dfrac{18}{x} = 6 \\[1em] \Rightarrow \dfrac{x^2 - 18}{x} = 6 \\[1em] \Rightarrow x^2 - 18 = 6x \\[1em] \Rightarrow x^2 - 6x - 18 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -6 , c = -18

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times -18}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 + 72}}{2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{108}}{2} \\[1em] \Rightarrow x = \dfrac{6 + \sqrt{108}}{2} \text{ or } \dfrac{6 - \sqrt{108}}{2} \\[1em] \text{ Also } \sqrt{108} = 10.392 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{6 + 10.392}{2} \text { or } \dfrac{6 - 10.392}{2} \\[1em] \Rightarrow x = \dfrac{16.392}{2} \text{ or } \dfrac{-4.392}{2} \\[1em] \Rightarrow x = 8.196 \text{ or } -2.195 \\[1em] \Rightarrow x = 8.2 \text{ or } -2.2 \text{ (correct to two significant figures) }$

Hence roots of the given equations are 8.2 , -2.2.

Question 11

Solve the equation 2x² - 10x + 5 = 0 and give your answer correct to 3 significant figures.

Answer

Comparing equation 2x² - 10x + 5 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -10 and c = 5

By formula,

⇒ x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2} \\[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \dfrac{10 \pm \sqrt{60}}{4} \\[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \\[1em] = \dfrac{2(5 \pm \sqrt{15})}{4} \\[1em] = \dfrac{5 \pm \sqrt{15}}{2} \\[1em] = \dfrac{5 \pm 3.87}{2} \\[1em] = \dfrac{5 + 3.87}{2}, \dfrac{5 - 3.87}{2} \\[1em] = \dfrac{8.87}{2}, \dfrac{1.13}{2} \\[1em] = 4.44, 0.565$

Hence, x = 4.44, 0.565

Exercise 5.4

Question 1

Find the discriminant of the following quadratic equations and hence find the nature of roots :

(i) 3x² - 5x - 2 = 0

(ii) 2x² - 3x + 5 = 0

(iii) 16x² - 40x + 25 = 0

(iv) 2x² + 15x + 30 = 0

Answer

(i) The given equation is 3x² - 5x - 2 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -5 , c = -2

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$(-5)^2 - 4 \times 3 \times -2 \\[0.5em] = 25 + 24 \\[0.5em] = 49 \gt 0$

Discriminant = 49; Hence, the given equation has two distinct real roots.

(ii) The given equation is 2x² - 3x + 5 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -3 , c = 5
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-3)^2 - 4 \times 2 \times 5 \\[0.5em] = 9 - 40 \\[0.5em] = -31 \lt 0$

Discriminant = -31; Hence, the given equation has no real roots.

(iii) The given equation is 16x² - 40x + 25 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 16 , b = -40 , c = 25
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-40)^2 - 4 \times 16 \times 25 \\[0.5em] = 1600 - 1600 \\[0.5em] = 0$

Discriminant = 0; Hence, the given equation has two equal real roots.

(iv) The given equation is 2x² + 15x + 30 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 15 , c = 30
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(15)^2 - 4 \times 2 \times 30 \\[0.5em] = 225 - 240 \\[0.5em] = -15 \lt 0$

Discriminant = -15; Hence, the given equation has no real roots.

Question 2

Discuss the nature of the roots of the following quadratic equations :

(i) $3x^2 - 4\sqrt{3}x + 4 = 0$

(ii) $x^2 - \dfrac{1}{2}x + 4 = 0$

(iii) $-2x^2 + x + 1 = 0$

(iv) $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$

Answer

(i) The given equation is $3x^2 - 4\sqrt{3}x + 4 = 0$ .

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = $-4\sqrt{3}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-4\sqrt{3})^2 - 4 \times 3 \times 4 \\[0.5em] = 48 - 48 \\[0.5em] = 0$

Hence the given equation has two equal real roots.

(ii) The given equation is $x^2 - \dfrac{1}{2}x + 4 = 0$ .
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\dfrac{1}{2}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$\Big(-\dfrac{1}{2}\Big)^2 - 4 \times 1 \times 4 \\[1em] = \dfrac{1}{4} - 16 \\[1em] = \dfrac{1 - 64}{4} \text{ Taking L.C.M. } \\[1em] = -\dfrac{63}{4} \lt 0$

Hence the given equation has no real roots.

(iii)The given equation is -2x² + x + 1 = 0.
Comparing it with ax² + bx + c = 0, we get
a = -2 , b = 1 , c = 1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(1)^2 - 4 \times -2 \times 1 \\[0.5em] = 1 + 8 \\[0.5em] = 9 \gt 0 \\[0.5em]$

Hence, the given equation has two distinct real roots.

(iv) The given equation is $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$ .
Comparing it with ax² + bx + c = 0, we get
a = $2\sqrt{3}$ , b = -5 , c = $\sqrt{3}$
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-5)^2 - 4 \times 2\sqrt{3} \times \sqrt{3} \\[0.5em] = 25 - 24 \\[0.5em] = 1 \gt 0$

Hence, the given equation has two distinct real roots.

Question 3

Find the nature of roots of the following quadratic equations :

(i) $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$

(ii) $x^2 - 2\sqrt{3}x - 1 = 0$

If real roots exist, find them.

Answer

(i) The given equation is $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\dfrac{1}{2}$ , c = $-\dfrac{1}{2}$

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$\Big(-\dfrac{1}{2}\Big)^2 - 4 \times 1 \times -\dfrac{1}{2} \\[1em] = \dfrac{1}{4} + 2 \\[1em] = \dfrac{1 + 8}{4} \\[1em] = \dfrac{9}{4} \\[1em]$

Discriminant = $\dfrac{9}{4}$
Since Discriminant > 0, hence the given equation has two distinct real roots.

The roots of the equation are given by:

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-\Big(-\dfrac{1}{2}\Big) ± \sqrt{\Big(-\dfrac{1}{2}\Big)^2 - 4\times 1 \times -\dfrac{1}{2}}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{1}{4} + 2}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{9}{4}}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{9}{4}}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \sqrt{\dfrac{9}{4}}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} + \dfrac{3}{2}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \dfrac{3}{2}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{4}{2}}{2} \text{ or } \dfrac{-\dfrac{2}{2}}{2} \\[1em] \Rightarrow x = \dfrac{4}{4} \text{ or } -\dfrac{2}{4} \\[1em] \Rightarrow x = 1 \text{ or } -\dfrac{1}{2}$

Hence roots of the given equations are 1 , $-\dfrac{1}{2}$ .

(ii) The given equation is $x^2 - 2\sqrt{3}x - 1 = 0$
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-2\sqrt{3}$ , c = -1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-2\sqrt{3})^2 - 4 \times 1 \times -1 \\[1em] = 12 + 4 \\[1em] = 16 \\[1em]$

Discriminant = 16 ,
Since Discriminant > 0, hence given equation have real and distinct roots.

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-2\sqrt{3}) ± \sqrt{(-2\sqrt{3})^2 - 4\times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} ± \sqrt{12 + 4}}{2} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} ± \sqrt{16}}{2} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} + 4}{2} \text{ or } \dfrac{2\sqrt{3} - 4}{2} \\[1em] \Rightarrow x = \sqrt{3} + 2 \text{ or } \sqrt{3} - 2 \\[1em]$

Hence roots of the given equations are , $\sqrt{3} + 2 , \sqrt{3} - 2$ .

Question 4

Without solving the following quadratic equations, find the value of 'p' for which the given equations have real and equal roots :

(i) px² - 4x + 3 = 0

(ii) x² + (p - 3)x + p = 0

Answer

(i) The given equation is px² - 4x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = p , b = -4 , c = 3

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-4)^2 - 4 \times p \times 3 \\[1em] = 16 - 12p$

For equal roots, discriminant = 0

$\Rightarrow 16 - 12p = 0 \\[1em] \Rightarrow 16 = 12p \\[1em] \Rightarrow p = \dfrac{16}{12} \\[1em] p = \dfrac{4}{3}$

Hence the value of p is $\dfrac{4}{3}$ .

(ii) The given equation is x² + (p - 3)x + p = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = (p - 3) , c = p

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (p - 3)^2 - 4 \times 1 \times p \\[1em] = p^2 + 9 - 6p - 4p \\[1em] = p^2 + 9 - 10p$

For equal roots, discriminant = 0

$\Rightarrow p^2 + 9 - 10p = 0 \\[1em] \Rightarrow p^2 - 10p + 9 = 0 \\[1em] \Rightarrow p^2 - 9p - p + 9 = 0 \\[1em] \Rightarrow p(p - 9) - 1(p - 9) = 0 \\[1em] \Rightarrow (p - 1)(p - 9) = 0 \\[1em] \Rightarrow (p - 1) = 0 \text{ or } p - 9 = 0 \\[1em] \Rightarrow p = 1 \text{ or } p = 9 .$

Hence the value of p is 1, 9.

Question 5

Find the values of k for which each of the following quadratic equation has equal roots :

(i) x² + 4kx + (k² - k + 2) = 0

(ii) (k - 4)x² + 2(k - 4)x + 4 = 0

Answer

(i) The given equation is x² + 4kx + (k² - k + 2) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 4k , c = (k² - k + 2)

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (4k)^2 - 4 \times 1 \times (k^2 - k + 2) \\[1em] = 16k^2 - 4(k^2 - k + 2) \\[1em] = 16k^2 - 4k^2 + 4k - 8 \\[1em] = 12k^2 + 4k - 8$

For equal roots, discriminant = 0

$\Rightarrow 12k^2 + 4k - 8 = 0 \\[1em] \Rightarrow 12k^2 + 12k - 8k - 8 = 0 \\[1em] \Rightarrow 12k(k + 1) - 8(k + 1) = 0 \\[1em] \Rightarrow (k + 1) - (12k - 8) = 0 \\[1em] \Rightarrow (k + 1)(12k - 8) = 0 \\[1em] \Rightarrow k + 1 = 0 \text{ or } 12k - 8 = 0 \\[1em] \Rightarrow k = -1 \text{ or } k = \dfrac{8}{12} \\[1em] \Rightarrow k = -1 \text{ or } k = \dfrac{2}{3}$

Hence, the value of k is -1, $\dfrac{2}{3}$ .

(ii) The given equation is (k - 4)x² + 2(k - 4)x + 4 = 0.

Comparing with ax² + bx + c = we obtain,
a = k - 4 , b = 2(k - 4) , c = 4

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (2k - 8)^2 - 4 \times k - 4 \times 4 \\[1em] = 4k^2 + 64 - 32k - 16(k - 4) \\[1em] = 4k^2 - 32k - 16k + 64 + 64 \\[1em] = 4k^2 - 48k + 128$

For equal roots, discriminant = 0

$\Rightarrow 4k^2 - 48k + 128 = 0 \\[1em] \Rightarrow 4(k^2 - 12k + 32) = 0 \\[1em] \Rightarrow k^2 - 12k + 32 = 0 \\[1em] \Rightarrow k^2 - 8k - 4k + 32 = 0 \\[1em] \Rightarrow k(k - 8) - 4(k - 8) = 0 \\[1em] \Rightarrow k - 4 = 0 \text{ or } k - 8 = 0 \\[1em] \Rightarrow k = 4 \text{ or } k = 8 \\[1em]$

k ≠ 4 , as that will make a = (k - 4) = 0 and thus roots will become = ∞ .

Hence, the value of k is 8.

Question 6

Find the value(s) of m for which each of the following quadratic equation has real and equal roots :

(i) (3m + 1)x² + 2(m + 1)x + m = 0

(ii) x² + 2(m - 1)x + (m + 5) = 0

Answer

(i) The given equation is (3m + 1)x² + 2(m + 1)x + m = 0.

Comparing with ax² + bx + c = we obtain,
a = 3m + 1 , b = 2(m + 1) , c = m

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m + 2)^2 - 4 \times 3m + 1 \times m \\[0.5em] = 4m^2 + 4 + 8m - 4m(3m + 1) \\[0.5em] = 4m^2 + 4 + 8m - 12m^2 - 4m \\[0.5em] = 4m^2 - 12m^2 + 8m - 4m + 4 \\[0.5em] = -8m^2 + 4m + 4 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow -8m^2 + 4m + 4 = 0 \\[0.5em] \Rightarrow -4(2m^2 - m - 1) = 0 \\[0.5em] \Rightarrow 2m^2 - m - 1 = 0 \\[0.5em] \Rightarrow 2m^2 - 2m + m - 1 = 0 \\[0.5em] \Rightarrow 2m(m - 1) + 1(m - 1) = 0 \\[0.5em] \Rightarrow (2m + 1)(m - 1) = 0 \\[0.5em] \Rightarrow 2m + 1 = 0 \text{ or } m - 1 = 0 \\[0.5em] \Rightarrow m = -\dfrac{1}{2} \text{ or } m = 1$

Hence, the value of m is $-\dfrac{1}{2}$ and 1.

(ii) The given equation is x² + 2(m - 1)x + (m + 5) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 2(m - 1) , c = m + 5

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m - 2)^2 - 4 \times 1 \times m + 5 \\[0.5em] = 4m^2 + 4 - 8m - 4(m + 5) \\[0.5em] = 4m^2 + 4 - 8m - 4m - 20 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow 4m^2 - 12m - 16 = 0 \\[0.5em] \Rightarrow 4(m^2 - 3m - 4) = 0 \\[0.5em] \Rightarrow m^2 - 3m - 4 = 0 \\[0.5em] \Rightarrow m^2 - 4m + m - 4 = 0 \\[0.5em] \Rightarrow m(m - 4) + 1(m - 4) = 0 \\[0.5em] \Rightarrow (m - 4)(m + 1) = 0 \\[0.5em] \Rightarrow m - 4 = 0 \text{ or } m + 1 = 0 \\[0.5em] \Rightarrow m = 4 \text{ or } m = -1$

Hence, the value of m is 4, -1.

Question 7

Find the values of k for which each of the following quadratic equation equal roots :

(i) 9x² + kx + 1 = 0

(ii) x² - 2kx + 7k - 12 = 0 Also, find the roots for those values of k in each case.

Answer

(i) The given equation is 9x² + kx + 1 = 0.

Comparing with ax² + bx + c = we obtain,
a = 9 , b = k , c = 1

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (k)^2 - 4 \times 9 \times 1 \\[0.5em] = k^2 - 36 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow k^2 - 36 = 0 \\[0.5em] \Rightarrow k^2 - (6)^2 = 0 \\[0.5em] \Rightarrow (k + 6)(k - 6) = 0 \\[0.5em] \Rightarrow k + 6 = 0 \text{ or } k - 6 = 0 \\[0.5em] \Rightarrow k = -6 \text{ or } k = 6 \\[0.5em]$

When k = 6 the equation becomes 9x² + 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(6) ± \sqrt{(6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \dfrac{-6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \dfrac{-6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \dfrac{-6 +0}{18} \text{ or } \dfrac{-6 - 0}{18} \\[1em] \Rightarrow x = -\dfrac{1}{3} \text{ or } -\dfrac{1}{3} \\[1em]$

When k = -6 the equation becomes 9x² - 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \dfrac{6 +0}{18} \text{ or } \dfrac{6 - 0}{18} \\[1em] \Rightarrow x = \dfrac{1}{3} \text{ or } \dfrac{1}{3} \\[1em]$

Hence, the values of k are 6, -6 ; when k = 6, roots are $-\dfrac{1}{3}, -\dfrac{1}{3}$ and when k = -6, roots are $\dfrac{1}{3}, \dfrac{1}{3}$ .

(ii) The given equation is x² - 2kx + 7k - 12 = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = -2k , c = 7k - 12

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (-2k)^2 - 4 \times 1 \times (7k - 12) \\[0.5em] = 4k^2 - 4(7k - 12) \\[0.5em] = 4k^2 - 28k + 48$

For equal roots, discriminant = 0

$\Rightarrow 4k^2 - 28k + 48 = 0 \\[0.5em] \Rightarrow 4(k^2 - 7k + 12) = 0 \\[0.5em] \Rightarrow k^2 - 7k + 12 = 0 \\[0.5em] \Rightarrow k^2 - 4k - 3k + 12 = 0 \\[0.5em] \Rightarrow k(k - 4) - 3(k - 4) = 0 \\[0.5em] \Rightarrow (k - 3)(k - 4) = 0 \\[0.5em] \Rightarrow k - 3 = 0 \text{ or } k - 4 = 0 \\[0.5em] \Rightarrow k = 3 \text{ or } k = 4 \\[0.5em]$

When k = 3 the equation becomes x² - 6x + 9 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times 9}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 36}}{2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{0}}{12} \\[1em] \Rightarrow x = \dfrac{6 +0}{2} \text{ or } \dfrac{6 - 0}{2} \\[1em] \Rightarrow x = 3 \text{ or } 3 \\[1em]$

When k = 4 the equation becomes x² - 8x + 16 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-8) ± \sqrt{(-8)^2 - 4\times 1 \times 16}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{8 ± \sqrt{64 - 64}}{2} \\[1em] \Rightarrow x = \dfrac{8 ± \sqrt{0}}{2} \\[1em] \Rightarrow x = \dfrac{8 + 0}{2} \text{ or } \dfrac{8 - 0}{2} \\[1em] \Rightarrow x = 4 \text{ or } 4 \\[1em]$

Hence the values of k are 3, 4 ; when k = 3, roots are 3, 3 and when k = 4, roots are 4, 4.

Question 8

Find the value(s) of p for which the quadratic equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 has equal roots. Also find these roots.

Answer

The given equation is (2p + 1)x² - (7P + 2)x + (7p - 3) = 0.

Comparing with ax² + bx + c = we obtain,
a = 2p + 1 , b = -(7p + 2) , c = 7p - 3

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-(7p + 2))^2 - 4 \times (2p + 1) \times (7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(2p + 1)(7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(14p^2 - 6p + 7p - 3) \\[1em] = 49p^2 - 56p^2 + 28p - 4p + 4 + 12 \\[1em] = -7p^2 + 24p + 16$

For equal roots, discriminant = 0

$\Rightarrow -7p^2 + 24p + 16 \\[1em] \Rightarrow 7p^2 - 24p - 16 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow 7p^2 - 28p + 4p - 16 = 0 \\[1em] \Rightarrow 7p(p - 4) + 4(p - 4) = 0 \\[1em] \Rightarrow (7p + 4)(p - 4) = 0 \\[1em] \Rightarrow 7p + 4 = 0 \text{ or } p - 4 = 0 \\[1em] \Rightarrow 7p = -4 \text{ or } p = 4 \\[1em] \Rightarrow p = -\dfrac{4}{7} \text{ or } p = 4 \\[1em]$

When p = $-\dfrac{4}{7}$ the equation becomes $(2 \times -\dfrac{4}{7} + 1)x^2 - (7 \times -\dfrac{4}{7} + 2)x + (7 \times -\dfrac{4}{7} - 3) = 0$ so the roots are :

$\Rightarrow -\dfrac{1}{7}x^2 + 2x - 7 = 0 \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \text{ (On multiplying complete equation by -7) } \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \\[1em] \Rightarrow x^2 - 7x - 7x + 49 = 0 \\[1em] \Rightarrow x(x - 7) - 7(x - 7) = 0 \\[1em] \Rightarrow (x - 7)(x - 7) = 0 \\[1em] \Rightarrow x - 7 = 0 \text{ or } x - 7 = 0 \\[1em] \Rightarrow x = 7 \text{ or } x = 7 \\[1em]$

When p = 4 the equation becomes $(2 \times 4 + 1)x^2 - (7 \times 4 + 2)x + (7 \times 4 - 3) = 0$ so the roots are :

$\Rightarrow 9x^2 - 30x + 25 = 0 \\[1em] \Rightarrow 9x^2 - 15x - 15x + 25 = 0 \\[1em] \Rightarrow 3x(3x - 5) - 5(3x - 5) = 0 \\[1em] \Rightarrow (3x - 5)(3x - 5) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } 3x - 5 = 0 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = \dfrac{5}{3} \\[1em]$

(Ans.) 4, $-\dfrac{4}{7}$ ; when p = $-\dfrac{4}{7}$ , roots are 7, 7 and when p = 4, roots are $\dfrac{5}{3} , \dfrac{5}{3}.$

Question 9

Find the value(s) of p for which the quadratic equation 2x² + 3x + p = 0 has real roots.

Answer

For real roots, Discriminant ≥ 0

or , b² - 4ac ≥ 0

The above equation is 2x² + 3x + p = 0
Comparing with ax² + bx + c = we obtain,
a = 2 , b = 3 , c = p

Putting values in b² - 4ac ≥ 0 we get,

$= 3^2 - 4 \times 2 \times p \ge 0 \\[0.5em] = 9 - 8p \ge 0 \\[0.5em] \Rightarrow 9 \ge 8p \\[0.5em] \Rightarrow 8p \le 9 \\[0.5em] \Rightarrow p \le \dfrac{9}{8} \\[0.5em]$

Hence the value of p is ≤ $\dfrac{9}{8}$ .

Question 10

Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.

Answer

For real roots, Discriminant ≥ 0

or , b² - 4ac ≥ 0

The above equation is x² + kx + 4 = 0
Comparing with ax² + bx + c = we obtain,
a = 1 , b = k , c = 4

Putting values in b² - 4ac ≥ 0 we get,

$= k^2 - 4 \times 1 \times 4 \ge 0 \\[0.5em] = k^2 - 16 \ge 0 \\[0.5em] \Rightarrow k^2 - 4^2 \ge 0 \\[0.5em] \Rightarrow (k - 4)(k + 4) \ge 0 \\[0.5em] \Rightarrow k \ge 4 \\[0.5em]$

Hence least positive value for which equation has real roots is 4 .

Question 11

Find the values of p for which the equation 3x² - px + 5 = 0 has real roots.

Answer

For real roots,

Discriminant ≥ 0

or, b² - 4ac ≥ 0

The above equation is 3x² - px + 5 = 0
Comparing with ax² + bx + c = we obtain,
a = 3 , b = -p , c = 5

Putting values in b² - 4ac ≥ 0 we get,

$= (-p)^2 - 4 \times 3 \times 5 \ge 0 \\[1em] = p^2 - 60 \ge 0 \\[1em] \Rightarrow p^2 - (\sqrt{60})^2 \ge 0 \\[1em] \Rightarrow (p + \sqrt{60})(p -\sqrt{60}) \ge 0 \\[1em] \Rightarrow (p + 2\sqrt{15})(p - 2\sqrt{15}) \ge 0 \\[1em] \text{(Using) } (a + b)(a - b) \ge 0 = a \ge b \text{ or } a \le -b \text{ we get, } \\[1em] \Rightarrow p \ge 2\sqrt{15} \text{ or } p \le -2\sqrt{15} \\[1em]$

Hence the values of p are $p \le -2\sqrt{15} \text{ or } p\ge 2\sqrt{15}$ .

Exercise 5.5

Question 1(i)

Find two consecutive natural numbers such that the sum of their squares is 61.

Answer

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

$\Rightarrow x^2 + (x + 1)^2 = 61 \\[1em] \Rightarrow x^2 + (x^2 + 1 + 2x) = 61 \\[1em] \Rightarrow x^2 + x^2 + 1 + 2x = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 - 61 = 0 \\[1em] \Rightarrow 2x^2 + 2x - 60 = 0 \\[1em] \Rightarrow 2(x^2 + x - 30) = 0 \\[1em] \Rightarrow x^2 + x - 30 = 0 \\[1em] \Rightarrow x^2 + 6x - 5x - 30 = 0 \\[1em] \Rightarrow x(x + 6) - 5(x + 6) = 0 \\[1em] \Rightarrow (x + 6)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -6 \text { or } x = 5$

Since the numbers are natural number so x ≠ -6.

∴ x = 5 , x + 1 = 6.

Hence, the required natural numbers are 5 , 6.

Question 1(ii)

Find two consecutive integers such that the sum of their squares is 61.

Answer

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

∴ When x = -6 , x + 1 = -5 and when x = 5 , x + 1 = 6.

Hence, required integers are 5,6 or -6,-5

Question 2(i)

If the product of two positive consecutive even integers is 288, find the integers.

Answer

Let the required two positive consecutive even integers be x , x + 2

Given, product of two consecutive even integers = 288

$\Rightarrow x(x + 2) = 288 \\[1em] \Rightarrow x^2 + 2x = 288 \\[1em] \Rightarrow x^2 + 2x - 288 = 0 \\[1em] \Rightarrow x^2 + 18x - 16x - 288 = 0 \\[1em] \Rightarrow x(x + 18) - 16(x + 18) = 0 \\[1em] \Rightarrow (x - 16)(x + 18) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 16 = 0 \text{ or } x + 18 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 16 \text{ or } x = -18$

Since the numbers are natural number so x ≠ -18.

∴ x = 16 , x + 2 = 18.

Hence, required integers are 16, 18.

Question 2(ii)

If the product of two consecutive even integers is 224 , find the integers.

Answer

Let the required two consecutive even integers be x , x + 2

Given, product of two consecutive even integers = 224

$\Rightarrow x(x + 2) = 224 \\[1em] \Rightarrow x^2 + 2x = 224 \\[1em] \Rightarrow x^2 + 2x - 224 = 0 \\[1em] \Rightarrow x^2 + 16x - 14x - 224 = 0 \\[1em] \Rightarrow x(x + 16) - 14(x + 16) = 0 \\[1em] \Rightarrow (x + 16)(x - 14) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 16 = 0 \text{ or } x - 14 = 0. \\[1em] \Rightarrow x = -16 \text{ or } x = 14 \\[1em]$

∴ When x = 14 , x + 2 = 16 and when x = -16 , x + 2 = -14.

Hence, required integers are 14 , 16 or -16, -14 .

Question 2(iii)

Find two consecutive even natural numbers such that the sum of their squares is 340.

Answer

Let the required two consecutive even integers be x , x + 2

Given, sum of squares of two consecutive even natural numbers = 340

$\Rightarrow x^2 + (x + 2)^2 = 340 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 340 \\[1em] \Rightarrow 2x^2 + 4x + 4 -340 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 336 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 168) = 0 \\[1em] \Rightarrow x^2 + 2x - 168 = 0 \\[1em] \Rightarrow x^2 + 14x - 12x - 168 = 0 \\[1em] \Rightarrow x(x + 14) - 12(x + 14) = 0 \\[1em] \Rightarrow (x - 12)(x + 14) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 12 = 0 \text{ or } x + 14 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 12 \text{ or } x = -14$

Since the numbers are natural number so x ≠ -14

∴ x = 12 , x + 2 = 14

Hence, required natural numbers are 12 , 14 .

Question 2(iv)

Find two consecutive odd integers such that sum of their squares is 394.

Answer

Let the required two consecutive odd integers be x , x + 2

Given, sum of squares of two consecutive odd integers = 394

$\Rightarrow x^2 + (x + 2)^2 = 394 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 394 \\[1em] \Rightarrow 2x^2 + 4x + 4 -394 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 390 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 195) = 0 \\[1em] \Rightarrow x^2 + 2x - 195 = 0 \\[1em] \Rightarrow x^2 + 15x - 13x - 195 = 0 \\[1em] \Rightarrow x(x + 15) - 13(x + 15) = 0 \\[1em] \Rightarrow (x + 15)(x - 13) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 15 = 0 \text{ or } x - 13 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -15 \text{ or } x = 13$

∴ When x = -15 , x + 2 = -13 and when x = 13 , x + 2 = 15.

Hence required integers are -15 , -13 or 13, 15 .

Question 3(i)

The sum of two numbers is 9 and the sum of their squares is 41. Find the numbers.

Answer

Let the first number be x.

Since the sum of two numbers is 9, so other number is 9 - x.

Given, the sum of squares of numbers = 41

⇒ x² + (9 - x)² = 41

⇒ x² + x² + 81 - 18x = 41

⇒ 2x² - 18x + 81 - 41 = 0

⇒ 2x² - 18x + 40 = 0

⇒ 2(x² - 9x + 20) = 0

⇒ x² - 9x + 20 = 0

⇒ x² - 4x - 5x + 20 = 0

⇒ x(x - 4) - 5(x - 4) = 0

⇒ (x - 4)(x - 5) = 0

⇒ x - 4 = 0 or x - 5 = 0

⇒ x = 4 or x = 5

∴ x = 5, 9 - x = 4

∴ x = 4, 9 - x = 5

Hence, the numbers are 4 and 5.

Question 3(ii)

The difference of two natural numbers is 7 and their product is 450. Find the numbers.

Answer

Let first number be x.

Since the difference of two numbers is 7, so other number is x + 7.

Given, the products of numbers = 450

⇒ x(x + 7) = 450

⇒ x² + 7x = 450

⇒ x² + 7x - 450 = 0

⇒ x² + 25x - 18x - 450 = 0

⇒ x(x + 25) - 18(x + 25) = 0

⇒ (x + 25)(x - 18) = 0

⇒ x + 25 = 0 or x - 18 = 0

⇒ x = -25 or x = 18.

Since, the numbers are natural numbers,

∴ x ≠ -25

∴ x = 18, x + 7 = 25

Hence, the numbers are 18 and 25.

Question 4

Five times a certain whole number is equal to three less than twice the square of the number. Find the number.

Answer

Let the number be x

Given, 5 times the number = 3 less than twice the square of the number

$\Rightarrow 5x = 2x^2 - 3 \\[0.5em] \Rightarrow 5x - 2x^2 + 3 = 0 \\[0.5em] \Rightarrow 2x^2 - 5x - 3 = 0 \text{ (on multiplying the equation by -1) } \\[0.5em] \Rightarrow 2x^2 - 6x + x - 3 = 0 \\[0.5em] \Rightarrow 2x(x - 3) + 1(x - 3) = 0 \\[0.5em] \Rightarrow (2x + 1)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x + 1 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = -\dfrac{1}{2} \text{ or } x = 3 \\[0.5em]$

Since the number is a whole number x ≠ $-\dfrac{1}{2}$

∴ x = 3

Hence, the required whole number is 3.

Question 5

Sum of two natural numbers is 8 and the difference of their reciprocal is $\dfrac{2}{15}.$ Find the numbers.

Answer

Let the first number be x

Since, the sum of two numbers is 8, so other number is 8 - x.

Given, the difference of reciprocal of numbers = $\dfrac{2}{15}$

$\Rightarrow \dfrac{1}{x} - \dfrac{1}{8 - x} = \dfrac{2}{15} \\[0.5em] \Rightarrow \dfrac{8 - x - x}{x(8 - x)} = \dfrac{2}{15} \text{ (On taking L.C.M.) } \\[0.5em] \Rightarrow 15(8 - 2x) = 2x(8 - x) \text{ (On cross multiplication ) } \\[0.5em] \Rightarrow 120 - 30x = 16x - 2x^2 \\[0.5em] \Rightarrow 120 - 30x - 16x + 2x^2 = 0 \\[0.5em] \Rightarrow 2x^2 - 46x + 120 = 0 \\[0.5em] \Rightarrow 2(x^2 - 23x + 60) = 0 \\[0.5em] \Rightarrow x^2 - 20x - 3x + 60 = 0 \\[0.5em] \Rightarrow x(x - 20) - 3(x - 20) = 0 \\[0.5em] \Rightarrow (x - 3)(x - 20) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ (or) } x - 20 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = 3 \text{ or } x = 20.$

If x = 20 , 8 - x = -12 , Since both are natural numbers hence x ≠ 20.

∴ x = 3 , 8 - x = 5

Hence, the required natural numbers are 3 , 5.

Question 6

The difference of the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.

Answer

Let the larger number be x .

Given, the square of the smaller number is 4 times the larger number.

Hence, square of smaller number = 4x

Given, the difference of squares of two numbers is = 45

$\Rightarrow x^2 - 4x = 45 \\[0.5em] \Rightarrow x^2 - 4x - 45 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[0.5em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[0.5em] \Rightarrow (x + 5)(x - 9) = 0 \\[0.5em] \Rightarrow x + 5 = 0 \text{ or } x - 9 = 0 \\[0.5em] \Rightarrow x = -5 \text{ or } x = 9$

If x = -5 , $\sqrt{4x} = \sqrt{-20}$ , which is not valid as there is no real value of square root of a negative value.

∴ x = 9 , $\sqrt{4x} = \sqrt{36} = 6 \text{ or } -6$ .

Hence, the two numbers are 9, 6 or 9, -6.

Question 7

There are three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154. What are the integers ?

Answer

Let the numbers be x , x + 1 , x + 2.

Given, sum of the square of the first and the product of the other two is = 154

$\Rightarrow x^2 + (x + 1)(x + 2) = 154 \\[1em] \Rightarrow x^2 + (x^2 + 2x + x + 2) = 154 \\[1em] \Rightarrow 2x^2 + 3x + 2 - 154 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 152 = 0 \\[1em] \Rightarrow 2x^2 + 19x - 16x - 152 = 0 \\[1em] \Rightarrow x(2x + 19) - 8(2x + 19) = 0 \\[1em] \Rightarrow (x - 8)(2x + 19) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } 2x + 19 = 0 \\[1em] x = 8 \text{ or } x = -\dfrac{19}{2}$

Since the integers are positive hence , x ≠ $-\dfrac{19}{2}$

∴ x = 8 , x + 1 = 9 , x + 2 = 10.

Hence , the required numbers are 8, 9, 10.

Question 8(i)

Find three succcessive even natural numbers, the sum of whose squares is 308.

Answer

Let the required numbers be x, x + 2, x + 4.

Given, the sum of squares of these numbers = 308

$\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 308 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 308 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 308 \\[1em] \Rightarrow 3x^2 + 12x - 288 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 96) = 0 \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x - 8)(x + 12) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } x + 12 = 0 \\[1em] x = 8 \text{ or } x = -12$

Since the numbers are natural hence , x ≠ -12

∴ x = 8 , x + 2 = 10 , x + 4 = 12.

Hence , the required numbers are 8, 10, 12.

Question 8(ii)

Find three consecutive odd integers , the sum of whose squares is 83.

Answer

Let the required numbers be x , x + 2 , x + 4.

Given, the sum of squares of these numbers = 308

$\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 83 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 83 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 83 \\[1em] \Rightarrow 3x^2 + 12x - 63 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 21) = 0 \\[1em] \Rightarrow x^2 + 4x - 21 = 0 \\[1em] \Rightarrow x^2 + 7x - 3x - 21 = 0 \\[1em] \Rightarrow x(x + 7) - 3(x + 7) = 0 \\[1em] \Rightarrow (x + 7)(x - 3) = 0 \\[1em] \Rightarrow x + 7 = 0 \text{ or } x - 3 = 0 \\[1em] x = -7 \text{ or } x = 3$

∴ When x = -7 , x + 2 = -5 , x + 4 = -3 and when x = 3 , x + 2 = 5 , x + 4 = 7.

Hence , the required numbers are -7, -5, -3 and 3, 5, 7.

Question 9

In a certain positive fraction , the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\dfrac{1}{14}$ . Find the fraction.

Answer

Let the numerator of the fraction be x

Given, the denominator is greater than numerator by 3 hence, denominator = x + 3

Fraction = $\dfrac{x}{x + 3}$

Given, if 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\dfrac{1}{14}$

$\Rightarrow \dfrac{x}{x + 3} - \dfrac{x - 1}{x + 3 - 1} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x}{x + 3} - \dfrac{x - 1}{x + 2} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x(x + 2) - (x - 1)(x + 3)}{(x + 3)(x + 2)} = \dfrac{1}{14} \text{ (On taking L.C.M) } \\[1em] \Rightarrow \dfrac{x^2 + 2x - (x^2 + 3x - x - 3)}{(x^2 + 2x + 3x + 6)} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x^2 - x^2 + 2x - 2x + 3}{x^2 + 5x + 6} = \dfrac{1}{14} \\[1em] \Rightarrow 3 \times 14 = x^2 + 5x + 6 \text{ (Cross multiplying) } \\[1em] \Rightarrow x^2 + 5x + 6 = 42 \\[1em] \Rightarrow x^2 + 5x + 6 - 42 = 0 \\[1em] \Rightarrow x^2 + 5x - 36 = 0 \\[1em] \Rightarrow x^2 + 9x - 4x - 36 = 0 \\[1em] \Rightarrow x(x + 9) - 4(x + 9) = 0 \\[1em] \Rightarrow (x - 4)(x + 9) = 0 \\[1em] x = 4 \text{ or } x = -9 \\[1em]$

If x = -9 , Fraction = $\dfrac{x}{x + 3} = \dfrac{9}{6}$ In this case numerator > denominator which is incorrect according to the question hence, x ≠ -9.

∴ x = 4 , Fraction = $\dfrac{x}{x + 3} = \dfrac{4}{7}$

Hence, the fraction is $\dfrac{4}{7}$ .

Question 10

The sum of numerator and denominator of a certain positive fraction is 8 . If 2 is added to both the numerator and denominator, the fraction is increased by $\dfrac{4}{35}.$ Find the fraction.

Answer

Let the denominator be = x so, numerator = 8 - x.

Fraction = $\dfrac{8 - x}{x}$

Given , if 2 is added to both the numerator and denominator, the fraction is increased by $\dfrac{4}{35}.$

$\Rightarrow \dfrac{8 - x + 2}{x + 2} - \dfrac{8 - x}{x} = \dfrac{4}{35} \\[1em] \Rightarrow \dfrac{x(10 - x) - (x + 2)(8 - x)}{x(x + 2)} = \dfrac{4}{35} \text{ (On taking L.C.M) }\\[1em] \Rightarrow \dfrac{10x - x^2 - (8x - x^2 + 16 - 2x)}{x^2 + 2x} = \dfrac{4}{35} \\[1em] \Rightarrow 35(-x^2 + x^2 - 8x + 10x + 2x - 16) = 4(x^2 + 2x) \\[1em] \Rightarrow 35(4x - 16) = 4x^2 + 8x \\[1em] \Rightarrow 140x - 560 = 4x^2 + 8x \\[1em] \Rightarrow 4x^2 + 8x - 140x + 560 = 0 \\[1em] \Rightarrow 4x^2 - 132x + 560 = 0 \\[1em] \Rightarrow 4(x^2 - 33x + 140) = 0 \\[1em] \Rightarrow x^2 - 33x + 140 = 0 \\[1em] \Rightarrow x^2 - 28x - 5x + 140 = 0 \\[1em] \Rightarrow x(x - 28) - 5(x - 28) = 0 \\[1em] \Rightarrow (x - 28)(x - 5) = 0 \\[1em] \Rightarrow x - 28 = 0 \text{ or } x - 5 = 0 \\[1em] x = 28 \text{ or } x = 5$

If x = 28 , 8 - x = -20 which will make fraction = $-\dfrac{20}{28}$ negative hence, x ≠ 28

∴ x = 5 , 8 - x = 3 ,fraction = $\dfrac{3}{5}$

Hence, the fraction is $\dfrac{3}{5}$ .

Question 11

A two digit number contains the Larger digit at ten's place. The product of the digits is 27 and the difference between two digits is 6 . Find the number.

Answer

Let the unit's digit be x,
so, ten's digit be = x + 6.

Number = 10(x + 6) + x = 10x + 60 + x = 11x + 60

Given, product of digits is 27

$\Rightarrow x(x + 6) = 27 \\[1em] \Rightarrow x^2 + 6x = 27 \\[1em] \Rightarrow x^2 + 6x - 27 = 0 \\[1em] \Rightarrow x^2 + 9x - 3x - 27 = 0 \\[1em] \Rightarrow x(x + 9) - 3(x + 9) = 0 \\[1em] \Rightarrow (x - 3)(x + 9) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 9 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -9$

When x = -9,
Number = 11x + 60
= 11(-9) + 60
= -99 + 60
= -39
In this case the ten's digit is smaller than unit's digit hence x ≠ -9

When x = 3,
Number = 11(x) + 60
= 11(3) + 60
= 33 + 60
= 93

Hence, the required number is 93 .

Question 12

A two digit positive number is such that the product of its digit is 6. If 9 is added to the number , the digits interchange their place . Find the number.

Answer

Let the digit at unit's place be x .

Since, the product of digits is 6 , it's ten's digit = $\dfrac{6}{x}$

$\therefore \text{ Number } = 10 \times \dfrac{6}{x} + x \\[0.5em] = \dfrac{60}{x} + x \\[0.5em] = \dfrac{60 + x^2}{x} \\[0.5em] = \dfrac{x^2 + 60}{x} \\[0.5em]$

Given, if 9 is added to the number , the digits interchange their place

On interchanging the digits, number becomes = $10 \times x + \dfrac{6}{x}$

Acccording to given,

$\Rightarrow \dfrac{x^2 + 60}{x} + 9 = 10 \times x + \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2 + 9x + 60}{x} = 10x + \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2 + 9x + 60}{x} = \dfrac{10x^2 + 6}{x} \\[1em] \Rightarrow x^2 + 9x + 60 =10x^2 + 6 \\[1em] \Rightarrow 10x^2 - x^2 - 9x + 6 - 60 = 0 \\[1em] \Rightarrow 9x^2 - 9x - 54 = 0 \\[1em] \Rightarrow 9(x^2 - x - 6) = 0 \\[1em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[1em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[1em] \Rightarrow (x + 2)(x - 3) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 3$

Since the number is positive hence x ≠ -2.

If x = 3 ,

$\text{ Number } = \dfrac{x^2 + 60}{x} \\[1em] = \dfrac{3^2 + 60}{3} \\[1em] = \dfrac{69}{3} \\[1em] = 23$

Hence, the required number is 23.

Question 13

A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of rectangle.

Answer

Length of rectangle = x

Since the area of rectangle = 105 cm², breadth = $\dfrac{105}{x}$ cm.

∴ Perimeter = 2(length + breadth)

$= 2(x + \dfrac{105}{x}) \\[0.5em] = 2(\dfrac{x^2 + 105}{x}) \\[0.5em]$

Given, Perimeter = 44 cm

$\Rightarrow 2(\dfrac{x^2 + 105}{x}) = 44 \\[0.5em] \Rightarrow \dfrac{x^2 + 105}{x} = 22 \\[0.5em] \Rightarrow x^2 + 105 = 22x \\[0.5em] \Rightarrow x^2 - 22x + 105 = 0 \\[0.5em] \Rightarrow x^2 - 15x - 7x + 105 = 0 \\[0.5em] \Rightarrow x(x - 15) - 7(x - 15) = 0 \\[0.5em] \Rightarrow (x - 7)(x - 15) = 0 \\[0.5em] \Rightarrow x - 7 = 0 \text{ or } x - 15 = 0 \\[0.5em] \Rightarrow x = 7 \text{ or } x = 15$

Breadth = $\dfrac{105}{x}$ cm
Equation in x for perimeter, 2(x + $\dfrac{105}{x}$ ) = 44
Length = 7 cm , Breadth = 15 cm

Question 14

A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres , assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Answer

Given,

Length of rectangular garden = 10 m

Breadth of rectangular garden = 16 m

Width of walk = x

So, length of garden and walk combined = (10 + x + x) m

Breadth of garden and walk combined = (16 + x + x) m

∴ Area of garden and walk combined = Length $\times$ Breadth = (10 + 2x)(16 + 2x) m²

Given, area of walk = 120m²

Area of walk = Area of combined - Area of garden

$\Rightarrow (10 + 2x)(16 + 2x) - 10 \times 16 = 120 \\[0.5em] \Rightarrow 160 + 20x + 32x + 4x^2 - 160 = 120 \\[0.5em] \Rightarrow 4x^2 + 52x = 120 \\[0.5em] \Rightarrow 4x^2 + 52x - 120 = 0 \\[0.5em] \Rightarrow 4(x^2 + 13x - 30) = 0 \\[0.5em] \Rightarrow x^2 + 13x - 30 = 0 \\[0.5em] \Rightarrow x^2 + 15x - 2x - 30 = 0 \\[0.5em] \Rightarrow x(x + 15) - 2(x + 15) = 0 \\[0.5em] \Rightarrow (x - 2)(x + 15) = 0 \\[0.5em] \Rightarrow x - 2 \text{ or } x + 15 = 0 \\[0.5em] x = 2 \text{ or } x = -15$

Since, width cannot be negative hence x ≠ -15.

The equation in x = (10 + 2x)(16 + 2x) - 10 x 16 = 120.
Value of x = 2m.

Question 15

The length of a rectangle exceeds its breadth by 5m. If the breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m². Find its dimensions.

Answer

Let breadth of rectangle be x meters

Since , length of rectangle exceeds breadth by 5 meters so, length = (x + 5) meters

Given, if breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m

∴ Area of new rectangle = Length $\times$ Breadth = (x + 5 - 9)(2x)

$\Rightarrow (x + 5 - 9)(2x) - x(x + 5) = 140 \\[0.5em] \Rightarrow (x - 4)(2x) - (x^2 + 5x) = 140 \\[0.5em] \Rightarrow 2x^2 - 8x - x^2 - 5x = 140 \\[0.5em] \Rightarrow x^2 - 13x - 140 = 0 \\[0.5em] \Rightarrow x^2 - 20x + 7x - 140 = 0 \\[0.5em] \Rightarrow x(x - 20) + 7(x - 20) = 0 \\[0.5em] \Rightarrow (x + 7)(x - 20) = 0 \\[0.5em] \Rightarrow x + 7 = 0 \text{ or } x - 20 = 0 \\[0.5em] x = -7 \text{ or } x = 20$

Since, breadth cannot be negative hence x ≠ -7

∴ x = 20 and x + 5 = 25

Length of rectangle = 25 m , Breadth of rectangle = 20 m.

Question 16

The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of the plot as x meters. Use the perimeter 180 m to write the value of the breadth in terms of x . Use the values of length, breadth and the area to write an equation in x . Solve the equation to calculate the length and breadth of the plot.

Answer

Length of rectangular plot = x meters

Perimeter = 2(length + breadth)

$\Rightarrow 2(x + \text{Breadth}) = 180 \\[0.5em] \Rightarrow x + \text{Breadth} = 90 \\[0.5em] \Rightarrow \text{Breadth} = 90 - x$

Area of the rectangle = Length $\times$ Breadth

Given, Area of rectangle = 1800 m²

$\Rightarrow x(90 - x) = 1800 \\[0.5em] \Rightarrow 90x - x^2 = 1800 \\[0.5em] \Rightarrow x^2 - 90x + 1800 = 0 \\[0.5em] \Rightarrow x^2 - 30x - 60x + 1800 = 0 \\[0.5em] \Rightarrow x(x - 30) - 60(x - 30) = 0 \\[0.5em] \Rightarrow (x - 30)(x - 60) = 0 \\[0.5em] \Rightarrow x - 30 = 0 \text{ or } x - 60 = 0 \\[0.5em] \Rightarrow x = 30 \text{ or } x = 60$

Breadth = (90 - x) meters
Equation in x : x(90 - x) = 1800
Length of rectangle = 60 m , Breadth of rectangle = 30 m

Question 17

The lengths of parallel sides of a trapezium are (x + 9) cm and (2x - 3) cm , and the distance between them is (x + 4) cm . If its area is 540 cm² , find x .

Answer

Given ,

Length of first parallel side = (x + 9) cm

Length of second parallel side = (2x - 3) cm

Distance between parallel side = (x + 4) cm

Area of trapezium = 540 cm²

Area of trapezium is given by,

$=\dfrac{1}{2} \times \text{ (sum of parallel sides) } \times \text{ (distance between them) } \\[1em] \Rightarrow \dfrac{1}{2} \times (x + 9 + 2x - 3) \times (x + 4) = 540 \\[1em] \Rightarrow \dfrac{1}{2} \times (3x + 6)(x + 4) = 540 \\[1em] \Rightarrow \dfrac{1}{2} \times (3x^2 + 12x + 6x + 24) = 540 \\[1em] \Rightarrow \dfrac{1}{2} \times (3x^2 + 18x + 24) = 540 \\[1em] \Rightarrow 3x^2 + 18x + 24 = 540 \times 2 \text{ (On cross multiplication) } \\[1em] \Rightarrow 3x^2 + 18x + 24 = 1080 \\[1em] \Rightarrow 3x^2 + 18x + 24 - 1080 = 0 \\[1em] \Rightarrow 3x^2 + 18x - 1056 = 0 \\[1em] \Rightarrow 3(x^2 + 6x - 352) = 0 \\[1em] \Rightarrow x^2 + 6x - 352 = 0 \\[1em] \Rightarrow x^2 + 22x - 16x - 352 = 0 \\[1em] \Rightarrow x(x + 22) - 16(x + 22) = 0 \\[1em] \Rightarrow (x - 16)(x + 22) = 0 \\[1em] \Rightarrow x - 16 = 0 \text{ or } x + 22 = 0 \\[1em] x = 16 \text{ or } x = -22$

If x = -22 , Length = x + 9 = -22 + 9 = -13 , Breadth = (2x - 3) = -44 - 3 = -47
Since length and breadth cannot be negative hence , x ≠ -22

∴ x = 16

The value of x is 16.

Question 18

If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m , find its area.

Answer

Taking length = l and breadth = b

Perimeter of rectangle = 2(l + b)

Length of diagonal of a rectangle = $\sqrt{l^2 + b^2}$

Given,

Perimeter = 68 m

$\Rightarrow 2(l + b) = 68 \\[0.5em] \Rightarrow l + b = 34 \\[0.5em] \Rightarrow l = 34 - b \qquad \text{ Equation (a)}$

Given,

Diagonal of a rectangle = 26 m

$\Rightarrow \sqrt{l^2 + b^2} = 26$

On squaring both sides,

$\Rightarrow l^2 + b^2 = 26^2$

Putting values of l from equation a,

$\Rightarrow (34 - b)^2 + b^2 = 676 \\[1em] \Rightarrow 1156 + b^2 - 68b + b^2 = 676 \\[1em] \Rightarrow 2b^2 - 68b + 1156 - 676 = 0 \\[1em] \Rightarrow 2b^2 - 68b + 480 = 0 \\[1em] \Rightarrow 2(b^2 - 34b + 240) = 0 \\[1em] \Rightarrow b^2 - 34b + 240 = 0 \\[1em] \Rightarrow b^2 - 24b - 10b + 240 = 0 \\[1em] \Rightarrow b(b - 24) - 10(b - 24) = 0 \\[1em] \Rightarrow (b - 24)(b - 10) = 0 \\[1em] \Rightarrow b - 24 = 0 \text{ or } b - 10 = 0 \\[1em] b = 24 \text{ or } b = 10$

∴ If b = 24 ,l = 34 - b = 34 - 24 = 10

If b = 10 , l = 34 - b = 34 - 10 = 24

Area of rectangle = Length $\times$ Breadth = 24 $\times$ 10 = 240 m²

Hence, the area of rectangle is 240 m².

Question 19

If the sum of two smaller sides of a right-angled triangle is 17 cm and the perimeter is 30 cm, then find the area of the triangle.

Answer

Let one of the two smaller sides be x cm, then the other side is (17 - x) cm.

Length of hypotenuse = perimeter - sum of other two sides = 30cm - 17cm = 13cm

In right angled triangle

Perpendicular² + Base² = Hypotenuse²

∴ x² + (17 - x)² = 13²

$\Rightarrow x^2 + 289 + x^2 - 34x = 169 \\[1em] \Rightarrow 2x^2 - 34x + 289 - 169 = 0 \\[1em] \Rightarrow 2x^2 - 34x + 120 = 0 \\[1em] \Rightarrow 2(x^2 - 17x + 60) = 0 \\[1em] \Rightarrow x^2 - 17x + 60 = 0 \\[1em] \Rightarrow x^2 - 12x - 5x + 60 = 0 \\[1em] \Rightarrow x(x - 12) - 5(x - 12) = 0 \\[1em] \Rightarrow (x - 5)(x - 12) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 12 = 0 \\[1em] \Rightarrow x = 5 \text{ or } x = 12$

If x = 5 , 17 - x = 12 and if x = 12 , 17 - x = 5.

Hence, two small sides are 5 , 12.

Area of right angled triangle = $\dfrac{1}{2} \times \text{base} \times \text{height}$

∴ $\dfrac{1}{2} \times 5\text{cm} \times 12\text{cm} = 30 cm^2$

Hence the area of triangle is 30cm².

Question 20

The hypotenuse of a grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.

Answer

Let the shortest side be x metres

So, hypotenuse = (2x + 1) metres and third side = (x + 7) metres

Hypotenuse² = Perpendicular² + Base²

∴ (2x + 1)² = x² + (x + 7)²

$\Rightarrow 4x^2 + 1 + 4x = x^2 + x^2 + 49 + 14x \\[1em] \Rightarrow 4x^2 + 1 + 4x = 2x^2 + 49 + 14x \\[1em] \Rightarrow 4x^2 - 2x^2 + 1 - 49 + 4x - 14x = 0 \\[1em] \Rightarrow 2x^2 - 48 - 10x = 0 \\[1em] \Rightarrow 2(x^2 - 24 - 5x) = 0 \\[1em] \Rightarrow x^2 - 5x - 24 = 0 \\[1em] \Rightarrow x^2 - 8x + 3x - 24 = 0 \\[1em] \Rightarrow x(x - 8) + 3(x - 8) = 0 \\[1em] \Rightarrow (x + 3)(x - 8) = 0 \\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 8 = 0 \\[1em] x = -3 \text{ or } x = 8$

Since no side of a triangle can be negative hence, x ≠ -3

If x = 8 , (2x + 1) = 17 , (x + 7) = 15

Hence, the hypotenuse of the triangle is 17 metres while the shorter sides are 8 metres and 15 metres.

Question 21

Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side, find the lengths of the rods.

Answer

Let the length of hypotenuse be x cm,

So, longer side = (x - 2) cm and shortest side = (x - 4) cm

Hypotenuse² = Sum of the squares of other two sides

∴ x² = (x - 2)² + (x - 4)²

$\Rightarrow x^2 = x^2 + 4 - 4x + x^2 + 16 - 8x \\[1em] \Rightarrow x^2 = 2x^2 - 12x + 20 \\[1em] \Rightarrow 2x^2 - x^2 - 12x + 20 = 0 \\[1em] \Rightarrow x^2 - 12x + 20 = 0 \\[1em] \Rightarrow x^2 - 10x - 2x + 20 = 0 \\[1em] \Rightarrow x(x - 10) - 2(x - 10) = 0 \\[1em] \Rightarrow (x - 2)(x - 10) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 10 = 0 \\[1em] x = 2 \text{ or } x = 10$

x ≠ 2 as that will make shortest side negative (x - 4) = -2.

If x = 10 , x - 2 = 8 , x - 4 = 6.

Hence, the dimensions of triangle are Hypotenuse = 10 cm , Longer side = 8 cm , Shortest side = 6 cm.

Question 22

In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.

Answer

Let the number of rows be x

So, the number of students in each row = x + 4

Given, total students = 480

$\Rightarrow x(x + 4) = 480 \\[1em] \Rightarrow x^2 + 4x = 480 \\[1em] \Rightarrow x^2 + 4x - 480 = 0 \\[1em] \Rightarrow x^2 + 24x - 20x - 480 = 0 \\[1em] \Rightarrow x(x + 24) - 20(x + 24) = 0 \\[1em] \Rightarrow (x - 20)(x + 24) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 24 = 0 \\[1em] x = 20 \text{ or } x = -24.$

Since number of rows cannot be negative hence, x ≠ -24.

If x = 20 , x + 4 = 24.

Hence the number of students in each row are 24.

Question 23

In an auditorium , the number of rows was equal to number of seats in each row. If the number of rows is doubled and the number of seats in each row is reduced by 5, then the total number of seats is increased by 375. How many rows were there ?

Answer

Let number of rows be = $x$ = number of seats in each row

Hence, total number of seats = $x \times x = x^2$

Given, if the number of rows is doubled and the number of seats in each row is reduced by 5, then the total number of seats is increased by 375

$\therefore 2x \times (x - 5) - x^2 = 375$

$\Rightarrow 2x(x - 5) - x^2 = 375 \\[1em] \Rightarrow 2x^2 - 10x - x^2 = 375 \\[1em] \Rightarrow x^2 - 10x - 375 = 0 \\[1em] \Rightarrow x^2 - 25x + 15x - 375 = 0 \\[1em] \Rightarrow x(x - 25) + 15(x - 25) = 0 \\[1em] \Rightarrow (x - 25)(x + 15) = 0 \\[1em] \Rightarrow x - 25 = 0 \text{ or } x + 15 = 0 \\[1em] \Rightarrow x = 25 \text{ or } x = -15$

Since number of rows cannot be negative hence, x ≠ -15.

Hence the number rows were 25.

Question 24

At an annual function of a school, each student gives gift to every other student. If the number of gifts is 1980, find the number of students.

Answer

Let the number of students be x

If each student gives gift to every other student so each student gives gift to (x - 1) students

So, x students gives gifts to total = x(x - 1) students.

According to given,

$\Rightarrow x(x - 1) = 1980 \\[1em] \Rightarrow x^2 - x = 1980 \\[1em] \Rightarrow x^2 - x - 1980 = 0 \\[1em] \Rightarrow x^2 - 45x + 44x - 1980 \\[1em] \Rightarrow x(x - 45) + 44(x - 45) \\[1em] \Rightarrow (x - 45)(x + 44) \\[1em] \Rightarrow x - 45 = 0 \text{ or } x + 44 = 0 \\[1em] x = 45 \text{ or } x = -44$

Since number of students cannot be negative hence, x ≠ -44.

Hence, the number of students are 45.

Question 25

A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be 'x' km/h, form an equation and solve it to evaluate x.

Answer

Uniform speed of bus = x km/h

Due to heavy rain the speed reduces to = (x - 10) km/h

Given, due to decrease in speed it takes two hours longer to cover the distance

Since, Time = $\dfrac{\text{Distance}}{\text{Speed}}$

$\therefore \dfrac{240}{x - 10} - \dfrac{240}{x} = 2 \\[1em] \Rightarrow \dfrac{240x - 240(x - 10)}{x(x - 10)} = 2 \text{ (On taking L.C.M.) } \\[1em] \Rightarrow 240x - 240x + 2400 = 2x(x - 10) \\[1em] \Rightarrow 2400 = 2x^2 - 20x \\[1em] \Rightarrow 2x^2 - 20x - 2400 = 0 \\[1em] \Rightarrow 2(x^2 - 10x - 1200) = 0 \\[1em] \Rightarrow x^2 - 10x - 1200 = 0 \\[1em] \Rightarrow x^2 - 40x + 30x - 1200 = 0 \\[1em] \Rightarrow x(x - 40) + 30(x - 40) = 0 \\[1em] \Rightarrow (x + 30)(x - 40) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 40 = 0 \\[1em] x = -30 \text{ or } x = 40$

Since, speed cannot be negative hence, x ≠ -30.

The equation in x is $\rightarrow \dfrac{240}{x - 10} - \dfrac{240}{x} = 2$
Hence, the value of uniform speed is 40 km/h.

Question 26

The speed of an express train is x km/h and the speed of an ordinary train is 12 km /h less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km , find the speed of the express train.

Answer

Speed of an express train is x km/h

So, the speed of ordinary train is (x - 12) km/h

Given, ordinary train takes one hour longer than express train to cover 240 km

Since, Time = $\dfrac{\text{Distance}}{\text{Speed}}$

$\therefore \dfrac{240}{x - 12} - \dfrac{240}{x} = 1 \\[1em] \Rightarrow \dfrac{240x - 240(x - 12)}{x(x - 12)} = 1 \\[1em] \Rightarrow 240x - 240x + 2880 = x(x - 12) \\[1em] \Rightarrow 2880 = x^2 - 12x \\[1em] \Rightarrow x^2 - 12x - 2880 = 0 \\[1em] \Rightarrow x^2 - 60x + 48x - 2880 = 0 \\[1em] \Rightarrow x(x - 60) + 48(x - 60) = 0 \\[1em] \Rightarrow (x - 60)(x + 48) = 0 \\[1em] \Rightarrow x - 60 = 0 \text{ or } x + 48 = 0 \\[1em] \Rightarrow x = 60 \text{ or } x = -48$

Since speed of train cannot be negative hence, x ≠ -48

Hence, the speed of express train is 60 km/h.

Question 27

A car covers a distance of 400 km at a certain speed . Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less . Find the original speed of the car.

Answer

Let the speed of car be x km/h

Given, if speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less

1 hour 40 minutes = $\dfrac{100}{60}$ hours

Since, Time = $\dfrac{\text{Distance}}{\text{Speed}}$

$\therefore \dfrac{400}{x} - \dfrac{400}{x + 12} = \dfrac{100}{60} \\[1em] \Rightarrow 100(\dfrac{4}{x} - \dfrac{4}{x + 12}) = \dfrac{100}{60} \\[1em] \Rightarrow \dfrac{4}{x} - \dfrac{4}{x + 12} = \dfrac{1}{60} \\[1em] \Rightarrow \dfrac{4(x + 12) - 4x}{x(x + 12)} = \dfrac{1}{60} \\[1em] \Rightarrow 60(4x + 48 - 4x) = x(x + 12) \\[1em] \Rightarrow 60 \times 48 = x^2 + 12x \\[1em] \Rightarrow x^2 + 12x - 2880 = 0 \\[1em] \Rightarrow x^2 + 60x - 48x - 2880 = 0 \\[1em] \Rightarrow x(x + 60) - 48(x + 60) = 0 \\[1em] \Rightarrow (x - 48)(x + 60) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x + 60 = 0 \\[1em] x = 48 \text{ or } x = -60$

Since speed of train cannot be negative hence, x ≠ -60

Hence, the speed of express train is 48 km/h.

Question 28

An aeroplane covered a distance of 400 km at an average speed of x km/h. On the return journey, the speed was increased by 40 km/h . Write down an expression for the time taken for :

(i) the onward journey

(ii) the return journey

If the return journey took 30 minutes less than the onward journey , write down an equation in x and find its value.

Answer

(i) Distance covered by plane = 400 km

Average speed of plane = x km/h

Time = $\dfrac{\text{ Distance}}{\text{Speed}} = \dfrac{400}{x} \text{ hrs}$

(ii) Distance covered by plane = 400 km

Average speed of plane = (x + 40) km/h

Time = $\dfrac{\text{ Distance}}{\text{Speed}} = \dfrac{400}{x + 40} \text{ hrs}$

According to question,

$\Rightarrow \dfrac{400}{x} - \dfrac{400}{x + 40} = \dfrac{30}{60} \\[1em] \Rightarrow \dfrac{400(x + 40) - 400x}{x(x + 40)} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{400x - 400x + 16000}{x(x + 40)} = \dfrac{1}{2} \\[1em] \Rightarrow 16000 \times 2 = x(x + 40) \\[1em] \Rightarrow 32000 = x^2 + 40x \\[1em] \Rightarrow x^2 + 40x - 32000 = 0 \\[1em] \Rightarrow x^2 + 200x - 160x - 32000 = 0 \\[1em] \Rightarrow x(x + 200) - 160(x + 200) = 0 \\[1em] \Rightarrow (x - 160)(x + 200) = 0 \\[1em] \Rightarrow x - 160 = 0 \text{ or } x + 200 = 0 \\[1em] x = 160 \text{ or } x = -200$

Since speed of aeroplane cannot be negative hence, x ≠ -200

∴ x = 160

Equation in x : $\dfrac{400}{x} - \dfrac{400}{x + 40} = \dfrac{1}{2}$
Hence, the speed of aerolane is 160 km/h.

Question 29

The distance by road between two towns A and B , is 216 km , and by rail it is 208 km. A car travels at a speed of x km/h, and the train travels at a speed which is 16 km/h faster than the car. Calculate :

(i) The time taken by car , to reach town B from A, in terms of x.

(ii) The time taken by the train , to reach town B from A, in terms of x.

(iii) If the train takes 2 hours less than the car, to reach town B , obtain an equation in x, and solve it.

(iv) Hence, find the speed of the train.

Answer

(i) Speed of car = x km/h

Distance between point A and B by road = 216 km

Time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{216}{x}$ hours

(ii) Speed of train = (x + 16) km/h

Distance between point A and B by rail = 208 km

Time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{208}{x + 16}$ hours

(iii) Given,

Train takes 2 hours less than car to reach town B

$\therefore \dfrac{216}{x} - \dfrac{208}{x + 16} = 2 \\[1em] \Rightarrow \dfrac{216(x + 16) - 208x}{x(x + 16)} = 2 \\[1em] \Rightarrow 216x -208x + 3456 = 2x(x + 16) \\[1em] \Rightarrow 8x + 3456 =2x^2 + 32x \\[1em] \Rightarrow 2x^2 + 32x -8x - 3456 = 0 \\[1em] \Rightarrow 2x^2 + 24x - 3456 = 0 \\[1em] \Rightarrow 2(x^2 + 12x - 1728) = 0 \\[1em] \Rightarrow x^2 + 12x - 1728 = 0 \\[1em] \Rightarrow x^2 + 48x - 36x - 1728 = 0 \\[1em] \Rightarrow x(x + 48) - 36(x + 48) = 0 \\[1em] \Rightarrow (x - 36)(x + 48) = 0 \\[1em] \Rightarrow x - 36 = 0 \text{ or } x + 48 = 0 \\[1em] \Rightarrow x = 36 \text{ or } x = -48$

Since speed of car cannot be negative hence x ≠ -48

∴ x = 36

Equation : $\dfrac{216}{x} - \dfrac{208}{x + 16} = 2$

(iv) Speed of train = (x + 16) = (36 + 16) = 52 km/h

Hence, speed of train is 52 km/h.

Question 30

An aeroplane flying with a wind of 30 km/h takes 40 minutes less to fly 3600 km , then what it would have taken to fly against the same wind. Find the plane's speed of flying in still air.

Answer

Let the speed of plane in still air be x km/h

Speed of wind = 30 km/h

∴ Speed of plane in wind = (x + 30) km/h and Speed of plane against wind = (x - 30) km/h.

40 minutes = $\dfrac{40}{60} \text{ hours } = \dfrac{2}{3} \text{ hours }$

According to question,

$\Rightarrow \dfrac{3600}{x - 30} - \dfrac{3600}{x + 30} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3600(x + 30) - 3600(x - 30)}{(x - 30)(x + 30)} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3600x + 108000 - 3600x + 108000}{x^2 - 30x + 30x - 900} = \dfrac{2}{3} \\[1em] \Rightarrow 216000 \times 3 = 2(x^2 - 900) \text{ (On cross multiplying) }\\[1em] \Rightarrow 648000 = 2(x^2 - 900) \\[1em] \Rightarrow 324000 = x^2 - 900 \text{ (Dividing the complete equation by 2) } \\[1em] \Rightarrow x^2 - 900 - 324000 = 0 \\[1em] \Rightarrow x^2 - 324900 = 0 \\[1em] \Rightarrow x^2 - (570)^2 = 0 \\[1em] \Rightarrow (x - 570)(x + 570) = 0 \\[1em] \Rightarrow x - 570 = 0 \text{ or } x + 570 = 0 \\[1em] x = 570 \text{ or } x = -570 \\[1em]$

Since speed of aeroplane cannot be negative hence, x ≠ -570.

The speed of aeroplane in still air is 570 km/h.

Question 31

A school bus transported an excursion party to a picnic spot 150 km away. While returning , it was raining and the bus had to reduce its speed by 5 km/h , and it took one hour longer to make the return trip. Find the time taken to return.

Answer

Let the speed of bus while reaching picnic spot be x km/h

Since, the speed of bus decrease by 5 km/h due to rain hence, speed of bus in rain = (x - 5) km/h

∴ Time taken to reach picnic spot = $\dfrac{150}{x}$ and Time taken to reach back to school = $\dfrac{150}{x - 5}$

According to given,

$\Rightarrow \dfrac{150}{x - 5} - \dfrac{150}{x} = 1 \\[1em] \Rightarrow \dfrac{150x - 150(x - 5)}{x(x - 5)} = 1 \\[1em] \Rightarrow \dfrac{150x - 150x + 750}{x^2 - 5x} = 1 \\[1em] \Rightarrow 750 = x^2 - 5x \text{ (On cross multiplying) } \\[1em] \Rightarrow x^2 - 5x - 750 = 0 \\[1em] \Rightarrow x^2 - 30x + 25x - 750 = 0 \\[1em] \Rightarrow x(x - 30) + 25(x - 30) = 0 \\[1em] \Rightarrow (x + 25)(x - 30) = 0 \\[1em] \Rightarrow x + 25 = 0 \text{ or } x - 30 = 0 \\[1em] x = -25 \text{ or } x = 30$

Since speed of bus cannot be negative hence, x ≠ -25.

∴ x = 30

If x = 30 , x - 5 = 25.

Time taken to return = $\dfrac{150}{x - 5}$ hours = 6 hours.

Hence, time taken on return trip is 6 hours.

Question 32

A boat can cover 10 km up the stream and 5 km down the stream in 6 hours . If the speed of the stream is 1.5 km/h, find the speed of the boat in still water.

Answer

Let the speed of boat in still water be x km/h

Speed of stream = 1.5 km/h

∴ Speed of boat upstream = (x - 1.5) km/h and Speed of boat downstream = (x + 1.5) km/h

According to given,

$\Rightarrow \dfrac{10}{x - 1.5} + \dfrac{5}{x + 1.5} = 6 \\[1em] \Rightarrow \dfrac{10(x + 1.5) + 5(x - 1.5)}{(x + 1.5)(x - 1.5)} = 6 \\[1em] \Rightarrow \dfrac{10x + 15 + 5x - 7.5}{x^2 - 1.5x + 1.5x - 2.25} = 6 \\[1em] \Rightarrow \dfrac{15x + 7.5}{x^2 - 2.25} = 6 \\[1em] \Rightarrow 15x + 7.5 = 6(x^2 - 2.25) \\[1em] \Rightarrow 15x + 7.5 = 6x^2 - 13.50 \\[1em] \Rightarrow 6x^2 - 13.5 - 7.5 - 15x = 0 \\[1em] \Rightarrow 6x^2 - 15x - 21 = 0 \\[1em] \Rightarrow 3(2x^2 - 5x - 7) = 0 \\[1em] \Rightarrow 2x^2 - 5x - 7 = 0 \\[1em] \Rightarrow 2x^2 - 7x + 2x - 7 = 0 \\[1em] \Rightarrow x(2x - 7) + 1(2x - 7) = 0 \\[1em] \Rightarrow (x + 1)(2x - 7) = 0 \\[1em] \Rightarrow x = -1 \text{ or } 2x - 7 = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x = \dfrac{7}{2} \\[1em] \Rightarrow x = -1 \text{ or } x = 3.5$

Since speed of bus cannot be negative hence, x ≠ -1

Hence speed of boat in still water is 3.5 km/h.

Question 33

Two pipes running together can fill a tank in $11\dfrac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would fill the tank.

Answer

The tank is filled by the two pipes together in $11\dfrac{1}{9}$ minutes i.e. in $\dfrac{100}{9}$ minutes,

∴ the part of tank filled in one minute = $\dfrac{9}{100}$

Let the time taken by two pipes to fill tank separately be x minutes and (x + 5) minutes.

∴ the part of tank filled by the first pipe in one minute = $\dfrac{1}{x}$ and

the part of tank filled by the second pipe in one minute = $\dfrac{1}{x + 5}$

According to the question

$\dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{9}{100} \\[1em] \Rightarrow \dfrac{x + 5 + x}{x(x + 5)} = \dfrac{9}{100} \\[1em] \Rightarrow \dfrac{2x + 5}{x^2 + 5x} = \dfrac{9}{100} \\[1em] \Rightarrow 100(2x + 5) = 9(x^2 + 5x) \\[1em] \Rightarrow 200x + 500 = 9x^2 + 45x \\[1em] \Rightarrow 9x^2 + 45x - 200x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 155x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 180x + 25x - 500 = 0 \\[1em] \Rightarrow 9x(x - 20) + 25(x - 20) = 0 \\[1em] \Rightarrow (9x + 25)(x - 20) = 0 \\[1em] \Rightarrow x = -\dfrac{25}{9} \text{ or } x = 20$

Since, time cannot be negative hence, x ≠ $-\dfrac{25}{9}$

∴ x = 20 , x + 5 = 25.

Time taken by each pipe is 20 minutes and 25 minutes.

Question 34

₹480 is divided equally among 'x' children. If the number of children were 20 more, then each would have got ₹12 less . Find 'x'.

Answer

Number of children = x

Money recieved by each = $\dfrac{480}{x}$

If number of children were 20 more then money received by each child = $\dfrac{480}{x + 20}$

Given, if there are 20 more children then each will get ₹12 less

$\therefore \dfrac{480}{x} - \dfrac{480}{x + 20} = 12 \\[1em] \Rightarrow \dfrac{40}{x} - \dfrac{40}{x + 20} = 1 \text{ (Dividing the equation by 12) } \\[1em] \Rightarrow \dfrac{40(x + 20) - 40x}{x(x + 20)} = 1 \\[1em] \Rightarrow \dfrac{40x + 800 - 40x}{x^2 + 20x} = 1 \\[1em] \Rightarrow 800 = x^2 + 20x \\[1em] \Rightarrow x^2 + 20x - 800 = 0 \\[1em] \Rightarrow x^2 + 40x - 20x - 800 = 0 \\[1em] \Rightarrow x(x + 40) - 20(x + 40) = 0 \\[1em] \Rightarrow (x - 20)(x + 40) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 40 = 0 \\[1em] x = 20 \text{ or } x = -40$

Since, number of children cannot be negative hence, x ≠ -40

∴ x = 20

Hence, number of children are 20.

Question 35

₹7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹100 more. Find the original number of children.

Answer

Let the number of children be x

₹7500 is divided equally among x children so , each received = ₹ $\dfrac{7500}{x}$

Given, if there were 20 less children, each would have received ₹ 100 more.

If there are 20 less children then each would receive = ₹ $\dfrac{7500}{x - 20}$

$\therefore \dfrac{7500}{x - 20} - \dfrac{7500}{x} = 100 \\[1em] \Rightarrow \dfrac{75}{x - 20} - \dfrac{75}{x} = 1 \text{ (Dividing the equation by 100) } \\[1em] \Rightarrow \dfrac{75x - 75(x - 20)}{x(x - 20)} = 1 \\[1em] \Rightarrow \dfrac{75x - 75x + 1500}{x(x - 20)} = 1 \\[1em] \Rightarrow \dfrac{1500}{x(x - 20)} = 1 \\[1em] \Rightarrow 1500 = x^2 - 20x \text{ (On cross multiplication) }\\[1em] \Rightarrow x^2 - 20x - 1500 = 0 \\[1em] \Rightarrow x^2 - 50x + 30x - 1500 = 0 \\[1em] \Rightarrow x(x - 50) + 30(x - 50) = 0 \\[1em] \Rightarrow (x + 30)(x - 50) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 50 = 0 \\[1em] x = -30 \text{ or } x = 50$

Since, number of children cannot be negative hence, x ≠ -30

∴ x = 50

Hence, number of children are 50.

Question 36

2x articles cost ₹ (5x + 54) and (x + 2) similar articles cost ₹(10x - 4); find x .

Answer

2x articles cost ₹(5x + 54)

So, cost of each article = ₹ $\Big(\dfrac{5x + 54}{2x}\Big)$

Similar (x + 2) articles cost ₹(10x - 4)

So, cost of each article = ₹ $\Big(\dfrac{10x - 4}{x + 2}\Big)$

$\therefore \dfrac{5x + 54}{2x} = \dfrac{10x - 4}{x + 2} \\[1em] \Rightarrow (5x + 54)(x + 2) = 2x(10x - 4) \\[1em] \Rightarrow 5x^2 + 10x + 54x + 108 = 20x^2 - 8x \\[1em] \Rightarrow 5x^2 - 20x^2 + 64x + 8x + 108 = 0 \\[1em] \Rightarrow -15x^2 + 72x + 108 = 0 \\[1em] \Rightarrow 15x^2 - 72x - 108 = 0 \text{ (On multiplying equation by -1) }\\[1em] \Rightarrow 3(5x^2 - 24x - 36) = 0 \\[1em] \Rightarrow 5x^2 - 24x - 36 = 0 \\[1em] \Rightarrow 5x^2 - 30x + 6x - 36 = 0 \\[1em] \Rightarrow 5x(x - 6) + 6(x - 6) = 0 \\[1em] \Rightarrow (5x + 6)(x - 6) = 0 \\[1em] \Rightarrow (5x + 6) = 0 \text{ or } x - 6 = 0 \\[1em] x = -\dfrac{6}{5} \text{ or } x = 6$

Value of x cannot be negative and in fraction as that will make number of articles in fraction which is not possible hence, x ≠ - $\dfrac{6}{5}$

∴ x = 6

Hence , value of x is 6.

Question 37

A trader buys x articles for a total cost of ₹600.

(i) Write down the cost of one article in terms of x.

If the cost per article were ₹5 more, the number of articles that can be bought for ₹600 would be four less.

(ii) Write down the equation in x for the above situation and solve it to find x.

Answer

(i) x articles cost ₹600.

So, cost of each article = $\dfrac{600}{x}$

(ii) Given,

If the cost per article were ₹5 more, the number of articles that can be bought for ₹600 would be four less.

$\therefore \dfrac{600}{x - 4} - \dfrac{600}{x} = 5 \\[1em] \Rightarrow \dfrac{600x - 600(x - 4)}{x(x - 4)} = 5 \\[1em] \Rightarrow 600x - 600x + 2400 = 5x(x - 4) \text{ (On cross multiplication) } \\[1em] \Rightarrow 5x^2 - 20x - 2400 = 0 \\[1em] \Rightarrow 5(x^2 - 4x - 480) = 0 \\[1em] \Rightarrow x^2 - 4x - 480 = 0 \\[1em] \Rightarrow x^2 - 24x + 20x - 480 = 0 \\[1em] \Rightarrow x(x - 24) + 20(x - 24) = 0 \\[1em] \Rightarrow (x - 24)(x + 20) = 0 \\[1em] x = 24 \text{ or } x = -20$

Since, no of articles cannot be negative hence, x ≠ -20

∴ x = 24

Hence, the number of articles that trader buys are 24.

Question 38

A shopkeeper buys a certain number of books for ₹960. If the cost per book was ₹8 less, the number of books that could be bought for ₹960 would be 4 more. Taking the original cost of each book to be ₹x, write an equation in x and solve it to find the original cost of each book.

Answer

Let the cost of each book be ₹x

Number of books that can be bought for ₹960 = $\dfrac{960}{x}$

If the cost of book is ₹8 less then number of books that can be bought for ₹960 = $\dfrac{960}{x - 8}$

Given, number of books would be 4 more if price would be ₹8 less

$\therefore \dfrac{960}{x - 8} - \dfrac{960}{x} = 4 \\[1em] \Rightarrow \dfrac{960x - 960(x - 8)}{x(x - 8)} = 4 \\[1em] \Rightarrow \dfrac{960x - 960x + 7680}{x^2 - 8x} = 4 \\[1em] \Rightarrow 7680 = 4(x^2 - 8x) \\[1em] \Rightarrow 4x^2 - 32x - 7680 = 0 \\[1em] \Rightarrow 4(x^2 - 8x - 1920) = 0 \\[1em] \Rightarrow x^2 - 8x - 1920 = 0 \\[1em] \Rightarrow x^2 - 48x + 40x - 1920 = 0 \\[1em] \Rightarrow x(x - 48) + 40(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x + 40) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x + 40 = 0 \\[1em] x = 48 \text{ or } x = -40$

Since, cost of book cannot be negative hence, x ≠ -40

∴ x = 48

Hence, the cost of each book is ₹48.

Question 39

A piece of cloth cost ₹300. If the piece was 5 metre longer and each metre of cloth cost ₹2 less, the cost of the piece would have remained unchanged. How long is the original piece of cloth and what is the rate per metre?

Answer

Let the length of original cloth be x metres

Since, cost of total piece = ₹300

So, cost of each metre = $\dfrac{300}{x}$

Given, new length is 5 metre more i.e. (x + 5) metres and cost of each metre is ₹2 less i.e. $\dfrac{300}{x} - 2$

Since total cost remains unchanged

$\therefore (x + 5) \times \big(\dfrac{300}{x} - 2 \big) = 300 \\[1em] \Rightarrow (x + 5)\big(\dfrac{300 - 2x}{x}\big) = 300 \\[1em] \Rightarrow (x + 5)(300 - 2x) = 300x \text{ (On cross multiplication) } \\[1em] \Rightarrow 300x - 2x^2 + 1500 - 10x - 300x = 0 \\[1em] \Rightarrow -2x^2 + 1500 - 10x = 0 \\[1em] \Rightarrow -2(x^2 + 5x - 750) = 0 \\[1em] \Rightarrow x^2 + 5x - 750 = 0 \\[1em] \Rightarrow x^2 + 30x - 25x - 750 = 0 \\[1em] \Rightarrow x(x + 30) - 25(x + 30) = 0 \\[1em] \Rightarrow (x + 30)(x - 25) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 25 = 0 \\[1em] x = -30 \text{ or } x = 25$

Since, length cannot be negative hence, x ≠ -30

∴ x = 25 , $\dfrac{300}{x}$ = 12

Hence, the length of original cloth is 25 metre and cost per metre is ₹12.

Question 40

The hotel bill for a number of people for overnight stay is ₹4800 . If there were 4 more the bill each person had to pay would have reduced by ₹200. Find the number of people staying overnight.

Answer

Let the number of people be x

Total bill = ₹4800

∴ Bill amount each person has to pay = $\dfrac{4800}{x}$

If there were 4 more people then bill each has to pay = $\dfrac{4800}{x + 4}$

According to question,

$\Rightarrow \dfrac{4800}{x} - \dfrac{4800}{x + 4} = 200 \\[1em] \Rightarrow \dfrac{4800(x + 4) - 4800x}{x(x + 4)} = 200 \\[1em] \Rightarrow 4800x - 4800x + 19200 = 200x(x + 4) \\[1em] \Rightarrow 19200 = 200x^2 + 800x \\[1em] \Rightarrow 200x^2 + 800x - 19200 = 0 \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \text{ (On Dividing the equation by 200) } \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x - 8)(x + 12) = 0 \\[1em] x = 8 \text{ or } x = -12$

Since, number of people cannot be negative hence, x ≠ -12

∴ x = 8

Hence, the number of people staying overnight are 8.

Question 41

A person was given ₹3000 for a tour. If he extends his tour programme by 5 days, he must cut down his daily expense by ₹20. Find the number of days of his tour programme.

Answer

Let number of days for which person plans trip be x

Total expense for trip = ₹3000

∴ Expense of each day = ₹ $\Big(\dfrac{3000}{x}\Big)$

Days extended = 5 , so total days = (x + 5)

Daily expense to be cut down is ₹20,

∴ New Expense of each day = ₹ $\Big(\dfrac{3000}{x} - 20\Big)$

According to question,

$\Rightarrow (x + 5)\Big(\dfrac{3000}{x} - 20\Big) = 3000 \\[1em] \Rightarrow (x + 5)\Big(\dfrac{3000 - 20x}{x}\Big) = 3000 \\[1em] \Rightarrow (x + 5)(3000 - 20x) = 3000x \text{ (Cross multiplying) } \\[1em] \Rightarrow 3000x - 20x^2 + 15000 - 100x = 3000x \\[1em] \Rightarrow -20x^2 + 3000x - 3000x -100x + 15000 = 0 \\[1em] \Rightarrow -20x^2 - 100x + 15000 = 0 \\[1em] \Rightarrow -20(x^2 + 5x - 750) = 0 \\[1em] \Rightarrow x^2 + 5x - 750 = 0 \\[1em] \Rightarrow x^2 + 30x - 25x - 750 = 0 \\[1em] \Rightarrow x(x + 30) - 25(x + 30) = 0 \\[1em] \Rightarrow (x - 25)(x + 30) = 0 \\[1em] \Rightarrow x - 25 = 0 \text{ or } x + 30 = 0 \\[1em] x = 25 \text{ or } x = -30$

Since, number of days cannot be negative hence, x ≠ -30

∴ x = 25

Hence, the number of days of tour programme are 25.

Question 42

The sum of the ages of Vivek and his youger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.

Answer

Let the age of Vivek be x years.

Since sum of ages of vivek and amit is 47, so amit's age = (47 - x) years

Product of ages = 550

∴ x(47 - x) = 550

$\Rightarrow 47x - x^2 = 550 \\[1em] \Rightarrow x^2 - 47x + 550 = 0 \\[1em] \Rightarrow x^2 - 25x - 22x + 550 = 0 \\[1em] \Rightarrow x(x - 25) - 22(x - 25) = 0 \\[1em] \Rightarrow (x - 22)(x - 25) = 0 \\[1em] \Rightarrow x - 22 = 0 \text{ or } x - 25 = 0 \\[1em] x = 22 \text{ or } x = 25.$

Vivek's age = 25 years and Amit's age = 22 years.

Question 43

Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Paul's age. Find their present ages.

Answer

Let paul's present age be x years

Father's age = 2x²

After 10 years,

Paul's age = (x + 10) years

Father's age = (2x² + 10)

According to question,

$\Rightarrow 2x^2 + 10 = 4(x + 10) \\[1em] \Rightarrow 2x^2 + 10 = 4x + 40 \\[1em] \Rightarrow 2x^2 - 4x + 10 - 40 = 0 \\[1em] \Rightarrow 2x^2 - 4x - 30 = 0 \\[1em] \Rightarrow 2(x^2 - 2x - 15)= 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3( x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0\\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 5 = 0 \\[1em] x = -3 \text{ or } x = 5$

Since, age cannot be negative hence x ≠ -3

∴ x = 5 , 2x² = 50

Paul's age is 5 years , while his father's age is 50 years.

Question 44

The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be 4 years more than 3 times the age of his son. Find their present ages.

Answer

Let son's present age be x years

Man's age = 2x²

After 8 years,

Son's age = (x + 8) years

Man's age = (2x² + 8)

According to question,

$\Rightarrow 2x^2 + 8 = 3(x + 8) + 4 \\[1em] \Rightarrow 2x^2 + 8 = 3x + 24 + 4 \\[1em] \Rightarrow 2x^2 + 8 = 3x + 28 \\[1em] \Rightarrow 2x^2 - 3x + 8 - 28 = 0 \\[1em] \Rightarrow 2x^2 - 3x - 20 = 0 \\[1em] \Rightarrow 2x^2 - 8x + 5x - 20 = 0 \\[1em] \Rightarrow 2x(x - 4) + 5(x - 4) = 0 \\[1em] \Rightarrow (x - 4)(2x + 5) = 0 \\[1em] \Rightarrow x - 4 = 0 \text{ or } 2x + 5 = 0 \\[1em] x = 4 \text{ or } x = -\dfrac{5}{2}$

Since, age cannot be negative hence x ≠ $-\dfrac{5}{2}$

∴ x = 4 , 2x² = 32

The present age of son is 4 years while the age of man is 32 years.

Question 45

Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Answer

Let present age of daughter be x years

Two years before daughter's age = (x - 2) years

Man's age two years before = 3(x - 2)²

So, present age of man = 3(x - 2)² + 2

Three year's later

Daughter's age = (x + 3) years

Man's age = (3(x - 2)² + 2 + 3) years

According to question,

$\Rightarrow (3(x - 2)^2 + 2 + 3) = 4(x + 3) \\[1em] \Rightarrow 3(x^2 + 4 - 4x) + 5 = 4x + 12 \\[1em] \Rightarrow 3x^2 + 12 - 12x + 5 = 4x + 12 \\[1em] \Rightarrow 3x^2 - 12x - 4x + 17 - 12 = 0 \\[1em] \Rightarrow 3x^2 - 16x + 5 = 0 \\[1em] \Rightarrow 3x^2 - 15x - x + 5 = 0 \\[1em] \Rightarrow 3x(x - 5) - 1(x - 5) = 0 \\[1em] \Rightarrow (3x - 1)(x - 5) = 0 \\[1em] \Rightarrow 3x - 1 = 0 \text{ or } x - 5 = 0 \\[1em] x = \dfrac{1}{3} \text{ or } x = 5$

Since, age cannot be in fraction so x ≠ $\dfrac{1}{3}$

∴ x = 5, 3(x - 2)² + 2 = 3(5 - 2)² + 2 = 3(3)² + 2 = 29

The present age of daughter is 5 years and of man is 29 years.

Question 46

The length (in cm) of the hypotenuse of a right angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.

Answer

Let the lengths of the two sides other than hypotenuse be x cm and y cm

According to question,

Hypotenuse = x + 2 (in terms of 1st side)

Hypotenuse = 2y + 1 (in terms of 2nd side)

∴ x + 2 = 2y + 1
x = 2y + 1 - 2
x = 2y - 1

Hypotenuse = 2y + 1

As the given triangle is right-angled, by using Pythagoras theorem, we get:

$x^2 + y^2 = (2y + 1)^2$

Putting value of x = 2y - 1 in above equation

$\Rightarrow (2y - 1)^2 + y^2 = (2y + 1)^2 \\[1em] \Rightarrow 4y^2 + 1 - 4y + y^2 = 4y^2 + 1 + 4y \\[1em] \Rightarrow 4y^2 - 4y^2 + y^2 + 1 - 1 - 4y -4y = 0 \\[1em] \Rightarrow y^2 - 8y = 0 \\[1em] \Rightarrow y(y - 8) = 0 \\[1em] \Rightarrow y = 0 \text{ or } y - 8 = 0 \\[1em] y = 0 \text{ or } y = 8$

y ≠ 0 , as that will make value of x negative and length cannot be negative.

∴ y = 8 , x = 2y - 1 = 15 , hypotenuse = 2y + 1 = 17

Perimeter = 8 + 15 + 17 = 40 cm

$\text{Area} = \dfrac{1}{2} \times \text{first side } \times \text{ second side} \\[1em] = \dfrac{1}{2} \times 8 \times 15 \text{ cm}^2 \\[1em] = 60 \text{ cm}^2$

Hence, the value of first side = 8cm, second side = 15 cm , Hypotenuse = 17cm , Perimeter = 40cm , Area = 60 cm².

Question 47

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of two squares.

Answer

Let the sides of the smaller and bigger square be x cm and y cm, respectively.

Area of smaller square = x² cm²

Area of larger square = y² cm²

According to question,

$y^2 - 2x^2 = 14 \\[1em] \Rightarrow y^2 = 14 + 2x^2 \qquad \text{[...Eq 1]}$

and

$3x^2 + 2y^2 = 203 \qquad \text{[...Eq 2]}$

Putting value of y² from Equation 1 in Equation 2, we get:

$\Rightarrow 3x^2 + 2(14 + 2x^2) = 203 \\[1em] \Rightarrow 3x^2 + 28 + 4x^2 = 203 \\[1em] \Rightarrow 7x^2 = 203 - 28 \\[1em] \Rightarrow 7x^2 = 175 \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x = \sqrt{25} \\[1em] x = 5 \text{ or } x = -5$

Since, side of a square cannot be negative hence, x ≠ -5.

∴ x = 5
y² = 14 + 2² = 64
⇒ y = 8

Hence, the length of larger square's side is 8cm and smaller square's is 5cm.

Multiple Choice Questions

Question 1

Which of the following is not a quadratic equation?

(x + 2)² = 2(x + 3)
x² + 3x = (-1)(1 - 3x)
(x + 2)(x - 1) = x² - 2x - 3
x³ - x² + 2x + 1 = (x + 1)³

Answer

Option 1:

$(x + 2)^2 = 2(x + 3) \\[0.5em] \Rightarrow x^2 + 4 + 4x = 2x + 6 \\[0.5em] \Rightarrow x^2 + 4x - 2x + 4 - 6 = 0 \\[0.5em] \Rightarrow x^2 + 2x - 2 = 0$

It is a quadratic equation as highest power of x is 2.

Option 2:

$x^2 + 3x = (-1)(1 - 3x) \\[0.5em] \Rightarrow x^2 + 3x = -1 + 3x \\[0.5em] \Rightarrow x^2 + 3x - 3x + 1 = 0 \\[0.5em] x^2 + 1 = 0$

It is a quadratic equation as the highest power of x is 2.

Option 3:

$(x + 2)(x - 1) = x^2 - 2x - 3 \\[0.5em] \Rightarrow x^2 - x + 2x - 2 = x^2 - 2x - 3 \\[0.5em] \Rightarrow x^2 - x^2 - x + 2x + 2x - 2 + 3 = 0 \\[0.5em] 3x + 1 = 0$

It is not a quadratic equation as highest power of x is not 2.

Option 4:

$x^3 - x^2 + 2x + 1 = (x + 1)^3 \\[0.5em] \Rightarrow x^3 - x^2 + 2x + 1 = x^3 + 1 + 3x(x + 1) \\[0.5em] \Rightarrow x^3 - x^2 + 2x + 1 = x^3 + 1 + 3x^2 + 3x \\[0.5em] \Rightarrow x^3 - x^3 - x^2 - 3x^2 + 2x - 3x + 1 - 1 = 0 \\[0.5em] \Rightarrow -4x^2 - x = 0 \\[0.5em] 4x^2 + x = 0$

It is a quadratic equation as highest power of x is 2.

∴ Option 3 is the correct option.

Question 2

If 3 is a root of the quadratic equation x² - px + 3 = 0 then p is equal to :

Answer

Given,

3 is a root of the quadratic equation x² - px + 3 = 0

∴ 3² - 3p + 3 = 0

⇒ 9 - 3p + 3 = 0

⇒ 12 - 3p = 0

⇒ 3p = 12

⇒ p = $\dfrac{12}{3}$ = 4.

Hence, Option 1 is the correct option.

Question 3

The roots of the equation x² - 3x - 10 = 0 are

2, -5
-2, 5
2, 5
-2, -5

Answer

Given,

$x^2 - 3x - 10 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 2x - 10 = 0 \\[0.5em] \Rightarrow x(x - 5) + 2(x - 5) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 5) = 0 \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 5 = 0 \\[0.5em] x = -2 \text{ or } x = 5 \\[0.5em]$

∴ Option 2 is the correct option

Question 4

If one root of a quadratic equation with rational coefficients is $\dfrac{3 - \sqrt{5}}{2}$ , then the other root is

$\dfrac{-3 - \sqrt{5}}{2}$
$\dfrac{-3 + \sqrt{5}}{2}$
$\dfrac{3 + \sqrt{5}}{2}$
$\dfrac{\sqrt{3} + 5}{2}$

Answer

Irrational roots occur in conjugate pair .

Hence if one root is $\dfrac{3 - \sqrt{5}}{2}$ then the other root is =

$\dfrac{3 + \sqrt{5}}{2}$

∴ Option 3 is the correct option

Question 5

If the equation 2x² - 5x + (k + 3) = 0 has equal roots then the value of k is

$\dfrac{9}{8}$
$-\dfrac{9}{8}$
$\dfrac{1}{8}$
$-\dfrac{1}{8}$

Answer

Given ,

2x² - 5x + (k + 3) = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -5 , c = k + 3

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-5)^2 - 4 \times 2 \times (k + 3) = 0 \\[0.5em] \Rightarrow 25 - 8(k + 3) = 0 \\[0.5em] \Rightarrow 25 - 8k - 24 = 0 \\[0.5em] \Rightarrow 1 - 8k = 0 \\[0.5em] \Rightarrow 8k = 1 \\[0.5em] k = \dfrac{1}{8}$

∴ Option 3 is the correct option.

Question 6

The value(s) of k for which the quadratic equation 2x² - kx + k = 0 has equal roots is (are)

0 only
4
8 only
0, 8

Answer

Given ,

2x² - kx + k = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -k , c = k

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-k)^2 - 4 \times 2 \times k = 0 \\[0.5em] \Rightarrow k^2 - 8k = 0 \\[0.5em] \Rightarrow k(k - 8) = 0 \\[0.5em] \Rightarrow k = 0 \text{ or } k - 8 = 0 \\[0.5em] k = 0 \text{ or } k = 8 \\[0.5em]$

∴ Option 4 is the correct option.

Question 7

If the equation 3x² - kx + 2k = 0 has equal roots, then the value(s) of k is (are)

6
0 only
24 only
0 or 24

Answer

Given ,

3x² - kx + 2k = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= 3 , b = -k , c = 2k

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-k)^2 - 4 \times 3 \times 2k = 0 \\[0.5em] \Rightarrow k^2 - 24k = 0 \\[0.5em] \Rightarrow k(k - 24) = 0 \\[0.5em] \Rightarrow k = 0 \text{ or } k - 24 = 0 \\[0.5em] k = 0 \text{ or } k = 24 \\[0.5em]$

∴ Option 4 is the correct option.

Question 8

If the equation (k + 1)x² - 2(k - 1)x + 1 = 0 has equal roots, then the values of k are

1, 3
0, 3
0, 1
0, $\dfrac{3}{4}$

Answer

Given ,

(k + 1)x² - 2(k - 1)x + 1 = 0 has equal roots

Comparing equation with ax² + bx + c = 0
a= (k + 1) , b = -2(k - 1) , c = 1

Since, equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-2(k - 1))^2 - 4 \times (k + 1) \times 1 = 0 \\[0.5em] \Rightarrow 4(k^2 + 1 - 2k) - 4(k + 1) = 0 \\[0.5em] \Rightarrow 4k^2 + 4 - 8k - 4k - 4 = 0 \\[0.5em] \Rightarrow 4k^2 - 12k = 0 \\[0.5em] \Rightarrow 4k(k - 3) = 0 \\[0.5em] \Rightarrow 4k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3.$

∴ Option 2 is the correct option.

Question 9

If the equation 2x² - 6x + p = 0 has real and different roots, then the values of p are given by

$p \lt \dfrac{9}{2}$
$p \le \dfrac{9}{2}$
$p \gt \dfrac{9}{2}$
$p \ge \dfrac{9}{2}$

Answer

Given ,

2x² - 6x + p = 0 has real and different roots

Comparing equation with ax² + bx + c = 0
a= 2 , b = -6 , c = p

Since, equation has real and different roots

∴ b² - 4ac > 0

$\Rightarrow (-6)^2 - 4 \times 2 \times p \gt 0 \\[0.5em] \Rightarrow (-6)^2 - 8p \gt 0 \\[0.5em] \Rightarrow 36 - 8p \gt 0 \\[0.5em] \Rightarrow 8p \lt 36 \\[0.5em] \Rightarrow p \lt \dfrac{36}{8} \\[0.5em] p \lt \dfrac{9}{2}$

∴ Option 1 is the correct option.

Question 10

The quadratic equation $2x^2 - \sqrt{5}x + 1 = 0$ has

two distinct real roots
two equal real roots
no real roots
more than two real roots

Answer

In order to find nature of roots we need to find the value of, b² - 4ac

Given,

$2x^2 - \sqrt{5}x + 1 = 0$

Comparing equation with ax² + bx + c = 0
a= 2 , b = - $\sqrt{5}$ , c = 1

Putting values in b² - 4ac

$(-\sqrt{5})^2 - 4 \times 2 \times 1 \\[0.5em] \Rightarrow 5 - 8 \\[0.5em] \Rightarrow -3$

Since, b² - 4ac = -3 < 0 , hence there are no real roots

∴ Option 3 is the correct option.

Question 11

If the roots of equation x² - 6x + k = 0 are real and distinct, then value of k is :

> -9
> -6
< 6
< 9

Answer

Given,

Roots of equation x² - 6x + k = 0 are real and distinct.

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-6)² - 4 × 1 × k > 0

⇒ 36 - 4k > 0

⇒ 4k < 36

⇒ k < $\dfrac{36}{4}$

⇒ k < 9.

Hence, Option 4 is the correct option.

Question 12

The roots of the quadratic equation px² - qx + r = 0 are real and equal if :

(a) p² = 4qr

(b) q² = 4pr

(d) p² > 4qr

Answer

By formula,

D = b² - 4ac

For equation, px² - qx + r = 0

D = (-q)² - 4 × p × r

We know that,

Roots of a quadratic equation are real and equal if discriminant = 0.

⇒ q² - 4pr = 0

⇒ q² = 4pr.

Hence, Option 2 is the correct option.

Question 13

If x² + kx + 6 = (x - 2)(x - 3) for all values of x, then the value of k is :

-5
-3
-2
5

Answer

Given,

⇒ x² + kx + 6 = (x - 2)(x - 3)

⇒ x² + kx + 6 = x² - 3x - 2x + 6

⇒ x² + kx + 6 = x² - 5x + 6

⇒ x² - x² + kx + 6 - 6 = -5x

⇒ kx = -5x

⇒ k = -5.

Hence, Option 1 is the correct option.

Question 14

The roots of quadratic equation x² - 1 = 0 are :

0
1
-1
±1

Answer

Solving,

⇒ x² - 1 = 0

⇒ (x + 1)(x - 1) = 0

⇒ x + 1 = 0 or x - 1 = 0

⇒ x = -1 or x = 1.

Hence, Option 4 is the correct option.

Assertion-Reason Type Questions

Question 1

Assertion (A): Every quadratic equation ax² + bx + c = 0, a ≠ 0, a, b and c are all real numbers has two real roots.

Reason (R): Every quadratic equation ax² + bx + c = 0, a ≠ 0, a, b and c are all real numbers has two real roots if b² - 4ac ≥ 0.

Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

The quadratic equation: ax² + bx + c = 0.

The expression b² - 4ac is called the discriminant (D).

When,

D > 0; two distinct real roots
D = 0; real and equal roots
D < 0; then roots are imaginary

thus, assertion (A) is false but reason(R) is true.

Hence, option 2 is the correct option.

Question 2

Assertion (A): The quadratic equation 4x² + 12x + 15 = 0, has no real roots.

Reason (R): The quadratic equation ax² + bx + c = 0, has real roots iff its 'discriminant' = b² - 4ac ≥ 0.

Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

For, the quadratic equation: ax² + bx + c = 0. The equation has real roots if

b² - 4ac ≥ 0

So, reason (R) is true.

Comparing equation 4x² + 12x + 15 = 0, with ax² + bx + c = 0, we get :

a = 4, b = 12, c = 15

D = b² - 4ac

= 12² - 4 x 4 x 15

= 144 - 240 = -96.

Since, D < 0, so the equation has no real roots.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Question 3

Assertion (A): The equation 9x² + 6x - k = 0 has real roots if k ≥ -1.

Reason (R): The quadratic equation ax² + bx + c = 0 has real roots if 'discriminant' = b² - 4ac > 0.

Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

We know that,

The quadratic equation ax² + bx + c = 0 has real roots if 'discriminant' = b² - 4ac ≥ 0.

So, reason (R) is false.

Given, 9x² + 6x - k = 0

Comparing above equation with ax² + bx + c = 0, we get :

a = 9, b = 6 and c = -k

If the equation has real roots, then D ≥ 0

⇒ b² - 4ac ≥ 0

⇒ 6² - 4 x 9 x (-k) ≥ 0

⇒ 36 + 36k ≥ 0

⇒ 36k ≥ -36

⇒ k ≥ - $\dfrac{36}{36}$

⇒ k ≥ -1

So, assertion (A) is true.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Question 4

Consider the polynomial 2x² - 3x + 5

Assertion (A): Factorisation of the above polynomial is not possible.

Reason (R): Discriminant 'b² - 4ac' is negative.

Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given,

Polynomial : 2x² - 3x + 5

Discriminant (D) = b² - 4ac

= (-3)² - 4 x 2 x 5

= 9 - 40

= -31.

So, reason (R) is true.

Since the discriminant is negative, this quadratic has no real roots and cannot be factorized into linear factors with real coefficients.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Question 5

Consider the following equation k²x² - 2kx + 1 = 0

Assertion (A): This equation has real roots for all non-zero values of k.

Reason (R): The discriminant of this equation is zero.

Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given,

Equation : k²x² - 2kx + 1 = 0

Comparing above equation with ax² + bx + c = 0, we get :

a = k², b = -2k and c = 1

Discriminant (D) = b² - 4ac

= (-2k)² - 4 x k² x 1

= 4k² - 4k²

= 0.

So, reason (R) is true.

Since, D = 0 this means the equation has one repeated real root.

So, assertion (A) is true and reason (R) correctly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Chapter Test

Question 1(i)

Solve the following equations by factorisation:

x² + 6x - 16 = 0

Answer

Given,

$\Rightarrow x^2 + 6x - 16 = 0 \\[1em] \Rightarrow x^2 + 8x - 2x - 16 = 0 \\[1em] \Rightarrow x(x + 8) - 2(x + 8) = 0 \\[1em] \Rightarrow (x - 2)(x + 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -8$

Hence, roots of given equation are 2, -8.

Question 1(ii)

Solve the following equations by factorisation:

3x² + 11x + 10 = 0

Answer

Given,

$\Rightarrow 3x^2 + 11x + 10 = 0 \\[1em] \Rightarrow 3x^2 + 6x + 5x + 10 = 0 \\[1em] \Rightarrow 3x(x + 2) + 5(x + 2) = 0 \\[1em] \Rightarrow (x + 2)(3x + 5) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } 3x + 5 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = -\dfrac{5}{3}$

Hence, roots of given equation are -2, - $\dfrac{5}{3}$ .

Question 2(i)

Solve the following equations by factorisation:

2x² + ax - a² = 0

Answer

Given,

$\Rightarrow 2x^2 + ax - a^2 = 0 \\[1em] \Rightarrow 2x^2 + 2ax - ax - a^2 = 0 \\[1em] \Rightarrow 2x(x + a) - a(x + a) = 0 \\[1em] \Rightarrow (x + a)(2x - a) = 0 \\[1em] \Rightarrow x + a = 0 \text{ or } 2x - a = 0 \\[1em] \Rightarrow x = -a \text{ or } x = \dfrac{a}{2}$

Hence, roots of given equation are $-a, \dfrac{a}{2}$ .

Question 2(ii)

Solve the following equations by factorisation:

$\sqrt{3}x^2 + 10x + 7\sqrt{3}$ = 0

Answer

Given,

$\Rightarrow \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \\[1em] \Rightarrow x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow x + \sqrt{3} = 0 \text{ or } \sqrt{3}x + 7 = 0 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\dfrac{7}{\sqrt{3}} \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\dfrac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = - \dfrac{7\sqrt{3}}{3}$

Hence, roots of given equation are $-\sqrt{3}, -\dfrac{7\sqrt{3}}{3}$ .

Question 3(i)

Solve the following equations by factorisation:

x(x + 1) +(x + 2)(x + 3) = 42

Answer

Given,

$\Rightarrow x(x + 1) + (x + 2)(x + 3) = 42 \\[1em] \Rightarrow x^2 + x + x^2 + 3x + 2x + 6 = 42 \\[1em] \Rightarrow 2x^2 + 6x + 6 = 42 \\[1em] \Rightarrow 2x^2 + 6x + 6 - 42 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 36 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 18) = 0 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] x = 3 \text{ or } x = -6$

Hence, roots of given equation are 3 , -6.

Question 3(ii)

Solve the following equations by factorisation:

$\dfrac{6}{x} - \dfrac{2}{x - 1} = \dfrac{1}{x - 2}$

Answer

Given,

$\dfrac{6}{x} - \dfrac{2}{x - 1} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{6(x - 1) - 2x}{x(x - 1)} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{6x - 6 - 2x}{x^2 - x} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{4x - 6}{x^2 - x} = \dfrac{1}{x - 2} \\[1em] \Rightarrow (4x - 6)(x - 2) = x^2 - x \\[1em] \Rightarrow (4x^2 - 8x - 6x + 12) = x^2 - x \\[1em] \Rightarrow 4x^2 - x^2 - 14x + x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 3 = 0 \\[1em] x = \dfrac{4}{3} \text{ or } x = 3.$

Hence, roots of given equation are $3 ,\dfrac{4}{3}$

Question 4(i)

Solve the following equations by factorisation:

$\sqrt{x + 15} = x + 3$

Answer

Given,

$\Rightarrow \sqrt{x + 15} = x + 3 \\[1em] \Rightarrow x + 15 = (x + 3)^2 \text{ (On squaring both sides) } \\[1em] \Rightarrow x + 15 = x^2 + 9 + 6x \\[1em] \Rightarrow x^2 + 6x - x + 9 - 15 = 0 \\[1em] \Rightarrow x^2 + 5x - 6 = 0 \\[1em] \Rightarrow x^2 + 6x - x - 6 = 0 \\[1em] \Rightarrow x(x + 6) - 1(x + 6) = 0 \\[1em] \Rightarrow (x - 1)(x + 6) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -6$

Since we squared the equation, so roots need to be checked

Putting x = -6 in equation

$\Rightarrow \sqrt{-6 + 15} = -6 + 3 \\[1em] \Rightarrow \sqrt{ 9 } = -3 \\[1em]$

L.H.S. = 3 and R.H.S. = -3

Since, L.H.S. ≠ R.H.S. hence, x = -6 is not root of the equation

Putting x = 1 in equation

$\Rightarrow \sqrt{1 + 15} = 1 + 3 \\[1em] \Rightarrow \sqrt{ 16 } = 4 \\[1em]$

L.H.S. = R.H.S. = 4

Hence, root of the equation is 1.

Question 4(ii)

Solve the following equations by factorisation:

$\sqrt{3x^2 - 2x - 1} = 2x - 2$

Answer

Given,

$\Rightarrow \sqrt{3x^2 - 2x - 1} = 2x - 2 \\[1em] \Rightarrow 3x^2 - 2x - 1 = (2x - 2)^2 \text{ (On squaring both sides) } \\[1em] \Rightarrow 3x^2 - 2x - 1 = 4x^2 + 4 - 8x \\[1em] \Rightarrow 3x^2 - 4x^2 - 2x + 8x - 1 - 4 = 0 \\[1em] \Rightarrow -x^2 + 6x - 5 = 0 \\[1em] \Rightarrow x^2 - 6x + 5 = 0 \text{ (On multiplying equation by -1) }\\[1em] \Rightarrow x^2 - 5x - x + 5 = 0 \\[1em] \Rightarrow x(x - 5) - 1(x - 5) = 0 \\[1em] \Rightarrow (x - 1)(x - 5) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 5 = 0 \\[1em] x = 1 \text{ or } x = 5$

Since we squared the equation, so roots need to be checked

Putting x = 1 in equation

$\Rightarrow \sqrt{3(1)^2 - 2(1) - 1} = 2(1) - 2 \\[1em] \Rightarrow \sqrt{ 0 } = 0 \\[1em]$

L.H.S. = R.H.S. = 0

Putting x = 5 in equation

$\Rightarrow \sqrt{3(5)^2 - 2(5) - 1} = 2(5) - 2 \\[1em] \Rightarrow \sqrt{ 64 } = 8 \\[1em]$

L.H.S. = R.H.S. = 8

Hence, roots of the equation are 1, 5.

Question 5(i)

Solve the following equations by using formula:

2x² - 3x - 1 = 0

Answer

The given equation is 2x² - 3x - 1 = 0

Comparing it with ax² + bx + c = 0
a = 2, b = -3, c = -1

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-3) ± \sqrt{(-3)^2 - 4 \times 2 \times -1 }}{2 \times 2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{9 + 8}}{4} \\[1em] \Rightarrow \dfrac{3 + \sqrt{17}}{4} \text{ or } \dfrac{3 - \sqrt{17}}{4} \\[1em]$

Hence, roots of the equation are $\dfrac{3 + \sqrt{17}}{4} , \dfrac{3 - \sqrt{17}}{4}$ .

Question 5(ii)

Solve the following equations by using formula:

$x \Big(3x + \dfrac{1}{2} \Big)$ = 6

Answer

The given equation is $x \Big(3x + \dfrac{1}{2} \Big)$ = 6

$\Rightarrow 3x^2 + \dfrac{1}{2}x = 6 \\[1em] \Rightarrow 6x^2 + x = 12 \\[1em] \Rightarrow 6x^2 + x - 12 = 0$

Comparing it with ax² + bx + c = 0
a= 6, b = 1, c = -12

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(1) ± \sqrt{(1)^2 - 4 \times 6 \times -12 }}{2 \times 6} \\[1em] \Rightarrow \dfrac{-1 ± \sqrt{1 + 288}}{12} \\[1em] \Rightarrow \dfrac{-1 + \sqrt{289}}{12} \text{ or } \dfrac{-1 - \sqrt{289}}{12} \\[1em] \Rightarrow \dfrac{-1 + 17}{12} \text{ or } \dfrac{-1 - 17}{12} \\[1em] \Rightarrow \dfrac{16}{12} \text{ or } -\dfrac{18}{12} \\[1em] \Rightarrow \dfrac{4}{3} \text{ or } -\dfrac{3}{2}$

Hence, roots of the equation are $\dfrac{4}{3} , -\dfrac{3}{2}$ .

Question 6(i)

Solve the following equations by using formula:

$\dfrac{2x + 5}{3x + 4} = \dfrac{x + 1}{x + 3}$

Answer

Given,

$\dfrac{2x + 5}{3x + 4} = \dfrac{x + 1}{x + 3} \\[1em] \Rightarrow (2x + 5)(x + 3) = (x + 1)(3x + 4) \\[1em] \Rightarrow (2x^2 + 6x + 5x + 15) = (3x^2 + 4x + 3x + 4) \\[1em] \Rightarrow 2x^2 - 3x^2 + 11x - 7x + 15 - 4 = 0 \\[1em] \Rightarrow -x^2 + 4x + 11 = 0 \\[1em] \Rightarrow x^2 - 4x - 11 = 0 \text{ ( On multiplying equation by -1) }$

Comparing it with ax² + bx + c = 0
a= 1, b = -4, c = -11

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-4) ± \sqrt{(-4)^2 - 4 \times 1 \times -11 }}{2 \times 1} \\[1em] \Rightarrow \dfrac{4 ± \sqrt{16 + 44}}{2} \\[1em] \Rightarrow \dfrac{4 + \sqrt{60}}{2} \text{ or } \dfrac{4 - \sqrt{60}}{2} \\[1em] \Rightarrow \dfrac{4 + 2\sqrt{15}}{2} \text{ or } \dfrac{4 - 2\sqrt{15}}{2} \\[1em] \Rightarrow 2 + \sqrt{15} \text{ or } 2 - \sqrt{15}$

Hence, roots of the equation are $2 + \sqrt{15} , 2 - \sqrt{15}$ .

Question 6(ii)

Solve the following equations by using formula:

$\dfrac{2}{x + 2} - \dfrac{1}{x + 1} = \dfrac{4}{x + 4} - \dfrac{3}{x + 3}$

Answer

The given equation is $\dfrac{2}{x + 2} - \dfrac{1}{x + 1} = \dfrac{4}{x + 4} - \dfrac{3}{x + 3}$

$\Rightarrow \dfrac{2(x + 1) - (x + 2)}{(x + 2)(x + 1)} = \dfrac{4(x + 3) - 3(x + 4)}{(x + 4)(x + 3)} \\[1em] \Rightarrow \dfrac{2x + 2 - x - 2}{x^2 + x + 2x + 2} = \dfrac{4x + 12 - 3x - 12}{x^2 + 3x + 4x + 12} \\[1em] \Rightarrow \dfrac{x}{x^2 + 3x + 2} = \dfrac{x}{x^2 + 7x + 12} \\[1em] \Rightarrow x(x^2 + 7x + 12) = x(x^2 + 3x + 2) \\[1em] \Rightarrow (x^3 + 7x^2 + 12x) = (x^3 + 3x^2 + 2x) \\[1em] \Rightarrow x^3 + 7x^2 + 12x - x^3 - 3x^2 - 2x = 0 \\[1em] \Rightarrow 4x^2 + 10x = 0 \\[1em] \Rightarrow 2x(2x + 5) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } 2x + 5 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -\dfrac{5}{2}$

Hence, roots of the equation are $0 ,-\dfrac{5}{2}$ .

Question 7(i)

Solve the following equations by using formula:

$\dfrac{3x - 4}{7} + \dfrac{7}{3x - 4} = \dfrac{5}{2}, x$ ≠ $\dfrac{4}{3}$

Answer

The given equation is $\dfrac{3x - 4}{7} + \dfrac{7}{3x - 4} = \dfrac{5}{2}$

$\Rightarrow \dfrac{(3x - 4)^2 + 7^2}{7(3x - 4)} = \dfrac{5}{2} \text{ (On taking L.C.M. ) }\\[1em] \Rightarrow \dfrac{9x^2 + 16 - 24x + 49}{21x - 28} = \dfrac{5}{2} \\[1em] \Rightarrow 2(9x^2 - 24x + 65) = 5(21x - 28) \\[1em] \Rightarrow 18x^2 - 48x + 130 = 105x - 140 \\[1em] \Rightarrow 18x^2 - 48x - 105x + 130 + 140 = 0 \\[1em] \Rightarrow 18x^2 - 153x + 270 = 0 \\[1em] \Rightarrow 9(2x^2 - 17x + 30) = 0 \\[1em] \Rightarrow 2x^2 - 17x + 30 = 0$

Comparing it with ax² + bx + c = 0
a= 2, b = -17, c = 30

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-17) ± \sqrt{(-17)^2 - 4 \times 2 \times 30}}{2 \times 2} \\[1em] \Rightarrow \dfrac{17 ± \sqrt{289 - 240}}{4} \\[1em] \Rightarrow \dfrac{17 + \sqrt{49}}{4} \text{ or } \dfrac{17 - \sqrt{49}}{4} \\[1em] \Rightarrow \dfrac{17 + 7}{4} \text{ or } \dfrac{17 - 7}{4} \\[1em] \Rightarrow \dfrac{24}{4} \text{ or } \dfrac{10}{4} \\[1em] 6 \text{ or } \dfrac{5}{2}$

Hence, roots of the given equation are $6, \dfrac{5}{2}$ .

Question 7(ii)

Solve the following equations by using formula:

$\dfrac{4}{x} - 3 = \dfrac{5}{2x + 3} , x$ ≠ $0, - \dfrac{3}{2}$ .

Answer

The given equation is $\dfrac{4}{x} - 3 = \dfrac{5}{2x + 3}$

$\Rightarrow \dfrac{4 - 3x}{x} = \dfrac{5}{2x + 3} \text{ (On taking L.C.M.) } \\[1em] \Rightarrow (4 - 3x)(2x + 3) = 5x \text{ (On Cross multiplication) } \\[1em] \Rightarrow 8x + 12 - 6x^2 - 9x = 5x \\[1em] \Rightarrow 6x^2 + 5x + 9x - 8x - 12 = 0 \\[1em] \Rightarrow 6x^2 + 6x - 12 = 0 \\[1em] \Rightarrow 6(x^2 + x - 2) = 0 \\[1em] x^2 + x - 2 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = 1, c = -2

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(1) ± \sqrt{(1)^2 - 4 \times 1 \times -2}}{2 \times 1} \\[1em] \Rightarrow \dfrac{-1 ± \sqrt{1 + 8}}{2} \\[1em] \Rightarrow \dfrac{-1 + \sqrt{9}}{2} \text{ or } \dfrac{-1 - \sqrt{9}}{2} \\[1em] \Rightarrow \dfrac{-1 + 3}{2} \text{ or } \dfrac{-1 - 3}{2} \\[1em] \Rightarrow \dfrac{2}{2} \text{ or } \dfrac{-4}{2} \\[1em] 1 \text{ or } -2$

Hence, roots of the given equation are 1, -2.

Question 8(i)

Solve the following equations by using formula:

x² + (4 - 3a)x - 12a = 0

Answer

The given equation is x² + (4 - 3a)x - 12a = 0

Comparing it with ax² + bx + c = 0
a= 1, b = (4 - 3a), c = -12a

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(4 - 3a) ± \sqrt{(4 - 3a)^2 - 4 \times 1 \times -12a}}{2 \times 1} \\[1em] \Rightarrow \dfrac{(3a - 4) ± \sqrt{16 + 9a^2 - 24a + 48a}}{2} \\[1em] \Rightarrow \dfrac{3a - 4 ± \sqrt{16 + 9a^2 + 24a}}{2} \\[1em] \Rightarrow \dfrac{3a - 4 ± \sqrt{(4 + 3a)^2}}{2}\\[1em] \Rightarrow \dfrac{3a - 4 + |4 + 3a|}{2} \text{ or } \dfrac{3a - 4 - |4 + 3a|}{2} \\[1em] \Rightarrow \dfrac{6a}{2} \text{ or } -\dfrac{8}{2} \\[1em] 3a \text{ or } -4$

Hence, roots of the given equation are 3a, -4.

Question 8(ii)

Solve the following equations by using formula:

10ax² - 6x + 15ax - 9 = 0 , a ≠ 0.

Answer

The given equation is 10ax² - 6x + 15ax - 9 = 0

Comparing it with ax² + bx + c = 0
a= 10a, b = (15a - 6), c = -9

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(15a - 6) ± \sqrt{(15a - 6)^2 - 4 \times 10a \times -9}}{2 \times 10a} \\[1em] \Rightarrow \dfrac{(6 - 15a) ± \sqrt{225a^2 + 36 - 180a + 360a}}{20a} \\[1em] \Rightarrow \dfrac{6 - 15a ± \sqrt{225a^2 + 36 + 180a}}{20a} \\[1em] \Rightarrow \dfrac{6 - 15a ± \sqrt{(15a + 6)^2}}{20a}\\[1em] \Rightarrow \dfrac{6 - 15a + |15a + 6|}{20a} \text{ or } \dfrac{6 - 15a - |15a + 6|}{20a} \\[1em] \Rightarrow \dfrac{12}{20a} \text{ or } -\dfrac{30a}{20a} \\[1em] \dfrac{3}{5a} \text{ or } -\dfrac{3}{2}$

Hence, roots of the given equation are $\dfrac{3}{5a}, -\dfrac{3}{2}$ .

Question 9

Solve for x using the quadratic formula. Write your answer correct to two significant figures : (x - 1)² - 3x + 4 = 0 .

Answer

Given,

$(x - 1)^2 - 3x + 4 = 0 \\[0.5em] \Rightarrow x^2 + 1 - 2x - 3x + 4 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 5 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = -5, c = 5

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-5) ± \sqrt{(-5)^2 - 4 \times 1 \times 5}}{2 \times 1} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{25 - 20}}{2} \\[1em] \Rightarrow \dfrac{5 + \sqrt{5}}{2} \text{ or } \dfrac{5 - \sqrt{5}}{2} \\[1em] \Rightarrow \dfrac{5 + 2.2}{2} \text{ or } \dfrac{5 - 2.2}{2} \\[1em] \Rightarrow \dfrac{7.2}{2} \text{ or } \dfrac{2.8}{2} \\[1em] 3.6 \text{ or } 1.4$

Hence, roots of the given equation are 3.6, 1.4.

Question 10

Discuss the nature of the roots of the following equations :

(i) $3x^2 - 7x + 8 = 0$

(ii) $x^2 - \dfrac{1}{2}x - 4 = 0$

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

(iv) $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$

In case real roots exist , then find them.

Answer

(i) The given equation is 3x² - 7x + 8 = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -7, c = 8

$\therefore \text{Discriminant }= b^2 - 4ac \\[0.5em] = (-7)^2 - 4 \times 3 \times 8 \\[0.5em] = 49 - 96 \\[0.5em] = -47 \lt 0$

Since, Discriminant < 0 , hence equation has no real roots.

(ii) The given equation is $x^2 - \dfrac{1}{2}x - 4 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = $-\dfrac{1}{2}$ , c = -4

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-\dfrac{1}{2})^2 - 4 \times 1 \times -4 \\[1em] = \dfrac{1}{4} + 16 \\[1em] = \dfrac{65}{4} \gt 0$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-\dfrac{1}{2}) ± \sqrt{(-\dfrac{1}{2})^2 - 4 \times 1 \times -4}}{2 \times 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{1}{4} + 16}}{2} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{65}{4}}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \sqrt{\dfrac{65}{4}}}{2} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} + \dfrac{\sqrt{65}}{2}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \dfrac{\sqrt{65}}{2}}{2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{65}}{4} \text{ or } \dfrac{1 - \sqrt{65}}{4}$

Hence, roots of the given equation are $\dfrac{1 + \sqrt{65}}{4} , \dfrac{1 - \sqrt{65}}{4}$ .

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

The given equation is $5x^2 - 6\sqrt{5}x + 9 = 0$

Comparing it with ax² + bx + c = 0
a= 5, b = -6√5, c = 9

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-6\sqrt{5})^2 - 4 \times 5 \times 9 \\[1em] = 180 - 180 \\[1em] = 0$

Since, Discriminant = 0, hence equation has two equal and real roots.

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-6\sqrt{5}) ± \sqrt{(-6\sqrt{5})^2 - 4 \times 5 \times 9}}{2 \times 5} \\[1em] \Rightarrow \dfrac{6\sqrt{5} ± \sqrt{180 - 180}}{10} \\[1em] \Rightarrow \dfrac{6\sqrt{5} + \sqrt{0}}{10} \text{ or } \dfrac{6\sqrt{5} - \sqrt{0}}{10} \\[1em] \Rightarrow \dfrac{6\sqrt{5}}{10} \text{ or } \dfrac{6\sqrt{5}}{10} \\[1em] \Rightarrow \dfrac{3\sqrt{5}}{5} \text{ or } \dfrac{3\sqrt{5}}{5} \\[1em] \dfrac{3}{\sqrt{5}} \text{ or } \dfrac{3}{\sqrt{5}}$

Hence, roots of the given equation are $\dfrac{3}{\sqrt{5}}, \dfrac{3}{\sqrt{5}}$ .

(iv) $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$

The given equation is $\sqrt{3}x^2 - 2x - \sqrt{3} = 0$ .

Comparing it with ax² + bx + c = 0
a= $\sqrt{3}$ , b = -2, c = - $\sqrt{3}$

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3} \\[1em] = 4 + 12 \\[1em] = 16$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-2) ± \sqrt{(-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow \dfrac{2 ± \sqrt{4 + 12}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 + \sqrt{16}}{2\sqrt{3}} \text{ or } \dfrac{2 - \sqrt{16}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 + 4}{2\sqrt{3}} \text{ or } \dfrac{2- 4}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{6}{2\sqrt{3}} \text{ or } -\dfrac{2}{2\sqrt{3}} \\[1em] \sqrt{3} \text{ or } -\dfrac{1}{\sqrt{3}}$

Hence, roots of the given equation are $\sqrt{3}, -\dfrac{1}{\sqrt{3}}$ .

Question 11

Find the values of k so that the quadratic equation (4 - k)x² + 2(k + 2)x + (8k + 1) = 0 has equal roots.

Answer

The given equation is (4 - k)x² + 2(k + 2)x + (8k + 1) = 0

Comparing it with ax² + bx + c = 0
a= (4 - k), b = (2k + 4), c = (8k + 1)

Given,

Equation has real and equal roots

∴ b² - 4ac = 0

$\Rightarrow (2k + 4)^2 - 4 \times (4 - k) \times (8k + 1) = 0 \\[1em] \Rightarrow (4k^2 + 16 + 16k) - 4(32k + 4 - 8k^2 - k) = 0 \\[1em] \Rightarrow 4k^2 + 16 + 16k - 128k - 16 + 32k^2 + 4k = 0 \\[1em] \Rightarrow 4k^2 + 32k^2 + 16k - 128k + 4k + 16 - 16 = 0 \\[1em] \Rightarrow 36k^2 - 108k = 0 \\[1em] \Rightarrow 36k(k - 3) = 0 \\[1em] \Rightarrow 36k = 0 \text{ or } k - 3 = 0 \\[1em] k = 0 \text{ or } k = 3$

Hence , the value of k are 0, 3.

Question 12

Find the values of m so that the quadratic equation 3x² - 5x - 2m = 0 has two distinct real roots.

Answer

The given equation is 3x² - 5x - 2m = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -5, c = -2m

Given,

Equation has two real and distinct roots

∴ b² - 4ac > 0

$\Rightarrow (-5)^2 - 4 \times 3 \times -2m \gt 0 \\[1em] \Rightarrow 25 + 24m \gt 0 \\[1em] \Rightarrow 24m \gt -25 \\[1em] m \gt -\dfrac{25}{24}$

Hence, the required value is m > $-\dfrac{25}{24}$

Question 13

Find the value(s) of k for which each of the following quadratic equation has equal roots:

(i) 3kx² = 4(kx - 1)

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Also, find the roots for that value(s) of k in each case.

Answer

(i) Given,

$3kx^2 = 4(kx - 1) \\[0.5em] \Rightarrow 3kx^2 - 4(kx - 1) = 0 \\[0.5em] \Rightarrow 3kx^2 - 4kx + 4 = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0
a= 3k, b = -4k, c = 4

Given,

Equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-4k)^2 - 4 \times 3k \times 4 = 0 \\[0.5em] \Rightarrow 16k^2 - 48k = 0 \\[0.5em] \Rightarrow 16k(k - 3) = 0 \\[0.5em] \Rightarrow 16k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3$

But k cannot be equal to 0 as that will make a = 3k = 0 , which will make roots equal to infinity.

∴ k = 3

Hence equation is 9x² - 12x + 4 = 0

$\Rightarrow 9x^2 - 6x - 6x + 4 = 0 \\[0.5em] \Rightarrow 3x(3x - 2) - 2(3x - 2) = 0 \\[0.5em] \Rightarrow (3x - 2)(3x - 2) = 0 \\[0.5em] \Rightarrow 3x - 2 = 0 \text{ or } 3x - 2 = 0 \\[0.5em] x = \dfrac{2}{3} \text{ or } x = \dfrac{2}{3}$

Hence the value of k is 3 and the roots are $\dfrac{2}{3}, \dfrac{2}{3}.$

(ii) The given equation is (k + 4)x² +(k + 1)x + 1 = 0

Comparing it with ax² + bx + c = 0
a = (k + 4), b = (k + 1), c = 1

Given,

Equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (k + 1)^2 - 4 \times (k + 4) \times 1 = 0 \\[1em] \Rightarrow (k^2 + 1 + 2k) - 4(k + 4) = 0 \\[1em] \Rightarrow k^2 + 1 + 2k - 4k - 16 = 0 \\[1em] \Rightarrow k^2 - 2k - 15 = 0 \\[1em] \Rightarrow k^2 - 5k + 3k - 15 = 0 \\[1em] \Rightarrow k(k - 5) + 3(k - 5) = 0 \\[1em] \Rightarrow (k + 3)(k - 5) = 0 \\[1em] \Rightarrow k + 3 = 0 \text{ or } k - 5 = 0 \\[1em] k = -3 \text{ or } k = 5$

∴ When k = -3 , equation is x² - 2x + 1 = 0

$\Rightarrow x^2 - x - x + 1 = 0 \\[1em] \Rightarrow x(x - 1) - 1(x - 1) = 0 \\[1em] \Rightarrow (x - 1)(x - 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 1 = 0 \\[1em] x = 1 \text{ or } x = 1$

∴ When k = 5 , equation is 9x² + 6x + 1 = 0

$\Rightarrow 9x^2 + 3x + 3x + 1 = 0 \\[1em] \Rightarrow 3x(3x + 1) + 1(3x + 1) = 0 \\[1em] \Rightarrow (3x + 1)(3x + 1) = 0 \\[1em] \Rightarrow 3x + 1 = 0 \text{ or } 3x + 1 = 0 \\[1em] x = -\dfrac{1}{3} \text{ or } -\dfrac{1}{3}$

k = -3, 5
When k = -3 , roots are 1, 1
When k = 5 , roots are $-\dfrac{1}{3}, -\dfrac{1}{3}$

Question 14

Find two natural numbers which differ by 3 and whose squares have the sum 117.

Answer

Let first number be x

Since difference between two numbers is 3 hence, the other number is (x + 3).

Given , sum of the squares of number = 117

∴ x² + (x + 3)² = 117

$\Rightarrow x^2 + x^2 + 9 + 6x = 117 \\[1em] \Rightarrow 2x^2 + 6x + 9 - 117 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 108 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 54) = 0 \\[1em] \Rightarrow x^2 + 3x - 54 = 0 \\[1em] \Rightarrow x^2 + 9x - 6x - 54 = 0 \\[1em] \Rightarrow x(x + 9) - 6(x + 9) = 0 \\[1em] \Rightarrow (x - 6)(x + 9) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 9 = 0 \\[1em] x = 6 \text{ or } x = -9$

Since, numbers are natural hence, x ≠ -9.

∴ x = 6 , x + 3 = 9.

Hence, the required numbers are 6, 9.

Question 15

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

Answer

Let the larger number be x , so the smaller number is 16 - x.

Given, twice the square of the larger part exceeds the square of the smaller part by 164

$\therefore 2x^2 - (16 - x)^2 = 164 \\[1em] \Rightarrow 2x^2 - (256 + x^2 - 32x) = 164 \\[1em] \Rightarrow 2x^2 - 256 - x^2 + 32x - 164 = 0 \\[1em] \Rightarrow x^2 + 32x - 420 = 0 \\[1em] \Rightarrow x^2 + 42x - 10x - 420 = 0 \\[1em] \Rightarrow x(x + 42) - 10(x + 42) = 0 \\[1em] \Rightarrow (x - 10)(x + 42) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x + 42 = 0 \\[1em] x = 10 \text{ or } x = -42$

Since, numbers are natural hence, x ≠ -42.

∴ x = 10 , 16 - x = 6.

Hence, the required numbers are 10, 6.

Question 16

Two natural numbers are in the ratio 3 : 4 . Find the numbers if the difference between their squares is 175.

Answer

Since, the numbers are in the ratio 3 : 4, hence the numbers be 3x and 4x.

Given , difference between their squares = 175

∴ (4x)² - (3x)² = 175

$\Rightarrow 16x^2 - 9x^2 = 175 \\[1em] \Rightarrow 7x^2 = 175 \\[1em] \Rightarrow x^2 = \dfrac{175}{7} \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x^2 - 25 = 0 \\[1em] \Rightarrow (x - 5)(x + 5) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x + 5 = 0 \\[1em] x = 5 \text{ or } x = -5$

Since, numbers are natural hence, x ≠ -5.

∴ x = 5, 3x = 15 , 4x = 20.

Hence, the required numbers are 15, 20.

Question 17

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation and solve it to find the sides of the squares.

Answer

Area of a square = (side)²

∴ Area of first square = x² and Area of second square = (x + 4)²

Given, sum of areas of two squares is = 656 cm²

∴ x² + (x + 4)² = 656

$\Rightarrow x^2 + x^2 + 16 + 8x = 656 \\[1em] \Rightarrow 2x^2 + 8x + 16 - 656 = 0 \\[1em] \Rightarrow 2x^2 + 8x - 640 = 0 \\[1em] \Rightarrow 2(x^2 + 4x - 320) = 0 \\[1em] \Rightarrow x^2 + 4x - 320 = 0 \\[1em] \Rightarrow x^2 + 20x - 16x - 320 = 0 \\[1em] \Rightarrow x(x + 20) - 16(x + 20) = 0 \\[1em] \Rightarrow (x + 20)(x - 16) = 0 \\[1em] \Rightarrow x + 20 = 0 \text{ or } x - 16 = 0 \\[1em] x = -20 \text{ or } x = 16$

Since, length cannot be negative hence x ≠ -20.

∴ x = 16 , x + 4 = 20.

Hence, the sides of two squares are 16 cm and 20 cm.

Question 18

The length of a rectangular garden is 12m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Answer

Let the value of breadth be x metre

So, length = (x + 12) metre

Perimeter = 2(Length + Breadth) = 2(x + x + 12) = 2(2x + 12) = (4x + 24) metre

Area = Length $\times$ Breadth = x(x + 12) = (x² + 12x) metre²

Given, area is equal to 4 times the perimeter

∴ x² + 12x = 4(4x + 24)

$\Rightarrow x^2 + 12x = 16x + 96 \\[1em] \Rightarrow x^2 + 12x - 16x - 96 = 0 \\[1em] \Rightarrow x^2 - 4x - 96 = 0 \\[1em] \Rightarrow x^2 - 12x + 8x - 96 = 0 \\[1em] \Rightarrow x(x - 12) + 8(x - 12) = 0 \\[1em] \Rightarrow (x - 12)(x + 8) = 0 \\[1em] \Rightarrow x - 12 = 0 \text{ or } x + 8 = 0 \\[1em] x = 12 \text{ or } x = -8 \\[1em]$

Since, breadth cannot be negative hence, x ≠ -8

∴ x = 12 , x + 12 = 24

Hence, the length of the garden is 24m and breadth is 12m.

Question 19

A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30m barbed wire , he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.

Answer

Let x metres be the length of the side opposite to unfenced side, then length of each of two others sides = $\dfrac{1}{2}(30 -x).$

According to question,

$\Rightarrow x \times \dfrac{1}{2}(30 - x) = 100 \\[1em] \Rightarrow 15x - \dfrac{x^2}{2} = 100 \\[1em] \Rightarrow \dfrac{30x - x^2}{2} = 100 \text{ (On taking L.C.M.) }\\[1em] \Rightarrow 30x - x^2 = 200 \text { (On cross multiplying) } \\[1em] \Rightarrow x^2 - 30x + 200 = 0 \\[1em] \Rightarrow x^2 - 20x - 10x + 200 = 0 \\[1em] \Rightarrow x(x - 20) - 10(x - 20) = 0 \\[1em] \Rightarrow (x - 10)(x - 20) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x - 20 = 0 \\[1em] x = 10 \text{ or } x = 20$

∴ if x = 10, $\dfrac{1}{2}(30 - x) :$

$= \dfrac{1}{2}(30 - 10) \\[1em] = \dfrac{1}{2} \times 20 \\[1em] = 10$

∴ if x = 20, $\dfrac{1}{2}(30 - x) :$

$= \dfrac{1}{2}(30 - 20) \\[1em] = \dfrac{1}{2} \times 10 \\[1em] = 5$

Hence, the dimensions of garden are 10m x 10m or 20m x 5m.

Question 20

The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Answer

Let the shortest side be x metre

According to question,

Third side = (x + 1)m

Hypotenuse = (2x - 1)m

For right angled triangle,

(Hypotenuse)² = (Perpendicular)² + (Base)²

$\therefore (2x - 1)^2 = (x + 1)^2 + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = x^2 + 1 + 2x + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = 2x^2 + 2x + 1 \\[1em] \Rightarrow 4x^2 - 2x^2 - 4x - 2x + 1 - 1 = 0 \\[1em] \Rightarrow 2x^2 - 6x = 0 \\[1em] \Rightarrow 2x(x - 3) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } x - 3 = 0 \\[1em] x = 0 \text{ or } x = 3$

Since, side's length cannot be equal to 0 , hence x ≠ 0.

∴ x = 3, (x + 1) = 4, (2x - 1) = 5

Hence, the sides of triangle are
Shortest side = 3 m
Hypotenuse = 5 m
Third side = 4 m

Question 21

A wire, 112 cm long, is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.

Answer

Sum of length of other two sides + Hypotenuse = 112

or, Sum of length of other two sides = 112 - Hypotenuse

∴ Sum of length of other two sides = 112 - 50 = 62 cm

Let the length of perpendicular = x cm

So, the length of base = (62 - x) cm

For right angled triangle,

(Perpendicular)² + (Base)² = (Hypotenuse)²

$\therefore x^2 + (62 - x)^2 = (50)^2 \\[1em] \Rightarrow x^2 + 3844 + x^2 - 124x = 2500 \\[1em] \Rightarrow 2x^2 - 124x + 3844 - 2500 = 0 \\[1em] \Rightarrow 2x^2 - 124x + 1344 = 0 \\[1em] \Rightarrow 2(x^2 - 62x + 672) = 0 \\[1em] \Rightarrow x^2 - 62x + 672 = 0 \\[1em] \Rightarrow x^2 - 48x - 14x + 672 = 0 \\[1em] \Rightarrow x(x - 48) - 14(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x - 14) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x - 14 = 0 \\[1em] x = 48 \text{ or } x = 14$

∴ The length of other two sides are 48 cm and 14 cm.

$\text{Area} = \dfrac{1}{2} \times \text{Perpendicular} \times \text{Base} \\[1em] = \dfrac{1}{2} \times 48 \times 14 \\[1em] = 336$

Hence, the area of triangle is 336 cm^2.

Question 22

The speed of a boat in still water is 11 km/h. It can go 12 km upstream and return downstream to original point in 2 hours 45 minutes. Find the speed of the stream.

Answer

Let the speed of the stream be x km/h.

Speed of boat in -

Still water = 11 km/h

Upstream = (11 - x) km/h

Downstream = (11 + x) km/h

Given,

Boat can go 12 km upstream and return downstream to original point in 2 hours 45 minutes.

Converting 2 hours 45 minutes to hours:

2 hours 45 minutes = ((2 x 60) + 45) minutes = 165 minutes.

165 minutes = $\dfrac{165}{60}$ hours

$\therefore \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = \dfrac{165}{60} \\[1em] \Rightarrow \dfrac{12(11 + x) + 12(11 - x)}{(11 + x)(11 - x)} = \dfrac{165}{60} \\[1em] \Rightarrow \dfrac{132 + 12x + 132 - 12x}{121 - x^2} = \dfrac{165}{60} \\[1em] \Rightarrow \dfrac{264}{121 - x^2} = \dfrac{11}{4}\\[1em] \Rightarrow 264 \times 4 = 11(121 - x^2) \\[1em] \Rightarrow 1056 = 1331 - 11x^2 \\[1em] \Rightarrow 11x^2 = 1331 - 1056 \\[1em] \Rightarrow 11x^2 = 275 \\[1em] \Rightarrow x^2 = \dfrac{275}{11} \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x^2 - 25 = 0 \\[1em] \Rightarrow (x - 5)(x + 5) = 0 \\[1em] \Rightarrow x = 5, -5$

Since speed cannot be negative hence, x ≠ -5.

Hence, speed of stream is 5 km/h.

Question 23

A man spent ₹2800 on buying a number of plants priced at ₹x each. Because of the number involved, the supplier reduced the price of each plant by one rupee. The man finally paid ₹2730 and received 10 more plants . Find x.

Answer

In first case,

Amount spent = ₹2800

Price of each plant = ₹x

No. of plants = $\dfrac{2800}{x}$

In second case,

Amount spent = ₹2730

Price of plant is reduced by ₹1 , so new price = ₹(x - 1)

Given, in this case 10 more plants were received hence, no. of plants = $\dfrac{2800}{x} + 10$

According to question,

$\Rightarrow \big(\dfrac{2800}{x} + 10\big)(x - 1) = 2730 \\[1em] \Rightarrow \big(\dfrac{2800 + 10x}{x}\big)(x - 1) = 2730 \\[1em] \Rightarrow \dfrac{(2800 + 10x)(x - 1)}{x} = 2730 \\[1em] \Rightarrow \dfrac{2800x - 2800 + 10x^2 - 10x}{x} = 2730 \\[1em] \Rightarrow 10x^2 + 2790x - 2800 = 2730x \text{ (On cross multiplication) } \\[1em] \Rightarrow 10x^2 + 60x - 2800 = 0 \\[1em] \Rightarrow x^2 + 6x - 280 = 0 \text{ (On dividing by 10) } \\[1em] \Rightarrow x^2 + 20x - 14x - 280 = 0 \\[1em] \Rightarrow x(x + 20) - 14(x + 20) = 0 \\[1em] \Rightarrow (x - 14)(x + 20) = 0 \\[1em] x = 14 \text{ or } x = -20$

Since price cannot be negative hence , x ≠ -20

Hence, the value of x is 14.

Question 24

Forty years hence, Mr. Pratap's age will be the square of what it was 32 years ago. Find his present age.

Answer

Let the present age of Mr. Pratap be x years.

After 40 years his age will be (x + 40) years

32 years before his age was (x - 32) years

According to question ,

$\Rightarrow (x - 32)^2 = (x + 40) \\[1em] \Rightarrow x^2 + 1024 - 64x = x + 40 \\[1em] \Rightarrow x^2 - 64x - x + 1024 - 40 = 0 \\[1em] \Rightarrow x^2 - 65x + 984 = 0 \\[1em] \Rightarrow x^2 - 24x - 41x + 984 = 0 \\[1em] \Rightarrow x(x - 24) - 41(x - 24) = 0 \\[1em] \Rightarrow (x - 41)(x - 24) = 0 \\[1em] \Rightarrow x - 41 = 0 \text{ or } x - 24 = 0 \\[1em] x = 41 \text{ or } x = 24$

But x ≠ 24 as it is less than 32.

∴ x = 41.

The present age of Mr. Pratap is 41 years.

Question 25

The total expenses of a trip for certain number of people is ₹ 18,000. If three more people join them, then the share of each reduces by ₹ 3,000. Take x to be the original number of people, form a quadratic equation in x and solve it to find the value of x.

Answer

Let no. of people be x.

Total expense = ₹ 18,000

Expense per person = ₹ $\dfrac{18000}{x}$

Given,

If three more people join them, then the share of each reduces by ₹ 3,000.

No, of people now = x + 3

Expense per person = ₹ $\dfrac{18000}{x + 3}$

According to question,

$\Rightarrow \dfrac{18000}{x} - \dfrac{18000}{x + 3} = 3000 \\[1em] \Rightarrow \dfrac{18000(x + 3) - 18000x}{x(x + 3)} = 3000 \\[1em] \Rightarrow \dfrac{18000x + 54000 - 18000x}{x^2 + 3x} = 3000 \\[1em] \Rightarrow \dfrac{54000}{x^2 + 3x} = 3000 \\[1em] \Rightarrow x^2 + 3x = \dfrac{54000}{3000} \\[1em] \Rightarrow x^2 + 3x = 18 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -6.$

Since, no. of people cannot be negative.

∴ x = 3.

Hence, original number of people = 3.

Question 26

A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed. If it takes 3 hours to complete the total journey then form a quadratic equation and solve it to find its original average speed.

Answer

Given,

A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed.

Total time taken to complete the journey = 3 hours

$\therefore \dfrac{72}{x} + \dfrac{81}{x + 6} = 3 \\[1em] \Rightarrow \dfrac{72(x + 6) + 81x}{x(x + 6)} = 3 \\[1em] \Rightarrow 72(x + 6) + 81x = 3x(x + 6) \\[1em] \Rightarrow 72x + 432 + 81x = 3x^2 + 18x \\[1em] \Rightarrow 3x^2 + 18x - 72x - 81x - 432 = 0 \\[1em] \Rightarrow 3x^2 - 135x - 432 = 0 \\[1em] \Rightarrow 3(x^2 - 45x - 144) = 0 \\[1em] \Rightarrow x^2 - 45x - 144 = 0 \\[1em] \Rightarrow x^2 - 48x + 3x - 144 = 0 \\[1em] \Rightarrow x(x - 48) + 3(x - 48) = 0 \\[1em] \Rightarrow (x + 3)(x - 48) = 0 \\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 48 = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 48.$

As speed cannot be negative in this case,

Hence, the original speed = 48 km/hr.

Chapter 5

Quadratic Equations in One Variable

Class - 10 ML Aggarwal Understanding ICSE Mathematics

Exercise 5.1

Question 1

Question 2

Question 3(i)

Question 3(ii)

Question 4(i)

Question 4(ii)

Question 5

Exercise 5.2

Question 1(i)

Question 1(ii)

Question 2(i)

Question 2(ii)

Question 3(i)

Question 3(ii)

Question 4(i)

Question 4(ii)

Question 5(i)

Question 5(ii)

Question 6

Question 7

Question 8(i)

Question 8(ii)

Question 9

Question 10(i)

Question 10(ii)

Question 11(i)

Question 11(ii)

Question 12(i)

Question 12(ii)

Question 13(i)

Question 13(ii)

Question 14(i)

Question 14(ii)

Question 15(i)

Question 15(ii)

Question 16(i)

Question 16(ii)

Question 17

Question 18(i)

Question 18(ii)

Question 19

Question 20

Question 21

Question 22

Question 23

Exercise 5.3

Question 1(i)

Question 1(ii)

Question 2(i)

Question 2(ii)

Question 3(i)

Question 3(ii)

Question 4(i)

Question 4(ii)

Question 5(i)

Question 5(ii)

Question 6(i)

Question 6(ii)

Question 7

Question 8

Question 9

Question 10

Question 11

Exercise 5.4

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Exercise 5.5