Chapter 5: Quadratic Equations

Exercise 5(A)

Question 1(a)

4x² - 9 = 0 implies x is equal to :

$\dfrac{3}{2}$
$\dfrac{9}{4}$
$-\dfrac{3}{2}$
$\pm \dfrac{3}{2}$

Answer

Given,

⇒ 4x² - 9 = 0

⇒ 4x² = 9

⇒ x² = $\dfrac{9}{4}$

⇒ x = $\sqrt{\dfrac{9}{4}} = \pm \dfrac{3}{2}$

Hence, Option 4 is the correct option.

Question 1(b)

(x - 3)(x + 5) = 0 gives x equal to :

3
3 or 5
3 or -5
3 and -5

Answer

Given,

⇒ (x - 3)(x + 5) = 0

⇒ (x - 3) or (x + 5) = 0

⇒ x - 3 = 0 or x + 5 = 0

⇒ x = 3 or x = -5.

Hence, Option 3 is the correct option.

Question 1(c)

If 4 is a root of the equation x² + kx - 4 = 0; the value of k is :

3
-3
2
-2

Answer

Since, 4 is the root of the equation x² + kx - 4 = 0.

∴ It will satisfy the equation x² + kx - 4 = 0.

∴ 4² + 4k - 4 = 0

⇒ 16 + 4k - 4 = 0

⇒ 4k + 12 = 0

⇒ 4k = -12

⇒ k = - $\dfrac{12}{4}$ = -3.

Hence, Option 2 is the correct option.

Question 1(d)

The equation 2x² - 3x + k = 0 is satisfied by x = 2; the value of k is :

-2
2
4
3

Answer

Given,

x = 2 satisfies the equation 2x² - 3x + k = 0.

∴ 2(2)² - 3(2) + k = 0

⇒ 2(4) - 6 + k = 0

⇒ 8 - 6 + k = 0

⇒ k + 2 = 0

⇒ k = -2.

Hence, Option 1 is the correct option.

Question 1(e)

If x² - 7x = 0; the value of x is :

0 and 7
7
0
0 or 7

Answer

Given,

⇒ x² - 7x = 0

⇒ x(x - 7) = 0

⇒ x = 0 or x - 7 = 0

⇒ x = 0 or x = 7.

Hence, Option 4 is the correct option.

Question 2

If $\sqrt{\dfrac{2}{3}}$ is a solution of equation 3x² + mx + 2 = 0, find the value of m.

Answer

Since, $\sqrt{\dfrac{2}{3}}$ is a solution of equation 3x² + mx + 2 = 0.

$\Rightarrow 3.\Big(\sqrt{\dfrac{2}{3}}\Big)^2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) + 2 = 0 \\[1em] \Rightarrow 3 \times \dfrac{2}{3} + 2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 2 + 2 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow 4 + m\Big(\sqrt{\dfrac{2}{3}}\Big) = 0 \\[1em] \Rightarrow m\Big(\sqrt{\dfrac{2}{3}}\Big) = -4 \\[1em] \Rightarrow m = -4 \times \sqrt{\dfrac{3}{2}} \\[1em] \Rightarrow m = -2\sqrt{2}\sqrt{3} \\[1em] \Rightarrow m = -2\sqrt{6}.$

Hence, value of m is $-2\sqrt{6}.$

Question 3

$\dfrac{2}{3}$ and 1 are the solutions of equation mx² + nx + 6 = 0. Find the values of m and n.

Answer

Since, $\dfrac{2}{3}$ is a solution of equation mx² + nx + 6 = 0.

Substituting $\dfrac{2}{3}$ in mx² + nx + 6 = 0,

$\Rightarrow m\Big(\dfrac{2}{3}\Big)^2 + n \times \dfrac{2}{3} + 6 = 0 \\[1em] \Rightarrow m \times \dfrac{4}{9} + \dfrac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \dfrac{4m}{9} + \dfrac{2n}{3} + 6 = 0 \\[1em] \Rightarrow \dfrac{4m + 6n + 54}{9} = 0 \\[1em] \Rightarrow 4m + 6n + 54 = 0 \\[1em] \Rightarrow 2(2m + 3n + 27) = 0 \\[1em] \Rightarrow 2m + 3n + 27 = 0 \\[1em] \Rightarrow 2m + 3n = -27 .......(i)$

Since, 1 is a solution of equation mx² + nx + 6 = 0.

Substituting 1 in mx² + nx + 6 = 0,

$\Rightarrow m(1)^2 + n(1) + 6 = 0 \\[1em] \Rightarrow m + n + 6 = 0 \\[1em] \Rightarrow m = -(n + 6) ........(ii)$

Sustituting above value of m in eq. 1 we get,

$\Rightarrow 2.-(n + 6) + 3n = -27 \\[1em] \Rightarrow -2(n + 6) + 3n = -27 \\[1em] \Rightarrow -2n - 12 + 3n = -27 \\[1em] \Rightarrow n - 12 = -27 \\[1em] \Rightarrow n = -27 + 12 \\[1em] \Rightarrow n = -15.$

Substituting value of n in (ii) we get,

$\Rightarrow m = -(-15 + 6) \\[1em] \Rightarrow m = -(-9) \\[1em] \Rightarrow m = 9.$

Hence, the value of m = 9 and n = -15.

Exercise 5(B)

Question 1(a)

The roots of the quadratic equation x² - 6x - 7 = 0 are :

-1 and 7
1 and 7
-1 or 7
1 and -7

Answer

Given,

⇒ x² - 6x - 7 = 0

⇒ x² - 7x + x - 7 = 0

⇒ x(x - 7) + 1(x - 7) = 0

⇒ (x + 1)(x - 7) = 0

⇒ x + 1 = 0 or x - 7 = 0

⇒ x = -1 or x = 7.

Hence, Option 3 is the correct option.

Question 1(b)

The roots of the quadratic equation x(x + 8) + 12 = 0 are :

6 or 2
-6 or -2
6 and -2
-6 and -2

Answer

Given,

⇒ x(x + 8) + 12 = 0

⇒ x² + 8x + 12 = 0

⇒ x² + 2x + 6x + 12 = 0

⇒ x(x + 2) + 6(x + 2) = 0

⇒ (x + 6)(x + 2) = 0

⇒ x + 6 = 0 or x + 2 = 0

⇒ x = -6 or x = -2.

Hence, Option 2 is the correct option.

Question 1(c)

If one root of equation (p - 3)x² + x + p = 0 is 2, the value of p is :

-2
2
±2
1 and 2

Answer

Given,

One root of equation (p - 3)x² + x + p = 0 is 2.

∴ x = 2 satisfies the equation.

∴ (p - 3)(2)² + 2 + p = 0

⇒ 4(p - 3) + 2 + p = 0

⇒ 4p - 12 + 2 + p = 0

⇒ 5p - 10 = 0

⇒ 5p = 10

⇒ p = $\dfrac{10}{5}$ = 2.

Hence, Option 2 is the correct option.

Question 1(d)

If $x + \dfrac{1}{x}$ = 2.5, the value of x is :

4
5 and $\dfrac{1}{5}$
2 or $\dfrac{1}{2}$
2 and $\dfrac{1}{2}$

Answer

Given,

$\Rightarrow x + \dfrac{1}{x} = 2.5 \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{25}{10} \\[1em] \Rightarrow 10(x^2 + 1) = 25x \\[1em] \Rightarrow 10x^2 + 10 = 25x \\[1em] \Rightarrow 10x^2 - 25x + 10 = 0 \\[1em] \Rightarrow 10x^2 - 20x - 5x + 10 = 0 \\[1em] \Rightarrow 10x(x - 2) - 5(x - 2) = 0 \\[1em] \Rightarrow (x - 2)(10x - 5) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } 10x - 5 =0 \\[1em] \Rightarrow x = 2 \text{ or } 10x = 5 \\[1em] \Rightarrow x = 2 \text{ or } x = \dfrac{5}{10} = \dfrac{1}{2}.$

Hence, Option 3 is the correct option.

Question 1(e)

For quadratic equation $2x + \dfrac{5}{x}$ = 5 :

x ≠ 0
x = 1
x = 5
x = 2

Answer

For quadratic equation :

2x + $\dfrac{5}{x}$ = 5

If x = 0, $\dfrac{5}{x}$ will not be defined.

∴ x ≠ 0.

Hence, Option 1 is the correct option.

Question 2

Solve the following equation by factorisation:

(2x - 3)² = 49

Answer

⇒ (2x - 3)² = 7²

⇒ (2x - 3)² - 7² = 0

As, (a² - b²) = (a + b)(a - b)

⇒ (2x - 3 + 7)(2x - 3 - 7) = 0

⇒ (2x + 4)(2x - 10) = 0

⇒ 2x + 4 = 0 or 2x - 10 = 0 [Using Zero-product rule]

⇒ 2x = -4 or 2x = 10

⇒ x = -2 or x = 5.

Hence, x = -2 or x = 5.

Question 3

Solve the following equation by factorisation:

(x + 1)(2x + 8) = (x + 7)(x + 3)

Answer

⇒ (x + 1)(2x + 8) = (x + 7)(x + 3)

⇒ 2x² + 8x + 2x + 8 = x² + 3x + 7x + 21

⇒ 2x² - x² + 10x + 8 = 10x + 21

⇒ x² + 10x - 10x = 21 - 8

⇒ x² = 13

⇒ x = $\pm \sqrt{13}$

Hence, x = $+ \sqrt{13}$ or x = $- \sqrt{13}$ .

i.e., x = 3.61 or x = -3.61

Question 4

Solve the following equation by factorisation:

4(2x - 3)² - (2x - 3) - 14 = 0

Answer

4(2x - 3)² - (2x - 3) - 14 = 0

⇒ 4(4x² + 9 - 12x) - 2x + 3 - 14 = 0

⇒ 16x² + 36 - 48x - 2x - 11 = 0

⇒ 16x² - 50x + 25 = 0

⇒ 16x² - 40x - 10x + 25 = 0

⇒ 8x(2x - 5) - 5(2x - 5) = 0

⇒ (8x - 5)(2x - 5) = 0

⇒ 8x - 5 = 0 or 2x - 5 = 0 [Using Zero-product rule]

⇒ 8x = 5 or 2x = 5

⇒ x = $\dfrac{5}{8}$ or x = $\dfrac{5}{2}$ .

Hence, x = $\dfrac{5}{8} \text{ or } x = \dfrac{5}{2}$ .

Question 5

Solve the following equation by factorisation:

2x² - 9x + 10 = 0, when :

(i) x ∈ N

(ii) x ∈ Q.

Answer

2x² - 9x + 10 = 0

⇒ 2x² - 4x - 5x + 10 = 0

⇒ 2x(x - 2) - 5(x - 2) = 0

⇒ (2x - 5)(x - 2) = 0

⇒ 2x - 5 = 0 or x - 2 = 0 [Using Zero-product rule]

⇒ 2x = 5 or x = 2

⇒ x = $\dfrac{5}{2}$ or x = 2.

(i) Since, x ∈ N

Hence, value of x = 2.

(ii) Since, x ∈ Q

Hence, value of x = $\dfrac{5}{2}$ or 2.

Question 6

Solve the following equation by factorisation:

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

Answer

Given,

$\phantom {\Rightarrow} \dfrac{x - 3}{x + 3} + \dfrac{x + 3}{x - 3} = 2\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{(x - 3)(x - 3) + (x + 3)(x + 3)}{(x + 3)(x - 3)} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{x^2 - 3x - 3x + 9 + x^2 + 3x + 3x + 9}{x^2 + 3x - 3x - 9} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{x^2 - 6x + 9 + x^2 + 6x + 9}{x^2 - 9} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{2x^2 + 18}{x^2 - 9} = \dfrac{5}{2} \\[1em] \Rightarrow 2(2x^2 + 18) = 5(x^2 - 9) \\[1em] \Rightarrow 4x^2 + 36 = 5x^2 - 45 \\[1em] \Rightarrow 5x^2 - 4x^2 = 36 + 45 \\[1em] \Rightarrow x^2 = 81 \\[1em] \Rightarrow x = \sqrt{81} = \pm 9.$

Hence, value of x = +9 or -9.

Question 7

Solve the following equation by factorisation:

$\dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1}$

Answer

Given,

$\Rightarrow \dfrac{4}{x + 2} - \dfrac{1}{x + 3} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{4(x + 3) - (x + 2)}{(x + 2)(x + 3)} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{4x + 12 - x - 2}{x^2 + 3x + 2x + 6} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow \dfrac{3x + 10}{x^2 + 5x + 6} = \dfrac{4}{2x + 1} \\[1em] \Rightarrow (3x + 10)(2x + 1) = 4(x^2 + 5x + 6) \\[1em] \Rightarrow 6x^2 + 3x + 20x + 10 = 4x^2 + 20x + 24 \\[1em] \Rightarrow 6x^2 - 4x^2 + 23x - 20x + 10 -24 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 14 = 0 \\[1em] \Rightarrow 2x^2 + 7x - 4x - 14 = 0 \\[1em] \Rightarrow x(2x + 7) - 2(2x + 7) = 0 \\[1em] \Rightarrow (x - 2)(2x + 7) = 0 \\[1em] \Rightarrow (x - 2) = 0 \text{ or } 2x + 7 = 0 \\[1em] \Rightarrow x = 2 \text{ or } 2x = -7 \\[1em] \Rightarrow x = 2 \text{ or } x = -\dfrac{7}{2}.$

Hence, value of x = 2 or - $\dfrac{7}{2}$ .

Question 8

Solve the following equation by factorisation:

$\dfrac{5}{x - 2} - \dfrac{3}{x + 6} = \dfrac{4}{x}$

Answer

Given,

$\Rightarrow \dfrac{5(x + 6) - 3(x - 2)}{(x - 2)(x + 6)} = \dfrac{4}{x} \\[1em] \Rightarrow \dfrac{5x + 30 - 3x + 6}{x^2 + 6x - 2x - 12} = \dfrac{4}{x} \\[1em] \Rightarrow \dfrac{2x + 36}{x^2 + 4x - 12} = \dfrac{4}{x} \\[1em] \Rightarrow x(2x + 36) = 4(x^2 + 4x - 12) \\[1em] \Rightarrow 2x^2 + 36x = 4x^2 + 16x - 48 \\[1em] \Rightarrow 4x^2 - 2x^2 + 16x - 36x - 48 = 0 \\[1em] \Rightarrow 2x^2 - 20x - 48 = 0 \\[1em] \Rightarrow 2(x^2 - 10x - 24) = 0 \\[1em] \Rightarrow x^2 - 10x - 24 = 0 \\[1em] \Rightarrow x^2 - 12x + 2x - 24 = 0 \\[1em] \Rightarrow x(x - 12) + 2(x - 12) = 0 \\[1em] \Rightarrow (x + 2)(x - 12) = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 12.$

Hence, value of x = -2 or 12.

Question 9

Solve the following equation by factorisation:

$\Big(1 + \dfrac{1}{x + 1}\Big)\Big(1 - \dfrac{1}{x - 1}\Big) = \dfrac{7}{8}$

Answer

Given,

$\Rightarrow \Big(1 + \dfrac{1}{x + 1}\Big)\Big(1 - \dfrac{1}{x - 1}\Big) = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x + 1 + 1}{x + 1} \times \dfrac{x - 1 - 1}{x - 1} = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x + 2}{x + 1} \times \dfrac{x - 2}{x - 1} = \dfrac{7}{8} \\[1em] \Rightarrow \dfrac{x^2 - 4}{x^2 - 1} = \dfrac{7}{8} \\[1em] \Rightarrow 8(x^2 - 4) = 7(x^2 - 1) \\[1em] \Rightarrow 8x^2 - 32 = 7x^2 - 7 \\[1em] \Rightarrow 8x^2 - 7x^2 = - 7 + 32 \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x = \sqrt{25} \\[1em] \Rightarrow x = \pm 5.$

Hence, value of x = -5 or +5.

Question 10

Find the quadratic equation, whose solution set is :

(i) {3, 5}

(ii) {-2, 3}

Answer

(i) Since, {3, 5} is solution set.

It means 3 and 5 are roots of the equation,

∴ x = 3 or x = 5

⇒ x - 3 = 0 or x - 5 = 0

⇒ (x - 3)(x - 5) = 0

⇒ (x² - 5x - 3x + 15) = 0

⇒ x² - 8x + 15 = 0.

Hence, quadratic equation with solution set {3, 5} = x² - 8x + 15 = 0.

(ii) Since, {-2, 3} is solution set.

It means -2 and 3 are roots of the equation,

∴ x = -2 or x = 3

⇒ x + 2 = 0 or x - 3 = 0

⇒ (x + 2)(x - 3) = 0

⇒ (x² - 3x + 2x - 6) = 0

⇒ x² - x - 6 = 0.

Hence, quadratic equation with solution set {-2, 3} = x² - x - 6 = 0.

Question 11(i)

Solve : $\dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15}$ ; (x ≠ 6)

Answer

Given,

$\Rightarrow \dfrac{x}{3} + \dfrac{3}{6 - x} = \dfrac{2(6 + x)}{15} \\[1em] \Rightarrow \dfrac{x(6 - x) + 9}{3(6 - x)} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow \dfrac{6x - x^2 + 9}{18 - 3x} = \dfrac{12 + 2x}{15} \\[1em] \Rightarrow 15(6x - x^2 + 9) = (12 + 2x)(18 - 3x) \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 36x + 36x - 6x^2 \\[1em] \Rightarrow 90x - 15x^2 + 135 = 216 - 6x^2 \\[1em] \Rightarrow 15x^2 - 6x^2 - 90x + 216 - 135 = 0 \\[1em] \Rightarrow 9x^2 - 90x + 81 = 0 \\[1em] \Rightarrow 9(x^2 - 10x + 9) = 0 \\[1em] \Rightarrow x^2 - 10x + 9 = 0 \\[1em] \Rightarrow x^2 - 9x - x + 9 = 0 \\[1em] \Rightarrow x(x - 9) - 1(x - 9) = 0 \\[1em] \Rightarrow (x - 1)(x - 9) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 9.$

Hence, x = 1 or x = 9.

Question 11(ii)

Solve the equation 9x² + $\dfrac{3x}{4} + 2$ = 0, if possible, for real values of x.

Answer

Given,

9x² + $\dfrac{3x}{4} + 2$ = 0

$\Rightarrow \dfrac{36x^2 + 3x + 8}{4} = 0$

⇒ 36x² + 3x + 8 = 0

Comparing 36x² + 3x + 8 = 0 with ax² + bx + c = 0 we get,

a = 36, b = 3 and c = 8

D = b² - 4ac = (3)² - 4(36)(8) = 9 - 1152 = -1143.

Since, D < 0 hence, roots are imaginary.

There is no possibility of real roots.

Question 12

Find the value of x, if a + 7 = 0; b + 10 = 0 and 12x² = ax - b.

Answer

Given,

⇒ a + 7 = 0 and b + 10 = 0

⇒ a = -7 and b = -10

Substituting value of a and b in 12x² = ax - b,

⇒ 12x² = (-7)x - (-10)

⇒ 12x² = -7x + 10

⇒ 12x² + 7x - 10 = 0

⇒ 12x² + 15x - 8x - 10 = 0

⇒ 3x(4x + 5) - 2(4x + 5) = 0

⇒ (3x - 2)(4x + 5) = 0

⇒ 3x - 2 = 0 or 4x + 5 = 0 [Using Zero-product rule]

⇒ 3x = 2 or 4x = - 5

⇒ x = $\dfrac{2}{3} \text{ or } x = -\dfrac{5}{4}$ .

Hence, x = $\dfrac{2}{3} \text{ or } -\dfrac{5}{4}$ .

Question 13

Use the substitution 2x + 3 = y to solve for x, if 4(2x + 3)² - (2x + 3) - 14 = 0.

Answer

Substituting, 2x + 3 = y in 4(2x + 3)² - (2x + 3) - 14 = 0 we get,

⇒ 4y² - y - 14 = 0

⇒ 4y² - 8y + 7y - 14 = 0

⇒ 4y(y - 2) + 7(y - 2) = 0

⇒ (4y + 7)(y - 2) = 0

⇒ 4y + 7 = 0 or y - 2 = 0

⇒ 4y = -7 or y = 2

⇒ y = $-\dfrac{7}{4}$ or y = 2.

∴ 2x + 3 = $-\dfrac{7}{4}$ or 2x + 3 = 2

⇒ 2x = $-\dfrac{7}{4} - 3$ or 2x = 2 - 3

⇒ 2x = $\dfrac{-7 - 12}{4}$ or 2x = -1

⇒ 2x = $-\dfrac{19}{4} \text{ or } x = -\dfrac{1}{2}$

⇒ x = $-\dfrac{19}{8} \text{ or } x = -\dfrac{1}{2}$

Hence, x = $-\dfrac{19}{8} \text{ or } x = -\dfrac{1}{2}$ .

Question 14

If x ≠ 0 and a ≠ 0, solve :

$\dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax}$

Answer

Given,

$\Rightarrow \dfrac{x}{a} - \dfrac{a + b}{x} = \dfrac{b(a + b)}{ax} \\[1em] \Rightarrow \dfrac{x^2 - a(a + b)}{ax} = \dfrac{ab + b^2}{ax} \\[1em] \Rightarrow \dfrac{x^2 - a^2 - ab}{ax} \times ax = ab + b^2 \\[1em] \Rightarrow x^2 - a^2 - ab = ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + ab + ab + b^2 \\[1em] \Rightarrow x^2 = a^2 + 2ab + b^2 \\[1em] \Rightarrow x^2 = (a + b)^2 \\[1em] \Rightarrow x^2 - (a + b)^2 = 0 \\[1em] \Rightarrow (x - (a + b))(x + (a + b)) = 0 \\[1em] \Rightarrow x = a + b \text{ or } x = -(a + b).$

Hence, x = a + b or -(a + b).

Question 15(i)

Solve :

$\Big(\dfrac{1200}{x} + 2\Big)(x - 10) - 1200 = 60.$

Answer

Given,

$\Rightarrow \Big(\dfrac{1200}{x} + 2\Big)(x - 10) - 1200 = 60 \\[1em] \Rightarrow \dfrac{(1200 + 2x)(x - 10)}{x} = 60 + 1200 \\[1em] \Rightarrow \dfrac{1200x - 12000 + 2x^2 - 20x }{x} = 1260 \\[1em] \Rightarrow 2x^2 + 1180x - 12000 = 1260x \\[1em] \Rightarrow 2x^2 + 1180x -1260x - 12000 = 0 \\[1em] \Rightarrow 2x^2 - 80x - 12000 = 0 \\[1em] \Rightarrow 2(x^2 - 40x - 6000) = 0 \\[1em] \Rightarrow x^2 - 40x - 6000 = 0 \\[1em] \Rightarrow x^2 - 100x + 60x - 6000 = 0 \\[1em] \Rightarrow x(x - 100) + 60(x - 100) = 0 \\[1em] \Rightarrow (x + 60)(x - 100) = 0 \\[1em] \Rightarrow (x + 60) = 0 \text{ or } (x - 100) = 0 \\[1em] \Rightarrow x = -60 \text{ or } x = 100.$

Hence, x = -60 or 100.

Question 15(ii)

Solve :

$\dfrac{14}{x + 3} - 1 = \dfrac{5}{x + 1}$

Answer

Given,

$\Rightarrow \dfrac{14}{x + 3} - 1 = \dfrac{5}{x + 1} \\[1em] \Rightarrow \dfrac{14 - (x + 3)}{x + 3} = \dfrac{5}{x + 1} \\[1em] \Rightarrow \dfrac{14 - x - 3}{x + 3} = \dfrac{5}{x + 1} \\[1em] \Rightarrow \dfrac{11 - x}{x + 3} = \dfrac{5}{x + 1} \\[1em] \Rightarrow (11 - x)(x + 1) = 5(x + 3) \\[1em] \Rightarrow 11x + 11 - x^2 - x = 5x + 15 \\[1em] \Rightarrow -x^2 + 10x + 11 = 5x + 15 \\[1em] \Rightarrow -x^2 + 10x - 5x + 11 - 15 = 0 \\[1em] \Rightarrow -x^2 + 5x - 4 = 0 \\[1em] \Rightarrow x^2 - 5x + 4 = 0 \\[1em] \Rightarrow x^2 − 4x − x + 4 = 0 \\[1em] \Rightarrow x(x − 4) −1(x − 4) = 0 \\[1em] \Rightarrow (x − 1)(x − 4) = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = 4.$

Hence, x = 1, 4.

Question 15(iii)

Solve :

2x² + ax - a² = 0

Answer

Given,

⇒ 2x² + ax - a² = 0

⇒ 2x² + 2ax - ax - a² = 0

⇒ 2x(x + a) - a(x + a) = 0

⇒ (2x - a)(x + a) = 0

⇒ (2x - a) = 0 or (x + a) = 0

⇒ x = $\dfrac{a}{2}$ or x = -a

Hence, x = $\dfrac{a}{2}$ or -a.

Question 15(iv)

Solve :

$\sqrt{2x + 9} + x = 13$

Answer

Given,

$\Rightarrow \sqrt{2x + 9} + x = 13 \\[1em] \Rightarrow \sqrt{2x + 9} = 13 - x \\[1em] \text{Squaring both sides, we get :} \\[1em] \Rightarrow (\sqrt{2x + 9})^2 = (13 - x)^2 \\[1em] \Rightarrow 2x + 9 = 169 - 26x + x^2 \\[1em] \Rightarrow x^2 - 26x - 2x + 169 - 9 = 0 \\[1em] \Rightarrow x^2 - 28x + 160 = 0 \\[1em] \Rightarrow x^2 - 20x - 8x + 160 = 0 \\[1em] \Rightarrow x(x - 20) - 8(x - 20) = 0 \\[1em] \Rightarrow (x - 20)(x - 8) = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = 8$

Substituting value of x = 20, in L.H.S of equation $\sqrt{2x + 9} + x = 13$ , we get:

$\Rightarrow \sqrt{2(20) + 9} + 20 \\[1em] \Rightarrow \sqrt{40 + 9} + 20 \\[1em] \Rightarrow \sqrt{49} + 20 \\[1em] \Rightarrow 7 + 20 = 27$

L.H.S ≠ R.H.S

x = 20 is not valid

Substituting value of x = 8, in L.H.S of equation $\sqrt{2x + 9} + x = 13$ , we get:

$\Rightarrow \sqrt{2(8) + 9} + 8 \\[1em] \Rightarrow \sqrt{16 + 9} + 8 \\[1em] \Rightarrow \sqrt{25} + 8 \\[1em] \Rightarrow 5 + 8 \\[1em] \Rightarrow 13$

L.H.S = R.H.S

x = 8 is valid

Hence, x = 8.

Question 15(v)

Solve :

$\dfrac{2800}{x - 100} - \dfrac{2800}{x} = \dfrac{1}{2}$

Answer

$\Rightarrow \dfrac{2800}{x - 100} - \dfrac{2800}{x} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2800x - 2800(x - 100)}{x(x - 100)} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2800x - 2800x + 280000}{x^2 - 100x} = \dfrac{1}{2} \\[1em] \Rightarrow 2(280000) = x^2 - 100x \\[1em] \Rightarrow 560000 = x^2 - 100x \\[1em] \Rightarrow x^2 - 100x - 560000 = 0 \\[1em] \Rightarrow x^2 - 800x + 700x - 560000 = 0 \\[1em] \Rightarrow x(x - 800) + 700(x - 800) = 0 \\[1em] \Rightarrow (x + 700)(x - 800) = 0 \\[1em] \Rightarrow x = -700 \text { or } x = 800$

Hence, x = 800.

Question 15(vi)

Solve :

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

Answer

$\Rightarrow 3x^2 - 2\sqrt{6}x + 2 = 0 \\[1em] \Rightarrow 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \\[1em] \Rightarrow \sqrt{3}x (\sqrt{3}x - \sqrt2) - \sqrt2(\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2) (\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2)^2 = 0 \\[1em] \Rightarrow (\sqrt{3}x - \sqrt2) = 0 \\[1em] \Rightarrow \sqrt{3}x = \sqrt2 \\[1em] \Rightarrow x = \dfrac{\sqrt2}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{\sqrt2 \times \sqrt3}{\sqrt{3} \times \sqrt3} \\[1em] \Rightarrow x = \dfrac{\sqrt6}{3}.$

Hence, x = $\dfrac{\sqrt6}{3}$ .

Exercise 5(C)

Question 1(a)

If x² - 3x + 2 = 0, values of x correct to one decimal place are :

2.0 and 1.0
2.0 or 1.0
3.0 and 2.0
3.0 or 2.0

Answer

Comparing equation x² - 3x + 2 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -3 and c = 2.

By formula,

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

Substituting values we get :

$x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 2}}{2 \times 1} \\[1em] = \dfrac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \dfrac{3 \pm \sqrt{1}}{2} \\[1em] = \dfrac{3 \pm 1}{2} \\[1em] = \dfrac{3 + 1}{2} \text{ or } \dfrac{3 - 1}{2} \\[1em] = \dfrac{4}{2} \text{ or } \dfrac{2}{2} \\[1em] = 2.0 \text{ or } 1.0$

Hence, Option 2 is the correct option.

Question 1(b)

If x² - 4x - 5 = 0, values of x correct to two decimal places are :

5.00 or -1.00
5 or 1
5.0 and -1.0
-5.00 and -1.00

Answer

Comparing equation x² - 4x - 5 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -4 and c = -5.

By formula,

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

Substituting values we get :

$x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times -5}}{2 \times 1} \\[1em] = \dfrac{4 \pm \sqrt{16 + 20}}{2} \\[1em] = \dfrac{4 \pm \sqrt{36}}{2} \\[1em] = \dfrac{4 \pm 6}{2} \\[1em] = \dfrac{4 + 6}{2} \text{ or } \dfrac{4 - 6}{2} \\[1em] = \dfrac{10}{2} \text{ or } \dfrac{-2}{2} \\[1em] = 5.00 \text{ or } -1.00$

Hence, Option 1 is the correct option.

Question 1(c)

If x² - 8x - 9 = 0; values of x correct to one significant figure are :

9 and -1
9 or -1
9.0 or -1.0
9.00 or -1.00

Answer

Comparing equation x² - 8x - 9 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -8 and c = -9.

By formula,

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

Substituting values we get :

$x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times -9}}{2 \times 1} \\[1em] = \dfrac{8 \pm \sqrt{64 + 36}}{2} \\[1em] = \dfrac{8 \pm \sqrt{100}}{2} \\[1em] = \dfrac{8 \pm 10}{2} \\[1em] = \dfrac{8 + 10}{2} \text{ or } \dfrac{8 - 10}{2} \\[1em] = \dfrac{18}{2} \text{ or } \dfrac{-2}{2} \\[1em] = 9 \text{ or } -1$

Hence, Option 2 is the correct option.

Question 1(d)

If x² - 2x - 3 = 0; values of x correct to two significant figures are :

3.0 and 1.0
-1.0 and 3.0
3.0 or -1.0
3.00 or -1.00

Answer

Given,

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ x + 1 = 0 or x - 3 = 0

⇒ x = -1 or x = 3.

Rounding off to two significant figures, we get :

⇒ x = -1.0 or x = 3.0

Hence, Option 3 is the correct option.

Question 1(e)

The value (values) of x satisfying the equation x² - 6x - 16 = 0 is/are :

8 or -2
-8 or 2
8 and -2
-8 and 2

Answer

Comparing equation x² - 6x - 16 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -6 and c = -16.

By formula,

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

Substituting values we get :

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times -16}}{2 \times 1} \\[1em] = \dfrac{6 \pm \sqrt{36 + 64}}{2} \\[1em] = \dfrac{6 \pm \sqrt{100}}{2} \\[1em] = \dfrac{6 \pm 10}{2} \\[1em] = \dfrac{6 + 10}{2} \text{ or } \dfrac{6 - 10}{2} \\[1em] = \dfrac{16}{2} \text{ or } \dfrac{-4}{2} \\[1em] = 8 \text{ or } -2.$

Substituting x = 8 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ 8² - 6(8) - 16

⇒ 64 - 48 - 16

⇒ 0.

Substituting x = -2 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ (-2)² - 6(-2) - 16

⇒ 4 + 12 - 16

⇒ 0.

Since, L.H.S. = R.H.S., thus x = -2 is a solution of the equation.

Thus 8 and -2 are the solution of the equation x² - 6x - 16 = 0.

Hence, Option 3 is the correct option.

Question 2(i)

Solve the following equation for x and give your answer correct to one decimal place :

x² - 8x + 5 = 0

Answer

Comparing x² - 8x + 5 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -8 and c = 5.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \dfrac{8 \pm \sqrt{64 - 20}}{2} \\[1em] = \dfrac{8 \pm \sqrt{44}}{2} \\[1em] = \dfrac{8 \pm 2\sqrt{11}}{2} \\[1em] = 4 \pm \sqrt{11} \\[1em] = 4 + \sqrt{11} \text{ and } 4 - \sqrt{11} \\[1em] = 4 + 3.3 \text{ and } 4 - 3.3 \\[1em] = 7.3 \text{ and } 0.7$

Hence, x = 7.3 and 0.7

Question 2(ii)

Solve the following equation for x and give your answer correct to one decimal place :

5x² + 10x - 3 = 0

Answer

Comparing 5x² + 10x - 3 = 0 with ax² + bx + c = 0 we get,

a = 5, b = 10 and c = -3.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(10) \pm \sqrt{(10)^2 - 4.(5).(-3)}}{2(5)} \\[1em] = \dfrac{-10 \pm \sqrt{100 + 60}}{10} \\[1em] = \dfrac{-10 \pm \sqrt{160}}{10} \\[1em] = \dfrac{-10 \pm 4\sqrt{10}}{10} \\[1em] = \dfrac{-10 \pm 12.8}{10} \\[1em] = \dfrac{-10 + 12.8}{10} \text{ and } \dfrac{-10 - 12.8}{10} \\[1em] = \dfrac{2.8}{10} \text{ and } \dfrac{-22.8}{10} \\[1em] = 0.28 \text{ and } -2.28 \\[1em] \approx 0.3 \text{ and } -2.3$

Hence, x = 0.3 and -2.3

Question 3(i)

Solve the following equation for x and give your answer correct to two decimal places :

2x² - 10x + 5 = 0

Answer

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

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

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4.(2).(5)}}{2(2)} \\[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \dfrac{10 \pm \sqrt{60}}{4} \\[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \\[1em] = \dfrac{10 \pm 7.74}{4} \\[1em] = \dfrac{10 + 7.74}{4} \text{ and } \dfrac{10 - 7.74}{4} \\[1em] = \dfrac{17.74}{4} \text{ and } \dfrac{2.26}{4} \\[1em] = 4.44 \text{ and } 0.56$

Hence, x = 4.44 and 0.56

Question 3(ii)

Solve the following equation for x and give your answer correct to two decimal places :

4x + $\dfrac{6}{x}$ + 13 = 0

Answer

Given,

$\Rightarrow 4x + \dfrac{6}{x} + 13 = 0 \\[1em] \Rightarrow \dfrac{4x^2 + 6 + 13x}{x} = 0 \\[1em] \Rightarrow 4x^2 + 13x + 6 = 0$

Comparing 4x² + 13x + 6 = 0 with ax² + bx + c = 0 we get,

a = 4, b = 13 and c = 6.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(13) \pm \sqrt{(-13)^2 - 4.(4).(6)}}{2(4)} \\[1em] = \dfrac{-13 \pm \sqrt{169 - 96}}{8} \\[1em] = \dfrac{-13 \pm \sqrt{73}}{8} \\[1em] = \dfrac{-13 \pm 8.54}{8} \\[1em] = \dfrac{-13 + 8.54}{8} \text{ and } \dfrac{-13 - 8.54}{8} \\[1em] = \dfrac{-4.46}{8} \text{ and } \dfrac{-21.54}{8} \\[1em] = -0.56 \text{ and } -2.69$

Hence, x = -0.56 and -2.69

Question 3(iii)

Solve the following equation for x and give your answer correct to two decimal places :

4x² - 5x - 3 = 0

Answer

Comparing 4x² - 5x - 3 = 0 with ax² + bx + c = 0 we get,

a = 4, b = -5 and c = -3.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4.(4).(-3)}}{2(4)} \\[1em] = \dfrac{5 \pm \sqrt{25 + 48}}{8} \\[1em] = \dfrac{5 \pm \sqrt{73}}{8} \\[1em] = \dfrac{5 \pm 8.54}{8} \\[1em] = \dfrac{5 + 8.54}{8} \text{ and } \dfrac{5 - 8.54}{8} \\[1em] = \dfrac{13.54}{8} \text{ and } \dfrac{-3.54}{8} \\[1em] = 1.69 \text{ and } -0.44$

Hence, x = 1.69 and -0.44

Question 4(i)

Solve the following equation for x, giving your answer correct to 3 decimal places :

3x² - 12x - 1 = 0

Answer

Comparing 3x² - 12x - 1 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -12 and c = -1.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-12) \pm \sqrt{(-12)^2 - 4.(3).(-1)}}{2(3)} \\[1em] = \dfrac{12 \pm \sqrt{144 + 12}}{6} \\[1em] = \dfrac{12 \pm \sqrt{156}}{6} \\[1em] = \dfrac{12 \pm 12.49}{6} \\[1em] = \dfrac{12 + 12.49}{6} \text{ or } \dfrac{12 - 12.49}{6} \\[1em] = \dfrac{24.49}{6} \text{ or } \dfrac{-0.49}{6} \\[1em] = 4.082 \text{ or } -0.082$

Hence, x = 4.082 or -0.082

Question 4(ii)

Solve the following equation for x, giving your answer correct to 3 decimal places :

x² - 16x + 6 = 0

Answer

Comparing x² - 16x + 6 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -16 and c = 6.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-16) \pm \sqrt{(-16)^2 - 4.(1).(6)}}{2(1)} \\[1em] = \dfrac{16 \pm \sqrt{256 - 24}}{2} \\[1em] = \dfrac{16 \pm \sqrt{232}}{2} \\[1em] = \dfrac{16 \pm 15.232}{2} \\[1em] = \dfrac{16 + 15.232}{2} \text{ or } \dfrac{16 - 15.232}{2} \\[1em] = \dfrac{31.232}{2} \text{ or } \dfrac{0.768}{2} \\[1em] = 15.616 \text{ or } 0.384$

Hence, x = 15.616 or 0.384

Question 4(iii)

Solve the following equation for x, giving your answer correct to 3 decimal places :

2x² + 11x + 4 = 0

Answer

Comparing 2x² + 11x + 4 = 0 with ax² + bx + c = 0 we get,

a = 2, b = 11 and c = 4.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(11) \pm \sqrt{(11)^2 - 4.(2).(4)}}{2(2)} \\[1em] = \dfrac{-11 \pm \sqrt{121 - 32}}{4} \\[1em] = \dfrac{-11 \pm \sqrt{89}}{4} \\[1em] = \dfrac{-11 \pm 9.434}{4} \\[1em] = \dfrac{-11 + 9.434}{4} \text{ or } \dfrac{-11 - 9.434}{4} \\[1em] = \dfrac{-1.566}{4} \text{ or } \dfrac{-20.434}{4} \\[1em] = -0.392 \text{ or } -5.109$

Hence, x = -0.392 or -5.109

Question 5

Solve the following equation and give your answer correct to 3 significant figures :

5x² - 3x - 4 = 0

Answer

Comparing 5x² - 3x - 4 = 0 with ax² + bx + c = 0 we get,

a = 5, b = -3 and c = -4.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4.(5).(-4)}}{2(5)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 80}}{10} \\[1em] = \dfrac{3 \pm \sqrt{89}}{10} \\[1em] = \dfrac{3 \pm 9.434}{10} \\[1em] = \dfrac{3 + 9.434}{10} \text{ or } \dfrac{3 - 9.434}{10} \\[1em] = \dfrac{12.434}{10} \text{ or } \dfrac{-6.434}{10} \\[1em] = 1.2434 \text{ or } -0.6434 \\[1em] \approx 1.24 \text{ or } -0.643$

Hence, x = 1.24 or -0.643

Question 6

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

⇒ x² + 1 - 2x - 3x + 4 = 0

⇒ x² - 5x + 5 = 0

Comparing x² - 5x + 5 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -5 and c = 5.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \dfrac{5 \pm \sqrt{25 - 20}}{2} \\[1em] = \dfrac{5 \pm \sqrt{5}}{2} \\[1em] = \dfrac{5 \pm 2.2}{2} \\[1em] = \dfrac{5 + 2.2}{2} \text{ or } \dfrac{5 - 2.2}{2} \\[1em] = \dfrac{7.2}{2} \text{ or } \dfrac{2.8}{2} \\[1em] = 3.6 \text{ or } 1.4$

Hence, x = 3.6 or 1.4

Question 7

x = 3 is a solution of the quadratic equation (k + 2)x² - kx + 6 = 0, then other root is:

1
3
-3
-4

Answer

Given,

x = 3 is a solution of the quadratic equation (k + 2)x² - kx + 6 = 0.

Substituting x = 3 in equation (k + 2)x² - kx + 6 = 0,

⇒ (k + 2)(3)² - k(3) + 6 = 0

⇒ (k + 2)(9) - 3k + 6 = 0

⇒ 9k + 18 - 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = -24

⇒ k = $\dfrac{-24}{6}$

⇒ k = -4.

Substituting k = −4 in equation (k + 2)x² - kx + 6 = 0 :

⇒ (-4 + 2)x² - (-4)x + 6 = 0

⇒ -2x² + 4x + 6 = 0

⇒ -2(x² - 2x - 3) = 0

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ (x + 1) = 0 or (x - 3) = 0

⇒ x = -1 or x = 3.

Hence, option 4 is the correct option.

Exercise 5(D)

Question 1(a)

Equation 2x² - 3x + 1 = 0 has :

distinct and real roots
no real roots
equal roots
imaginary roots

Answer

Comparing equation 2x² - 3x + 1 = 0, with ax² + bx + c = 0, we get :

a = 2, b = -3 and c = 1.

By formula,

D = b² - 4ac

= (-3)² - 4 × 2 × 1

= 9 - 8

= 1; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ Roots are real and unequal.

Hence, Option 1 is the correct option.

Question 1(b)

Which of the following equations has two real and distinct roots ?

x² - 5x + 6 = 0
x² - 3x + 6 = 0
x² - 2x + 5 = 0
x² - 4x + 6 = 0

Answer

Comparing equation x² - 5x + 6 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -5 and c = 6.

By formula,

D = b² - 4ac

= (-5)² - 4 × 1 × 6

= 25 - 24

= 1; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ Roots are real and distinct.

Hence, Option 1 is the correct option.

Question 1(c)

If the roots of equation x² - 6x + k = 0 are real and distinct, the 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 1(d)

If the roots of x² - px + 4 = 0 are equal, the value (values) of p is/are :

4 and -4
4
-4
4 or -4

Answer

Comparing equation x² - px + 4 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -p and c = 4.

Since, roots are equal.

∴ D = 0

∴ b² - 4ac = 0

⇒ (-p)² - 4 × 1 × 4 = 0

⇒ p² - 16 = 0

⇒ p² - 4² = 0

⇒ (p - 4)(p + 4) = 0

⇒ (p - 4) = 0 or (p + 4) = 0

⇒ p = 4 or p = -4.

Hence, Option 4 is the correct option.

Question 1(e)

Which of the following equations has imaginary roots ?

x² + 10x - 3 = 0
2x² - 5x + 9 = 0
x² + 5x + 4 = 0
5x² - 8x - 1 = 0

Answer

Comparing equation x² + 10x - 3 = 0, with ax² + bx + c = 0, we get :

a = 1, b = 10 and c = -3.

By formula,

D = b² - 4ac

= (10)² - 4 × 1 × -3

= 100 + 12

= 112; which is positive.

Comparing equation 2x² - 5x + 9 = 0, with ax² + bx + c = 0, we get :

a = 2, b = -5 and c = 9.

By formula,

D = b² - 4ac

= (-5)² - 4 × 2 × 9

= 25 - 72

= -47; which is negative.

∴ Roots are imaginary.

Hence, Option 2 is the correct option.

Question 1(f)

One root of equation 3x² - mx + 4 = 0 is 1, the value of m is :

7
-7
$\dfrac{4}{3}$
$-\dfrac{4}{3}$

Answer

Since, 1 is the root of equation 3x² - mx + 4 = 0.

∴ x = 1, will satisfy the equation 3x² - mx + 4 = 0.

⇒ 3(1)² - m(1) + 4 = 0

⇒ 3 - m + 4 = 0

⇒ 7 - m = 0

⇒ m = 7.

Hence, Option 1 is the correct option.

Question 2(i)

Without solving, comment upon the nature of roots of the following equation :

7x² - 9x + 2 = 0

Answer

Comparing 7x² - 9x + 2 = 0 with ax² + bx + c = 0 we get,

a = 7, b = -9 and c = 2.

We know that,

Discriminant = D = b² - 4ac = (-9)² - 4(7)(2)

= 81 - 56 = 25; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.

Question 2(ii)

Without solving, comment upon the nature of roots of the following equation :

6x² - 13x + 4 = 0

Answer

Comparing 6x² - 13x + 4 = 0 with ax² + bx + c = 0 we get,

a = 6, b = -13 and c = 4.

We know that,

Discriminant = D = b² - 4ac = (-13)² - 4(6)(4)

= 169 - 96 = 73; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.

Question 2(iii)

Without solving, comment upon the nature of roots of the following equation :

25x² - 10x + 1 = 0

Answer

Comparing 25x² - 10x + 1 = 0 with ax² + bx + c = 0 we get,

a = 25, b = -10 and c = 1.

We know that,

Discriminant = D = b² - 4ac = (-10)² - 4(25)(1)

= 100 - 100 = 0;

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac = 0

∴ The roots are real and equal.

Question 2(iv)

Without solving, comment upon the nature of roots of the following equation :

x² + $2\sqrt{3}x - 9$ = 0

Answer

Comparing x² + $2\sqrt{3}$ x - 9 = 0 with ax² + bx + c = 0 we get,

a = 1, b = $2\sqrt{3}$ and c = -9.

We know that,

Discriminant = D = b² - 4ac = $(2\sqrt{3})$ ² - 4(1)(-9)

= 12 + 36 = 48; which is positive.

Since, a, b and c are real numbers; a ≠ 0 and b² - 4ac > 0

∴ The roots are real and unequal.

Question 3

The equation 3x² - 12x + (n - 5) = 0 has equal roots. Find the value of n.

Answer

Comparing 3x² - 12x + (n - 5) = 0 with ax² + bx + c = 0 we get,

a = 3, b = -12 and c = (n - 5).

Since equations have equal roots,

∴ D = 0

⇒ (-12)² - 4.(3).(n - 5) = 0

⇒ 144 - 12(n - 5) = 0

⇒ 144 - 12n + 60 = 0

⇒ 204 - 12n = 0

⇒ 12n = 204

⇒ n = $\dfrac{204}{12}$

⇒ n = 17

Hence, n = 17.

Question 4

Find the values of 'm', if the following equation has equal roots :

(m - 2)x² - (5 + m)x + 16 = 0

Answer

Comparing (m - 2)x² - (5 + m)x + 16 = 0 with ax² + bx + c = 0 we get,

a = (m - 2), b = -(5 + m) and c = 16.

Since equations have equal roots,

∴ D = 0

⇒ (-(5 + m))² - 4.(m - 2).(16) = 0

⇒ 25 + m² + 10m - 64(m - 2) = 0

⇒ 25 + m² + 10m - 64m + 128 = 0

⇒ m² - 54m + 153 = 0

⇒ m² - 51m - 3m + 153 = 0

⇒ m(m - 51) - 3(m - 51) = 0

⇒ (m - 3)(m - 51) = 0

⇒ (m - 3) = 0 or (m - 51) = 0

⇒ m = 3 or m = 51.

Hence, m = 3 or 51.

Question 5

Find the value of k for which the equation 3x² - 6x + k = 0 has distinct and real roots.

Answer

Comparing 3x² - 6x + k = 0 with ax² + bx + c = 0 we get,

a = 3, b = -6 and c = k.

Since equations have distinct and real roots,

∴ D > 0

⇒ (-6)² - 4.(3).(k) > 0

⇒ 36 - 12k > 0

⇒ 12k < 36

⇒ k < 3.

Hence, k < 3.

Question 6

Given that 2 is a root of the equation 3x² - p(x + 1) = 0 and that the equation px² - qx + 9 = 0 has equal roots, find the values of p and q.

Answer

Since, 2 is the root hence, it satisfies the equation 3x² - p(x + 1) = 0.

⇒ 3(2)² - p(2 + 1) = 0

⇒ 3(4) - 3p = 0

⇒ 3p = 12

⇒ p = 4.

Substituting value of p in px² - qx + 9 = 0

⇒ 4x² - qx + 9 = 0

Comparing 4x² - qx + 9 = 0 with ax² + bx + c = 0 we get,

a = 4, b = -q and c = 9.

Since equation has equal roots,

∴ D = 0

⇒ (-q)² - 4.(4).(9) = 0

⇒ q² - 144 = 0

⇒ q² = 144

⇒ q = 12 or -12.

Hence, p = 4 and q = 12 or -12.

Question 7(i)

Use quadratic formula to solve:

$\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$

Answer

Given,

$\Rightarrow \dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{x - 7 - x - 4}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow \dfrac{-11}{(x + 4)(x - 7)} = \dfrac{11}{30} \\[1em] \Rightarrow -11 \times 30 = 11(x + 4)(x - 7) \\[1em] \Rightarrow \dfrac{-11 \times 30}{11} = x^2 - 7x + 4x - 28 \\[1em] \Rightarrow -30 = x^2 - 3x - 28 \\[1em] \Rightarrow x^2 - 3x + 2 = 0$

Now we have,

x² - 3x + 2 = 0

Comparing x² - 3x + 2 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -3 and c = 2.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \dfrac{3 \pm \sqrt{1}}{2} \\[1em] = \dfrac{3 \pm 1}{2} \\[1em] = \dfrac{3 + 1}{2} \text{ or } \dfrac{3 - 1}{2} \\[1em] = \dfrac{4}{2} \text{ or } \dfrac{2}{2} \\[1em] = 2 \text{ or } 1.$

Hence, x = 1 or 2.

Question 7(ii)

Use quadratic formula to solve:

x² = 4x

Answer

Given,

x² = 4x

x² - 4x = 0

Comparing x² - 4x = 0 with ax² + bx + c = 0 we get,

a = 1, b = -4 and c = 0.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(0)}}{2(1)} \\[1em] = \dfrac{4 \pm \sqrt{16 - 0}}{2} \\[1em] = \dfrac{4 \pm 4}{2} \\[1em] = \dfrac{4 + 4}{2} \text{ or } \dfrac{4 - 4}{2} \\[1em] = \dfrac{8}{2} \text{ or } \dfrac{0}{2} \\[1em] = 4 \text{ or } 0.$

Hence, x = 0 or 4.

Question 7(iii)

Use quadratic formula to solve:

3y + $\dfrac{5}{16y}$ = 2

Answer

Given,

3y + $\dfrac{5}{16y}$ = 2

$\dfrac{48y^2 + 5}{16y}$ = 2

48y² + 5 = 32y

48y² - 32y + 5 = 0

Comparing 48y² - 32y + 5 = 0 with ax² + bx + c = 0 we get,

a = 48, b = -32 and c = 5.

We know that,

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

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-32) \pm \sqrt{(-32)^2 - 4(48)(5)}}{2(48)} \\[1em] = \dfrac{32 \pm \sqrt{1024 - 960}}{96} \\[1em] = \dfrac{32 \pm \sqrt{64}}{96} \\[1em] = \dfrac{32 \pm 8}{96} \\[1em] = \dfrac{32 + 8}{96} \text{ or } \dfrac{32 - 8}{96} \\[1em] = \dfrac{40}{96} \text{ or } \dfrac{24}{96} \\[1em] = \dfrac{5}{12} \text{ or } \dfrac{1}{4}.$

Hence, x = $\dfrac{5}{12} \text{ or } \dfrac{1}{4}$ .

Question 8

From each of the following equations, find the value of constant 'k' so that each equation has real and equal roots.

(i) kx(x - 2) + 6 = 0

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Answer

(i) kx(x - 2) + 6 = 0

Given,

kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

Comparing kx² - 2kx + 6 = 0 with ax² + bx + c = 0 we get,

a = k, b = -2k and c = 6.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (-2k)² - 4(k)(6) = 0

⇒ 4k² - 24k = 0

⇒ 4k(k - 6) = 0

⇒ 4k = 0 or (k - 6) = 0

⇒ k = 0 or k = 6

If k = 0, substituting in L.H.S. of kx² - 2kx + 6 = 0 we get,

= (0)x² - 2(0)x + 6

= 6

L.H.S. ≠ R.H.S.

Thus, k ≠ 0.

Substituting k = 0 in kx² - 2kx + 6 = 0 we get,

⇒ (6)x² - 2(6)x + 6 = 0

⇒ 6x² - 12x + 6 = 0

⇒ 6x² - 6x - 6x + 6 = 0

⇒ 6x(x - 1) - 6(x - 1) = 0

⇒ (6x - 6)(x - 1) = 0

⇒ (6x - 6) = 0 or (x - 1) = 0

⇒ x = $\dfrac{6}{6}$ or x = 1

⇒ x = 1 equation has real and equal roots

∴ k = 6

Hence, k = 6.

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Given,

(k + 4)x² + (k + 1)x + 1 = 0

Comparing (k + 4)x² + (k + 1)x + 1 = 0 with ax² + bx + c = 0 we get,

a = k + 4, b = k + 1 and c = 1.

Since roots are real and equal,

We know that,

∴ D = 0

⇒ b² - 4ac = 0

⇒ (k + 1)² - 4(k + 4)(1) = 0

⇒ k² + 2k + 1 - 4k - 16 = 0

⇒ k² - 2k - 15 = 0

⇒ k² - 5k + 3k - 15 = 0

⇒ k(k - 5) + 3(k - 5) = 0

⇒ (k + 3)(k - 5) = 0

⇒ k + 3 = 0 or k - 5 = 0

⇒ k = -3 or k = 5.

If k = 5, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (5 + 4)x² + (5 + 1)x + 1 = 0

⇒ 9x² + 6x + 1 = 0

⇒ (3x + 1)² = 0

⇒ 3x + 1 = 0

⇒ x = $\dfrac{-1}{3}$

Both roots are the same.

If k = -3, substituting in (k + 4)x² + (k + 1)x + 1 = 0 we get,

⇒ (-3 + 4)x² + (-3 + 1)x + 1 = 0

⇒ x² - 2x + 1 = 0

⇒ (x - 1)² = 0

⇒ x - 1 = 0

⇒ x = 1

Both roots are the same.

Hence, k = -3 or 5.

Exercise 5(E)

Question 1(a)

If x⁴ - 5x² + 4 = 0; the values of x are :

1 or 2
± 1 or ± 2
-1 and 2
-1 and -2

Answer

Given,

⇒ x⁴ - 5x² + 4 = 0

⇒ x⁴ - 4x² - x² + 4 = 0

⇒ x²(x² - 4) - 1(x² - 4) = 0

⇒ (x² - 1)(x² - 4) = 0

⇒ x² - 1 = 0 or x² - 4 = 0

⇒ x² = 1 or x² = 4

⇒ x = $\sqrt{1}$ or x = $\sqrt{4}$

⇒ x = ± 1 or x = ± 2.

Hence, Option 2 is the correct option.

Question 1(b)

For equation $\dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10}$ ; one value of x is :

$-\dfrac{5}{3}$
10
-10
5

Answer

Given,

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{x - 5 + x}{x(x - 5)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x - 5}{x^2 - 5x} = \dfrac{3}{10} \\[1em] \Rightarrow 10(2x - 5) = 3(x^2 - 5x) \\[1em] \Rightarrow 20x - 50 = 3x^2 - 15x \\[1em] \Rightarrow 3x^2 - 15x - 20x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 35x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 30x - 5x + 50 = 0 \\[1em] \Rightarrow 3x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (3x - 5)(x - 10) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow 3x = 5 \text{ or } x = 10 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = 10.$

Hence, Option 2 is the correct option.

Question 1(c)

Which of the following is correct for the equation $\dfrac{1}{x - 3} - \dfrac{1}{x + 5}$ = 1 ?

x ≠ 3 and x = -5
x = 3 and x ≠ -5
x ≠ 3 and x ≠ -5
x > 3 and x < 5

Answer

Given,

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

So, in above equation.

x - 3 and x + 5 cannot be equal to zero.

⇒ x - 3 ≠ 0

⇒ x ≠ 3

⇒ x + 5 ≠ 0

⇒ x ≠ -5.

Hence, Option 3 is the correct option.

Question 2

Solve :

2x⁴ - 5x² + 3 = 0

Answer

Let x² = y,

⇒ 2x⁴ - 5x² + 3 = 0

⇒ 2y² - 5y + 3 = 0

⇒ 2y² - 2y - 3y + 3 = 0

⇒ 2y(y - 1) - 3(y - 1) = 0

⇒ (2y - 3)(y - 1) = 0

⇒ 2y - 3 = 0 or y - 1 = 0 [Zero product rule]

⇒ 2y = 3 or y = 1

⇒ y = $\dfrac{3}{2}$ or y = 1.

∴ x² = $\dfrac{3}{2}$ or x² = 1

⇒ x = $\pm \sqrt{\dfrac{3}{2}} = \pm 1.22$ or x = $\sqrt{1} = \pm 1$

Hence, x = +1.22, -1.22, +1, -1.

Question 3

Solve :

x⁴ - 2x² - 3 = 0

Answer

Let x² = y,

⇒ x⁴ - 2x² - 3 = 0

⇒ y² - 2y - 3 = 0

⇒ y² - 3y + y - 3 = 0

⇒ y(y - 3) + 1(y - 3) = 0

⇒ (y + 1)(y - 3) = 0

⇒ y + 1 = 0 or y - 3 = 0 [Zero product rule]

⇒ y = -1 or y = 3

∴ x² = -1 or x² = 3

Since, square of a number cannot be negative,

∴ x² = 3

⇒ x = $\sqrt{3} = \pm 1.73$

Hence, x = +1.73, -1.73.

Question 4(i)

Solve :

(x² - x)² + 5(x² - x) + 4 = 0

Answer

Let x² - x = a

Substituting value in (x² - x)² + 5(x² - x) + 4 = 0 we get,

⇒ a² + 5a + 4 = 0

⇒ a² + 4a + a + 4 = 0

⇒ a(a + 4) + 1(a + 4) = 0

⇒ (a + 1)(a + 4) = 0

⇒ a + 1 = 0 or a + 4 = 0 [Zero product rule]

⇒ a = -1 or a = -4

∴ x² - x = -1 and x² - x = -4

Solving, x² - x = -1

⇒ x² - x = -1

⇒ x² - x + 1 = 0

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

a = 1, b = -1, c = 1

Discriminant = D = b² - 4ac = (-1)² - 4(1)(1) = 1 - 4 = -3.

-3 < 0

∴ No real solution.

Solving, x² - x = -4

⇒ x² - x + 4 = 0

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

a = 1, b = -1, c = 4

Discriminant = D = b² - 4ac = (-1)² - 4(1)(4) = 1 - 16 = -15.

-15 < 0

∴ No real solution.

Hence, there is no real solution.

Question 4(ii)

Solve :

(x² - 3x)² - 16(x² - 3x) - 36 = 0

Answer

Let x² - 3x = a

Substituting value in (x² - 3x)² - 16(x² - 3x) - 36 = 0 we get,

⇒ a² - 16a - 36 = 0

⇒ a² - 18a + 2a - 36 = 0

⇒ a(a - 18) + 2(a - 18) = 0

⇒ (a + 2)(a - 18) = 0

⇒ a + 2 = 0 or a - 18 = 0 [Zero product rule]

⇒ a = -2 or a = 18

∴ x² - 3x = -2 and x² - 3x = 18

Solving, x² - 3x = -2

⇒ x² - 3x = -2

⇒ x² - 3x + 2 = 0

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x - 1 = 0 or x - 2 = 0 [Zero product rule]

⇒ x = 1 or x = 2.

Solving, x² - 3x = 18

⇒ x² - 3x = 18

⇒ x² - 3x - 18 = 0

⇒ x² - 6x + 3x - 18 = 0

⇒ x(x - 6) + 3(x - 6) = 0

⇒ (x + 3)(x - 6) = 0

⇒ x + 3 = 0 or x - 6 = 0 [Zero product rule]

⇒ x = -3 or x = 6.

Hence, x = 1, 2, -3, 6.

Question 5(i)

Solve :

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

Answer

Let $\sqrt{\dfrac{x}{x - 3}} =$ a .......(i)

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} + \sqrt{\dfrac{x - 3}{x}} = \dfrac{5}{2} \\[1em] \Rightarrow a + \dfrac{1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow 2(a^2 + 1) = 5a \\[1em] \Rightarrow 2a^2 + 2 = 5a \\[1em] \Rightarrow 2a^2 - 5a + 2 = 0 \\[1em] \Rightarrow 2a^2 - 4a - a + 2 = 0 \\[1em] \Rightarrow 2a(a - 2) - 1(a - 2) = 0 \\[1em] \Rightarrow (2a - 1)(a - 2) = 0 \\[1em] \Rightarrow 2a - 1 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.$

Substituting value of a = $\dfrac{1}{2}$ in (i) we get,

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} = \dfrac{1}{2}$

Squaring both sides we get,

$\Rightarrow \dfrac{x}{x - 3} = \dfrac{1}{4} \\[1em] \Rightarrow 4x = x - 3 \\[1em] \Rightarrow 4x - x = -3 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1.$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} = 2$

Squaring both sides we get,

$\Rightarrow \dfrac{x}{x - 3} = 4 \\[1em] \Rightarrow x = 4(x - 3) \\[1em] \Rightarrow x = 4x - 12 \\[1em] \Rightarrow 4x - x = 12 \\[1em] \Rightarrow 3x = 12 \\[1em] \Rightarrow x = 4.$

Hence, x = -1, 4.

Question 5(ii)

Solve :

$\Big(\dfrac{2x - 3}{x - 1}\Big) - 4\Big(\dfrac{x - 1}{2x -3}\Big) = 3$

Answer

Let $\dfrac{2x - 3}{x - 1} =$ a .......(i)

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

Substituting value of a = -1 in (i) we get,

$\Rightarrow \dfrac{2x - 3}{x - 1} = -1 \\[1em] \Rightarrow 2x - 3 = -1(x - 1) \\[1em] \Rightarrow 2x - 3 = -x + 1 \\[1em] \Rightarrow 2x + x = 1 + 3 \\[1em] \Rightarrow 3x = 4 \\[1em] \Rightarrow x = \dfrac{4}{3} \\[1em] \Rightarrow x = 1\dfrac{1}{3}$

Substituting value of a = 4 in (i) we get,

$\Rightarrow \dfrac{2x - 3}{x - 1} = 4 \\[1em] \Rightarrow 2x - 3 = 4(x - 1) \\[1em] \Rightarrow 2x - 3 = 4x - 4 \\[1em] \Rightarrow 4x - 2x = -3 + 4 \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Hence, x = $1\dfrac{1}{3}, \dfrac{1}{2}$ .

Question 5(iii)

Solve :

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

Answer

Let $\Big(\dfrac{3x + 1}{x + 1}\Big) = a$ .......(i)

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) + \Big(\dfrac{x + 1}{3x + 1}\Big) = \dfrac{5}{2} \\[1em] \Rightarrow a + \dfrac{1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{5}{2} \\[1em] \Rightarrow 2(a^2 + 1) = 5a \\[1em] \Rightarrow 2a^2 + 2 - 5a = 0 \\[1em] \Rightarrow 2a^2 - 5a + 2 = 0 \\[1em] \Rightarrow 2a^2 - 4a - a + 2 = 0 \\[1em] \Rightarrow 2a(a - 2) - 1(a - 2) = 0 \\[1em] \Rightarrow (2a - 1)(a - 2) = 0 \\[1em] \Rightarrow 2a - 1 = 0 \text{ or } a - 2 = 0 \\[1em] \Rightarrow 2a = 1 \text{ or } a = 2 \\[1em] \Rightarrow a = \dfrac{1}{2} \text{ or } a = 2.$

Substituting value of a = $\dfrac{1}{2}$ in (i) we get,

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = \dfrac{1}{2} \\[1em] \Rightarrow 2(3x + 1) = x + 1 \\[1em] \Rightarrow 6x + 2 = x + 1 \\[1em] \Rightarrow 6x - x = 1 - 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\dfrac{1}{5}$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = 2 \\[1em] \Rightarrow 3x + 1 = 2(x + 1) \\[1em] \Rightarrow 3x + 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 - 1 \\[1em] \Rightarrow x = 1.$

Hence, x = 1, $-\dfrac{1}{5}$ .

Question 6

Solve :

$9(x^2 + \dfrac{1}{x^2}) - 9(x + \dfrac{1}{x}) - 52 = 0$

Answer

Let $x + \dfrac{1}{x} = a$ ........(i)

Squaring, both sides we get,

$\Rightarrow x^2 + \dfrac{1}{x^2} + 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 - 2 .......(ii)$

Substituting the values from equations (i) and (ii) we get,

⇒ 9(a² - 2) - 9a - 52 = 0

⇒ 9a² - 18 - 9a - 52 = 0

⇒ 9a² - 9a - 70 = 0

⇒ 9a² - 30a + 21a - 70 = 0

⇒ 3a(3a - 10) + 7(3a - 10) = 0

⇒ (3a + 7)(3a - 10) = 0

⇒ (3a + 7) = 0 or 3a - 10 = 0 [Zero product rule]

⇒ 3a = -7 or 3a = 10

⇒ $a = -\dfrac{7}{3} \text{ or } a = \dfrac{10}{3}$

Considering a = $\dfrac{10}{3}$ we get,

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

Considering a = $-\dfrac{7}{3}$ we get,

$\Rightarrow x + \dfrac{1}{x} = -\dfrac{7}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = -\dfrac{7}{3} \\[1em] \Rightarrow 3(x^2 + 1) = -7x \\[1em] \Rightarrow 3x^2 + 3 = -7x \\[1em] \Rightarrow 3x^2 + 7x + 3 = 0$

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

a = 3, b = 7 and c = 3.

We know that,

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

$\Rightarrow x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)} \\[1em] = \dfrac{-7 \pm \sqrt{49 - 36}}{6} \\[1em] = \dfrac{-7 \pm \sqrt{13}}{6}.$

Hence, x = 3, $\dfrac{1}{3}, \dfrac{-7 \pm \sqrt{13}}{6}.$

Question 7

Solve :

$(x^2 + \dfrac{1}{x^2}) - 3(x - \dfrac{1}{x}) - 2 = 0$

Answer

Let $x - \dfrac{1}{x} = a$ ........(i)

Squaring, both sides we get,

$\Rightarrow x^2 + \dfrac{1}{x^2} - 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 + 2 .......(ii)$

Substituting the values from equations (i) and (ii) we get,

$(x^2 + \dfrac{1}{x^2}) - 3(x - \dfrac{1}{x}) - 2 = 0$

⇒ a² + 2 - 3a - 2 = 0

⇒ a² - 3a = 0

⇒ a(a - 3) = 0

⇒ a = 0 or a - 3 = 0

⇒ a = 0 or a = 3.

Considering a = 0 we get,

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

Considering a = 3 we get,

$\Rightarrow x - \dfrac{1}{x} = 3 \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 3 \\[1em] \Rightarrow x^2 - 1 = 3x \\[1em] \Rightarrow x^2 - 3x - 1 = 0$

Comparing x² - 3x - 1 = 0 with ax² + bx + c = 0 we get,

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

We know that,

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

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 4}}{2} \\[1em] = \dfrac{3 \pm \sqrt{13}}{2}.$

Hence, x = 1, -1, $\dfrac{3 \pm \sqrt{13}}{2}$ .

Question 8

Solve :

(x² + 5x + 4)(x² + 5x + 6) = 120

Answer

Let x² + 5x = a we get,

⇒ (a + 4)(a + 6) = 120

⇒ a² + 6a + 4a + 24 = 120

⇒ a² + 10a + 24 - 120 = 0

⇒ a² + 10a - 96 = 0

⇒ a² + 16a - 6a - 96 = 0

⇒ a(a + 16) - 6(a + 16) = 0

⇒ (a - 6)(a + 16) = 0

⇒ (a - 6) = 0 or (a + 16) = 0

⇒ a = 6 or a = -16.

Considering a = 6 we get,

⇒ x² + 5x = 6

⇒ x² + 5x - 6 = 0

⇒ x² + 6x - x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x - 1)(x + 6) = 0

⇒ (x - 1) = 0 or (x + 6) = 0

⇒ x = 1 or x = -6.

Considering a = -16 we get,

⇒ x² + 5x = -16

⇒ x² + 5x + 16 = 0

In this case,

D = b² - 4ac = 5² - 4(1)(16) = 25 - 64 = -39 < 0.

It means roots are imaginary in this case.

Hence, x = 1, -6.

Question 9

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

Answer

Solving,

$\Rightarrow 3\Big(\dfrac{3x - 1}{2x + 3}\Big) - 2\Big(\dfrac{2x + 3}{3x - 1}\Big) = 5 \\[1em] \Rightarrow \dfrac{3(3x - 1)^2 - 2(2x + 3)^2}{(2x + 3)(3x - 1)} = 5 \\[1em] \Rightarrow 3(3x - 1)^2 - 2(2x + 3)^2 = 5(2x + 3)(3x - 1) \\[1em] \Rightarrow 3(9x^2 - 6x + 1) - 2(4x^2 + 12x + 9) = 5(6x^2 - 2x + 9x - 3) \\[1em] \Rightarrow 27x^2 - 18x + 3 - 8x^2 - 24x - 18 = 5(6x^2 + 7x - 3) \\[1em] \Rightarrow 19x^2 - 42x - 15 = 30x^2 + 35x - 15 \\[1em] \Rightarrow 30x^2 - 19x^2 + 35x + 42x - 15 + 15 = 0 \\[1em] \Rightarrow 11x^2 + 77x = 0 \\[1em] \Rightarrow 11x(x + 7) = 0 \\[1em] \Rightarrow 11x = 0 \text{ or } x + 7 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -7.$

Hence, x = 0 \text{ or } x = -7.

Question 10

Solve: 5^x+1 + 5^2-x = 5³ + 1

Answer

Given,

5^x+1 + 5^2-x = 5³ + 1

⇒ 5^x.5¹ + 5².5^-x = 125 + 1

⇒ 5^x.5¹ + $\dfrac{5^2}{5^x}$ = 126

Let 5^x be y.

⇒ 5y + $\dfrac{25}{\text{y}}$ = 126

⇒ 5y² + 25 = 126y

⇒ 5y² - 126y + 25 = 0

⇒ 5y² - 125y - y + 25 = 0

⇒ 5y(y - 25) - (y - 25) = 0

⇒ (y - 25)(5y - 1) = 0

⇒ (y - 25) = 0 or (5y - 1) = 0

⇒ y = 25 or y = $\dfrac{1}{5}$

⇒ y = 5² or y = 5^-1

Substituting the value of y,

⇒ 5^x = 5² or 5^x = 5^-1

⇒ x = 2 or -1

Hence, the value of x = 2 or -1.

Test Yourself

Question 1(a)

If (k + 2)x² - 2x + 1 = 0 has real roots then greater value of k(∈ Z) is :

1
3
-1
none of these

Answer

Given,

Equation : (k + 2)x² - 2x + 1 = 0

Comparing (k + 2)x² - 2x + 1 = 0 with ax² + bx + c = 0 we get,

a = (k + 2), b = -2 and c = 1.

Since, roots are real,

∴ D ≥ 0

⇒ b² - 4ac ≥ 0

⇒ (-2)² - 4(k + 2)(1) ≥ 0

⇒ 4 - 4(k + 2) ≥ 0

⇒ 4 - 4k - 8 ≥ 0

⇒ -4k - 4 ≥ 0

⇒ 4k ≤ -4

⇒ k ≤ -1.

Thus, greatest value of k = -1.

Hence, option 3 is the correct option.

Question 1(b)

Find the greatest value of k ∈ N for which the equation x² - 4x + k = 0 has distinct real roots.

-4
3
1
4

Answer

Given,

x² - 4x + k = 0

Comparing x² - 4x + k = 0 with ax² + bx + c = 0 we get,

a = 1, b = -4 and c = k.

Since roots are distinct and real,

∴ D > 0

⇒ b² - 4ac > 0

⇒ (-4)² - 4(1)(k) > 0

⇒ 16 - 4k > 0

⇒ 16 > 4k

⇒ $\dfrac{16}{4}$ > k

⇒ k < 4

Since k ∈ N,

Possible natural numbers less than 4 are: 1, 2, 3

The greatest value is: 3

Hence, Option 2 is the correct option.

Question 1(c)

If the quadratic equation kx² + kx + 1 = 0 has real and equal roots, the value of k is :

0
4
0 and 4
0 or 4

Answer

Comparing equation kx² + kx + 1 = 0, with ax² + bx + c = 0 , we get :

a = k, b = k and c = 1.

Since, quadratic equation has real and equal roots.

∴ D = 0

∴ b² - 4ac = 0

⇒ k² - 4 × k × 1 = 0

⇒ k² - 4k = 0

⇒ k(k - 4) = 0

⇒ k = 0 or (k - 4) = 0.

⇒ k = 0 or k = 4.

Hence, Option 4 is the correct option.

Question 1(d)

If x² - 4x = 5, the value of x is :

5
-1
5 or -1
5 and -1

Answer

Given,

⇒ x² - 4x = 5

⇒ x² - 4x - 5 = 0

⇒ x² - 5x + x - 5 = 0

⇒ x(x - 5) + 1(x - 5) = 0

⇒ (x + 1)(x - 5) = 0

⇒ (x + 1) = 0 or (x - 5) = 0

⇒ x = -1 or x = 5.

Hence, Option 3 is the correct option.

Question 1(e)

If x² - 7x = 0, the value of x is :

7
0
0 and 7
0 or 7

Answer

Given,

⇒ x² - 7x = 0

⇒ x(x - 7) = 0

⇒ x = 0 or x - 7 = 0

⇒ x = 0 or x = 7.

Hence, Option 4 is the correct option.

Question 1(f)

If x = 1 is a root of the equation $\sqrt{x} + kx - 2$ = 0; the value of k is :

1
-1
2
-2

Answer

Given,

x = 1 is a root of the equation $\sqrt{x} + kx - 2$ = 0.

$\Rightarrow \sqrt{1} + k(1) - 2 = 0 \\[1em] \Rightarrow 1 + k - 2 = 0 \\[1em] \Rightarrow k - 1 = 0 \\[1em] \Rightarrow k = 1.$

Hence, Option 1 is the correct option.

Question 1(g)

The equation $\sqrt{15 - 2x} = x$ .

Assertion (A): x = 3.

Reason (R): $\sqrt{15 - 2x} = x$

$\Rightarrow 15 - 2x = x^2 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x = -5 \text{ or } 3$

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

A is true, R is false.

Reason

Given,

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

The quadratic equation mentioned in the reason is $x^2 - 2x - 15 = 0$ whereas we see that the correct quadratic equation is $x^2 + 2x - 15 = 0$
∴ Reason (R) is false.

Our solution shows that one of the roots is 3.
∴ Assertion (A) is true.

Hence, option 1 is correct.

Question 1(h)

A quadratic equation 2x² + 5x - 3 = 0.

Assertion (A): The roots of equation 2x² + 5x - 3 = 0 are real and unequal.

Reason (R): For the equation ax² + bx + c = 0, the roots are real and unequal if b² - 4ac > 0.

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

Both A and R are true and R is correct reason for A.

Reason

Given, 2x² + 5x - 3 = 0

As we know that the roots of equation ax² + bx + c = 0 are real and unequal if b² - 4ac > 0.

⇒ b² - 4ac = 5² - 4 x 2 x (-3)

= 25 + 24 = 49 > 0

So, Assertion (A) is true.

And, Reason (R) is also true and it clearly explain assertion as a positive discriminant (b² - 4ac > 0) guarantees that the roots are real and unequal.

Hence, option 3 is correct.

Question 1(i)

One root of a quadratic equation is 3 + $\sqrt{2}$ .

Statement (1): The other root of the given quadratic equation is 3 - $\sqrt{2}$ .

Statement (2): If one root of the given quadratic equation is in the form of a surd, the other root is its conjugate.

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Both the statements are true.

Reason

We are given that one root of the quadratic equation is 3 + $\sqrt{2}$ . If the quadratic equation has real coefficients, the conjugate of a root that involves a surd (i.e., a square root or irrational number) must also be a root of the quadratic equation.

Therefore, 3 - $\sqrt{2}$ is other root of the given quadratic equation.

So, statement (1) is true.

The property of conjugate roots holds for quadratic equations with real coefficients, meaning if one root involves a surd, the other root will be its conjugate. In this case, since 3 + $\sqrt{2}$ is a surd, the other root must be 3 - $\sqrt{2}$ .

So, statement (2) is true.

Hence, option 1 is correct.

Question 2

If p - 15 = 0 and 2x² + px + 25 = 0; find the values of x.

Answer

Given,

⇒ p - 15 = 0

⇒ p = 15.

Substituting value of p in 2x² + px + 25 = 0 we get,

⇒ 2x² + 15x + 25 = 0

⇒ 2x² + 10x + 5x + 25 = 0

⇒ 2x(x + 5) + 5(x + 5) = 0

⇒ (2x + 5)(x + 5) = 0

⇒ 2x + 5 = 0 or x + 5 = 0

⇒ 2x = -5 or x = -5

⇒ x = - $\dfrac{5}{2}$ or x = -5.

Hence, x = -5 or $-\dfrac{5}{2}$ .

Question 3

Solve :

$\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{x + p + q}$

Answer

Given,

$\Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{x + p + q} \\[1em] \Big(\dfrac{1}{p} + \dfrac{1}{q}\Big) + \Big(\dfrac{1}{x} - \dfrac{1}{x + p + q}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{x + p + q - x}{x(x + p + q)}\Big) = 0 \\[1em] \Big(\dfrac{q + p}{pq}\Big) + \Big(\dfrac{p + q}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{1}{pq} + \dfrac{1}{x(x + p + q)}\Big) = 0 \\[1em] (p + q)\Big(\dfrac{x(x + p + q) + pq}{xpq(x + p + q)} \Big) = 0 \\[1em] (p + q)\Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] \therefore \Big(\dfrac{x^2 + x(p + q) + pq}{xpq(x + p + q)}\Big) = 0 \\[1em] x^2 + x(p + q) + pq = 0 \\[1em]$

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

a = 1, b = (p + q), c = pq

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

Substituting value in above equation we get,

$\Rightarrow x = \dfrac{-(p + q) \pm \sqrt{(p + q)^2 - 4.(1).pq}}{2(1)} \\[1em] = \dfrac{-(p + q) \pm \sqrt{p^2 + q^2 + 2pq - 4pq}}{2} \\[1em] = \dfrac{-(p + q) \pm \sqrt{(p - q)^2}}{2} \\[1em] = \dfrac{-(p + q) \pm (p - q)}{2} \\[1em] = \dfrac{-(p + q) + (p - q)}{2} \text{ or } \dfrac{-(p + q) - (p - q)}{2} \\[1em] = \dfrac{-p - q + p - q}{2} \text{ or } \dfrac{-p - q - p + q}{2} \\[1em] = \dfrac{-2q}{2} \text{ or } \dfrac{-2p}{2} \\[1em] = -q \text{ or } -p.$

Hence, x = -q or -p.

Question 4

Solve the quadratic equation 8x² - 14x + 3 = 0

(i) When x ∈ I (integers)

(ii) When x ∈ Q (rational numbers)

Answer

Given,

⇒ 8x² - 14x + 3 = 0

⇒ 8x² - 12x - 2x + 3 = 0

⇒ 4x(2x - 3) - 1(2x - 3) = 0

⇒ (4x - 1)(2x - 3) = 0

⇒ 4x - 1 = 0 or 2x - 3 = 0

⇒ 4x = 1 or 2x = 3

⇒ x = $\dfrac{1}{4}$ or x = $\dfrac{3}{2}$ .

(i) Since, there is no integer in the solution,

Hence, no solution.

(ii) Here, x ∈ Q

Hence, x = $\dfrac{1}{4}, \dfrac{3}{2}$ .

Question 5

Solve, using formula :

x² + x - (a + 2)(a + 1) = 0

Answer

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

a = 1, b = 1, c = -(a + 2)(a + 1)

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

Substituting values we get,

$\Rightarrow \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-(a + 2)(a + 1))}}{2(1)} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a + 2)(a + 1)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + a + 2a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 4(a^2 + 3a + 2)}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 8 + 1}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{-1 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{-1 \pm 2a + 3}{2} \\[1em] = \dfrac{-1 + (2a + 3)}{2} \text{ or } \dfrac{-1 - (2a + 3)}{2} \\[1em] = \dfrac{2a + 2}{2} \text{ or } \dfrac{-2a - 4}{2} \\[1em] = (a + 1) \text{ or } -(a + 2).$

Hence, x = (a + 1) or -(a + 2).

Question 6

If m and n are roots of the equation :

$\dfrac{1}{x} - \dfrac{1}{x - 2} = 3$ ; where x ≠ 0 and x ≠ 2; find m × n.

Answer

Solving,

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

Comparing 3x² - 6x + 2 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -6 and c = 2.

We know that,

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

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} \\[1em] x = \dfrac{6 \pm \sqrt{36 - 24}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{12}}{6} \\[1em] x = \dfrac{6 \pm \sqrt{4 \times 3}}{6} \\[1em] x = \dfrac{6 \pm 2\sqrt{3}}{6} \\[1em] x = \dfrac{2(3 \pm \sqrt{3})}{6} \\[1em] x = \dfrac{3 + \sqrt{3}}{3}, \dfrac{3 - \sqrt{3}}{3} \\[1em] \therefore m = \dfrac{3 + \sqrt{3}}{3} \text{ and } n = \dfrac{3 - \sqrt{3}}{3}.$

Solving m × n,

$m × n = \Big(\dfrac{3 + \sqrt{3}}{3}\Big)\Big(\dfrac{3 - \sqrt{3}}{3}\Big) \\[1em] = \dfrac{(3)^2 - (\sqrt{3})^2}{9} \\[1em] = \dfrac{9 - 3}{9} \\[1em] = \dfrac{6}{9} \\[1em] = \dfrac{2}{3}.$

Hence, m × n = $\dfrac{2}{3}$ .

Question 7

One root of the quadratic equation 8x² + mx + 15 = 0 is $\dfrac{3}{4}$ . Find the value of m. Also, find other root of equation.

Answer

Given, $\dfrac{3}{4}$ is root of 8x² + mx + 15 = 0

$\therefore 8\Big(\dfrac{3}{4}\Big)^2 + m \times \dfrac{3}{4} + 15 = 0 \\[1em] \Rightarrow 8 \times \dfrac{9}{16} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{9}{2} + \dfrac{3m}{4} + 15 = 0 \\[1em] \Rightarrow \dfrac{3m}{4} + \dfrac{30 + 9}{2} = 0 \\[1em] \Rightarrow \dfrac{3m}{4} = -\dfrac{39}{2} \\[1em] \Rightarrow m = -\dfrac{39}{2} \times \dfrac{4}{3} \\[1em] \Rightarrow m = -26.$

Substituting value of m in equation,

⇒ 8x² + mx + 15 = 0

⇒ 8x² - 26x + 15 = 0

⇒ 8x² - 20x - 6x + 15 = 0

⇒ 4x(2x - 5) - 3(2x - 5) = 0

⇒ (4x - 3)(2x - 5) = 0

⇒ 4x - 3 = 0 or 2x - 5 = 0

⇒ 4x = 3 or 2x = 5

⇒ x = $\dfrac{3}{4}$ or x = $\dfrac{5}{2}$ .

Hence, m = -26 and other root = $\dfrac{5}{2}$ .

Question 8

Show that one root of the quadratic equation x² + (3 - 2a)x - 6a = 0 is -3. Hence, find its other root.

Answer

Substituting x = -3 in x² + (3 - 2a)x - 6a = 0,

⇒ (-3)² + (3 - 2a)(-3) - 6a = 0

⇒ 9 - 9 + 6a - 6a = 0

⇒ 0 = 0.

Hence, -3 is one root of the quadratic equation.

We know that,

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

$= \dfrac{-(3 - 2a) \pm \sqrt{(3 - 2a)^2 - 4(1)(-6a)}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{9 + 4a^2 - 12a + 24a}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{4a^2 + 12a + 9}}{2} \\[1em] = \dfrac{2a - 3 \pm \sqrt{(2a + 3)^2}}{2} \\[1em] = \dfrac{(2a - 3) + (2a + 3)}{2} \text{ or } \dfrac{(2a - 3) - (2a + 3)}{2} \\[1em] = \dfrac{4a}{2} \text{ or } \dfrac{-6}{2} \\[1em] = 2a \text{ or } -3.$

Hence, the other root is 2a.

Question 9

Find the solution of the quadratic equation 2x² - mx - 25n = 0; if m + 5 = 0 and n - 1 = 0.

Answer

Given, m + 5 = 0 and n - 1 = 0

∴ m = -5 and n = 1.

Substituting values of m and n in 2x² - mx - 25n = 0 we get,

⇒ 2x² - (-5)x - 25(1) = 0

⇒ 2x² + 5x - 25 = 0

⇒ 2x² + 10x - 5x - 25 = 0

⇒ 2x(x + 5) - 5(x + 5) = 0

⇒ (2x - 5)(x + 5) = 0

⇒ (2x - 5) = 0 or x + 5 = 0

⇒ x = $\dfrac{5}{2}$ or x = -5.

Hence, x = $\dfrac{5}{2}$ or -5.

Question 10

Solve : (a + b)²x² - (a + b)x - 6 = 0; a + b ≠ 0.

Answer

Let (a + b)x = y

⇒ (a + b)²x² - (a + b)x - 6 = 0

⇒ y² - y - 6 = 0

⇒ y² - 3y + 2y - 6 = 0

⇒ y(y - 3) + 2(y - 3) = 0

⇒ (y + 2)(y - 3) = 0

⇒ (y + 2) = 0 or y - 3 = 0

⇒ y = -2 or y = 3.

∴ (a + b)x = -2 or (a + b)x = 3

⇒ x = $-\dfrac{2}{a + b} \text{ or } \dfrac{3}{a + b}$ .

Hence, x = $-\dfrac{2}{a + b} \text{ or } \dfrac{3}{a + b}.$

Question 11

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.

x² + 2(m - 1)x + (m + 5) = 0

Answer

Since, equation has equal roots, D = 0.

∴ b² - 4ac = 0

⇒ (2(m - 1))² - 4(1)(m + 5) = 0

⇒ (2m - 2)² - (4m + 20) = 0

⇒ 4m² + 4 - 8m - 4m - 20 = 0

⇒ 4m² - 12m - 16 = 0

⇒ 4(m² - 3m - 4) = 0

⇒ m² - 3m - 4 = 0

⇒ m² - 4m + m - 4 = 0

⇒ m(m - 4) + 1(m - 4) = 0

⇒ (m + 1)(m - 4) = 0

⇒ m + 1 = 0 or m - 4 = 0

⇒ m = -1 or m = 4.

Hence, m = -1, 4.

Question 12

Find the value of k for which equation 4x² + 8x - k = 0 has real roots.

Answer

Since equations has real roots, D ≥ 0

∴ b² - 4ac ≥ 0

⇒ 8² - 4(4)(-k) ≥ 0

⇒ 64 + 16k ≥ 0

⇒ 16k ≥ -64

Dividing both sides by 16 we get,

⇒ k ≥ -4

Hence, k ≥ -4.

Question 13

If -2 is a root of the equation 3x² + 7x + p = 1, find the value of p. Now find the value of k so that the roots of the equation x² + k(4x + k - 1) + p = 0 are equal.

Answer

Since, -2 is a root of the equation 3x² + 7x + p = 1,

∴ 3(-2)² + 7(-2) + p = 1

⇒ 3(4) - 14 + p = 1

⇒ 12 - 14 + p = 1

⇒ -2 + p = 1

⇒ p = 3.

Substituting value of p in x² + k(4x + k - 1) + p = 0 we get,

⇒ x² + k(4x + k - 1) + 3 = 0

⇒ x² + 4kx + k² - k + 3 = 0

Since, roots are equal, D = 0.

∴ b² - 4ac = 0

⇒ (4k)² - 4(1)(k² - k + 3) = 0

⇒ 16k² - 4k² + 4k - 12 = 0

⇒ 12k² + 4k - 12 = 0

⇒ 4(3k² + k - 3) = 0

⇒ 3k² + k - 3 = 0

$\Rightarrow k = \dfrac{-1 \pm \sqrt{(1)^2 - 4(3)(-3)}}{2(3)} \\[1em] = \dfrac{-1 \pm \sqrt{1 + 36}}{6} \\[1em] = \dfrac{-1 \pm \sqrt{37}}{6}.$

Hence, p = 3 and k = $\dfrac{-1 \pm \sqrt{37}}{6}$ .

Case Study Based Question

Question 1

Case study:
An Air India flight from Mumbai to Colombo was delayed by 60 minutes due to emergency landing of the plane in Vizag as a passengers in the flight suddenly had a cardiac arrest. Now to reach destination of 1800 km from Vizag to Colombo in time, so that passengers could catch their connecting flights, the speed of the plane was increased by 300 km/h than the usual speed.

An Air India flight from Mumbai to Colombo was delayed by 60 minutes due to emergency landing of the plane in Vizag as a passengers in the flight suddenly had a cardiac arrest. Now to reach destination of 1800 km from Vizag to Colombo in time, so that passengers could catch their connecting flights, the speed of the plane was increased by 300 km/h than the usual speed. Quadratic equations, Concise Mathematics Solutions ICSE Class 10.

Based on the above information, answer the following questions :

(i) Taking the usual speed of plane as x km/h, form quadratic equation for situation described above.

(ii) Find the nature of the roots of the quadratic equation.

(iii) Find the usual speed of the plane.

Answer

(i) Let the usual speed of the plane be x km/hr.

Distance from Vizag to Colombo = 1800 km

Time taken at usual speed = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{1800}{x}$ hr

Increased speed = x + 300

Time taken at increased speed = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{1800}{x + 300}$ hr

Since flight delayed by 60 minutes = 1 hour,

$\dfrac{1800}{x} - \dfrac{1800}{x + 300} = 1$

Solving,

$\Rightarrow \dfrac{1800(x + 300) - 1800x}{x(x + 300)} = 1 \\[1em] \Rightarrow \dfrac{1800x + 540000 - 1800x}{x(x + 300)} = 1 \\[1em] \Rightarrow 1800x + 540000 - 1800x = x(x + 300) \\[1em] \Rightarrow 540000 = x^2 + 300x \\[1em] \Rightarrow x^2 + 300x - 540000 = 0$

Hence, the required equation is x² + 300x - 540000 = 0.

(ii) Comparing x² + 300x - 540000 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 300 and c = -540000.

We know that,

D = b² - 4ac

= (300)² - 4(1)(-540000)

= 90000 + 2160000

= 2250000

∴ D > 0

Hence, roots are real and unequal.

(iii) Comparing x² + 300x - 540000 = 0 with ax² + bx + c = 0 we get,

a = 1, b = 300 and c = -540000.

We know that,

x = $\dfrac{-b \pm \sqrt{D}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(300) \pm \sqrt{2250000}}{2(1)} \\[1em] = \dfrac{-300 \pm 1500}{2} \\[1em] = \dfrac{-300 + 1500}{2} \text{ or } \dfrac{-300 - 1500}{2} \\[1em] = \dfrac{1200}{2} \text{ or } \dfrac{-1800}{2} \\[1em] = 600 \text{ or } -900.$

Speed cannot be negative,

∴ x = 600 km/h

Hence, usual speed of the plane = 600 km/h.

Chapter 5

Quadratic Equations

Class - 10 Concise Mathematics Selina

Exercise 5(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Exercise 5(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11(i)

Question 11(ii)

Question 12

Question 13

Question 14

Question 15(i)

Question 15(ii)

Question 15(iii)

Question 15(iv)

Question 15(v)

Question 15(vi)

Exercise 5(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2(i)

Question 2(ii)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 5

Question 6

Question 7

Exercise 5(D)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 1(f)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 3

Question 4

Question 5

Question 6

Question 7(i)

Question 7(ii)

Question 7(iii)

Question 8

Exercise 5(E)

Question 1(a)

Question 1(b)

Question 1(c)

Question 2

Question 3

Question 4(i)