Quadratic Equations — Exercise 5.2

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.

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.

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, $-\dfrac{4}{3}$.

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 $-\dfrac{2}{7}$, $\dfrac{2}{3}$.

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, $\dfrac{4}{3}$.

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 $-\dfrac{7}{3}$, $\dfrac{5}{2}$.

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 $-\dfrac{5}{2}$, $\dfrac{2}{3}$.

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, $\dfrac{3}{2}$.

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.

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, $\dfrac{19}{3}$.

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.

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

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 $-\dfrac{1}{a} , -\dfrac{1}{a}$

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.

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 $-\dfrac{b^2}{a^2}, -1.$

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 $-\sqrt{3}, -\dfrac{7\sqrt{3}}{3}$

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 $-\dfrac{2\sqrt{3}}{3}$, $\dfrac{\sqrt{3}}{4}.$

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.

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 $\dfrac{4}{5}$ , $\dfrac{5}{4}$.

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.

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.

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.

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.

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.

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.

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.

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.

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 $\dfrac{a + b}{ab} , \dfrac{2}{(a + b)}$.

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.

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.

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.

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.

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.

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.

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.

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 , $\dfrac{1}{2}$.

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.