Quadratic Equations — Exercise 5(A)

Find which of the following are the solutions of equation 6x² - x - 2 = 0 ?

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

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

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

Answer

Given,

⇒ 6x² - x - 2 = 0

⇒ 6x² - 4x + 3x - 2 = 0

⇒ 2x(3x - 2) + 1(3x - 2) = 0

⇒ (2x + 1)(3x - 2) = 0

⇒ 2x + 1 = 0 or 3x - 2 = 0      [Using Zero-product rule]

⇒ 2x = -1 or 3x = 2

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

Hence, $-\dfrac{1}{2}, \dfrac{2}{3}$ are the solutions of equation 6x² - x - 2 = 0.

Determine whether x = $-\dfrac{1}{3}$ and x = $\dfrac{2}{3}$ are the solutions of the equation 9x² - 3x - 2 = 0.

Answer

Given,

⇒ 9x² - 3x - 2 = 0

⇒ 9x² - 6x + 3x - 2 = 0

⇒ 3x(3x - 2) + 1(3x - 2) = 0

⇒ (3x + 1)(3x - 2) = 0

⇒ 3x + 1 = 0 or 3x - 2 = 0      [Using Zero-product rule]

⇒ 3x = -1 or 3x = 2

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

Hence, $\dfrac{-1}{3}, \dfrac{2}{3}$ are the solutions of the equation 9x² - 3x - 2 = 0.

Solve the following equation by factorization:

16x² = 25

Answer

Given,

⇒ 16x² = 25

⇒ 16x² - 25 = 0

⇒ (4x)² - (5)² = 0

⇒ (4x + 5)(4x - 5) = 0

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

⇒ 4x = -5 or 4x = 5

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

Hence, x = $\Big\lbrace\dfrac{5}{4}, \dfrac{-5}{4}\Big\rbrace$.

Solve the Following equation by factorization:

x² + 2x = 24

Answer

Given,

⇒ x² + 2x = 24

⇒ x² + 2x - 24 = 0

⇒ x² + 6x - 4x - 24 = 0

⇒ x(x + 6) - 4(x + 6) = 0

⇒ (x - 4)(x + 6) = 0

⇒ x - 4 = 0 or x + 6 = 0      [Using Zero-product rule]

⇒ x = 4 or x = -6.

Hence, x = {-6, 4}.

Solve the following equation by factorization:

x² - x = 156

Answer

Given,

⇒ x² - x = 156

⇒ x² - x - 156 = 0

⇒ x² - 13x + 12x - 156 = 0

⇒ x(x - 13) + 12(x - 13) = 0

⇒ (x + 12)(x - 13) = 0

⇒ x + 12 = 0 or x - 13 = 0      [Using Zero-product rule]

⇒ x = -12 or x = 13.

Hence, x = {13, -12}.

Solve the following equation by factorization:

x² - 11x = 42

Answer

Given,

⇒ x² - 11x = 42

⇒ x² - 11x - 42 = 0

⇒ x² - 14x + 3x - 42 = 0

⇒ x(x - 14) + 3(x - 14) = 0

⇒ (x + 3)(x - 14) = 0

⇒ x + 3 = 0 or x - 14 = 0      [Using Zero-product rule]

⇒ x = -3 or x = 14.

Hence, x = {14, -3}.

Solve the following equation by factorization:

x² - 7x + 10 = 0

Answer

Given,

⇒ x² - 7x + 10 = 0

⇒ x² - 5x - 2x + 10 = 0

⇒ x(x - 5) - 2(x - 5) = 0

⇒ (x - 2)(x - 5) = 0

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

⇒ x = 2 or x = 5

Hence, x = {2, 5}.

Solve the following equation by factorization:

x² + 18x = 40

Answer

Given,

⇒ x² + 18x = 40

⇒ x² + 18x - 40 = 0

⇒ x² + 20x - 2x - 40 = 0

⇒ x(x + 20) - 2(x + 20) = 0

⇒ (x - 2)(x + 20) = 0

⇒ x - 2 = 0 or x + 20 = 0      [Using Zero-product rule]

⇒ x = 2 or x = -20.

Hence, x = {-20, 2}.

Solve the following equation by factorization:

x² + 17 = 18x

Answer

Given,

⇒ x² + 17 = 18x

⇒ x² - 18x + 17 = 0

⇒ x² - 17x - x + 17 = 0

⇒ x(x - 17) - 1(x - 17) = 0

⇒ (x - 1)(x - 17) = 0

⇒ x - 1 = 0 or x - 17 = 0      [Using Zero-product rule]

⇒ x = 1 or x = 17.

Hence, x = {1, 17}.

Solve the following equation by factorization:

3x² = 5x

Answer

Given,

⇒ 3x² = 5x

⇒ 3x² - 5x = 0

⇒ x(3x - 5) = 0

⇒ x = 0 or (3x - 5) = 0      [Using Zero-product rule]

⇒ x = 0 or 3x = 5

⇒ x = 0 or x = $\dfrac{5}{3}$.

Hence, x = $\Big\lbrace0, \dfrac{5}{3}\Big\rbrace$.

Solve the following equation by factorization:

(x + 3)(x - 3) = 27

Answer

Given,

⇒ (x + 3)(x - 3) = 27

⇒ (x)² - (3)² = 27

⇒ x² - 9 = 27

⇒ x² = 27 + 9

⇒ x² = 36

⇒ x = $\sqrt{36}$

⇒ x = ± 6

Hence, x = {6, -6}.

Solve the following equation by factorization:

x² - 30x + 216 = 0

Answer

Given,

⇒ x² - 30x + 216 = 0

⇒ x² - 18x - 12x + 216 = 0

⇒ x(x - 18) - 12(x - 18) = 0

⇒ (x - 18)(x - 12) = 0

⇒ (x - 18) = 0 or (x - 12) = 0      [Using Zero-product rule]

⇒ x = 18 or x = 12.

Hence, x = {18, 12}.

Solve the following equation by factorization:

12x² + 29x + 14 = 0

Answer

Given,

⇒ 12x² + 29x + 14 = 0

⇒ 12x² + 21x + 8x + 14 = 0

⇒ 3x(4x + 7) + 2(4x + 7) = 0

⇒ (4x + 7)(3x + 2) = 0

⇒ 4x + 7 = 0 or 3x + 2 = 0      [Using Zero-product rule]

⇒ 4x = -7 or 3x = -2

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

Hence, x = $\Big\lbrace\dfrac{-7}{4}, \dfrac{-2}{3}\Big\rbrace$.

Solve the following equation by factorization:

2x² - 7x = 39

Answer

Given,

⇒ 2x² - 7x = 39

⇒ 2x² - 7x - 39 = 0

⇒ 2x² + 6x - 13x - 39 = 0

⇒ 2x(x + 3) - 13(x + 3) = 0

⇒ (x + 3)(2x - 13) = 0

⇒ x + 3 = 0 or 2x - 13 = 0      [Using Zero-product rule]

⇒ x = -3 or 2x = 13

⇒ x = -3 or x = $\dfrac{13}{2}$.

Hence, x = $\Big\lbrace\dfrac{13}{2}, -3\Big\rbrace$.

Solve the following equation by factorization:

10x² = 9x + 7

Answer

Given,

⇒ 10x² = 9x + 7

⇒ 10x² - 9x - 7 = 0

⇒ 10x² + 5x - 14x - 7 = 0

⇒ 5x(2x + 1) - 7(2x + 1) = 0

⇒ (2x + 1)(5x - 7) = 0

⇒ (2x + 1) = 0 or (5x - 7) = 0      [Using Zero-product rule]

⇒ 2x = -1 or 5x = 7

⇒ x = $-\dfrac{1}{2}$ or x = $\dfrac{7}{5}$.

Hence, x = $\Big\lbrace-\dfrac{1}{2}, \dfrac{7}{5}\Big\rbrace$.

Solve the following equation by factorization:

15x² - 28 = x

Answer

Given,

⇒ 15x² - 28 = x

⇒ 15x² - x - 28 = 0

⇒ 15x² - 21x + 20x - 28 = 0

⇒ 3x(5x - 7) + 4(5x - 7) = 0

⇒ (5x - 7)(3x + 4) = 0

⇒ (5x - 7) = 0 or (3x + 4) = 0      [Using Zero-product rule]

⇒ 5x = 7 or 3x = - 4

⇒ x = $\dfrac{7}{5}$ or x = $\dfrac{-4}{3}$

Hence, x = $\Big\lbrace\dfrac{-4}{3}, \dfrac{7}{5}\Big\rbrace$.

Solve the following equation by factorization:

8x² + 15 = 26x

Answer

Given,

⇒ 8x² + 15 = 26x

⇒ 8x² - 26x + 15 = 0

⇒ 8x² - 20x - 6x + 15 = 0

⇒ 4x(2x - 5) - 3(2x - 5) = 0

⇒ (2x - 5)(4x - 3) = 0

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

⇒ 2x = 5 or 4x = 3

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

Hence, x = $\Big\lbrace\dfrac{5}{2}, \dfrac{3}{4}\Big\rbrace$.

Solve the following equation by factorization:

3x² + 8 = 10x

Answer

Given,

⇒ 3x² + 8 = 10x

⇒ 3x² - 10x + 8 = 0

⇒ 3x² - 6x - 4x + 8 = 0

⇒ 3x(x - 2) - 4(x - 2) = 0

⇒ (x - 2)(3x - 4) = 0

⇒ (x - 2) = 0 or (3x - 4) = 0      [Using Zero-product rule]

⇒ x = 2 or 3x = 4

⇒ x = 2 or x = $\dfrac{4}{3}$.

Hence, x = $\Big\lbrace2, \dfrac{4}{3}\Big\rbrace$.

Solve the following equation by factorization:

x(6x - 11) = 35

Answer

Given,

⇒ x(6x - 11) = 35

⇒ 6x² - 11x - 35 = 0

⇒ 6x² + 10x - 21x - 35 = 0

⇒ 2x(3x + 5) - 7(3x + 5) = 0

⇒ (3x + 5)(2x - 7) = 0

⇒ (3x + 5) = 0 or (2x - 7) = 0      [Using Zero-product rule]

⇒ 3x = -5 or 2x = 7

⇒ x = $\dfrac{-5}{3}$ or x = $\dfrac{7}{2}$.

Hence, x = $\Big\lbrace\dfrac{-5}{3}, \dfrac{7}{2}\Big\rbrace$.

Solve the following equation by factorization:

6x(3x - 7) = 7(7 - 3x)

Answer

Given,

⇒ 6x(3x - 7) = 7(7 - 3x)

⇒ 18x² - 42x = 49 - 21x

⇒ 18x² - 42x + 21x - 49 = 0

⇒ 6x(3x - 7) + 7(3x - 7) = 0

⇒ (3x - 7)(6x + 7) = 0

⇒ (3x - 7) = 0 or (6x + 7) = 0      [Using Zero-product rule]

⇒ 3x = 7 or 6x = -7

⇒ x = $\dfrac{7}{3}$ or x = $\dfrac{-7}{6}$.

Hence, x = $\Big\lbrace\dfrac{7}{3}, \dfrac{-7}{6}\Big\rbrace$.

Solve the following equation by factorization:

2x² - 9x + 10 = 0, when (i) x ∈ N (ii) x ∈ Q.

Answer

Given,

⇒ 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, x = $\Big\lbrace2, \dfrac{5}{2}\Big\rbrace$.

Solve the following equation by factorization:

4x² - 9x - 100 = 0, when x ∈ Q

Answer

Given,

⇒ 4x² - 9x - 100 = 0

⇒ 4x² + 16x - 25x - 100 = 0

⇒ 4x(x + 4) - 25(x + 4) = 0

⇒ (4x - 25)(x + 4) = 0

⇒ 4x - 25 = 0 or x + 4 = 0      [Using Zero-product rule]

⇒ 4x = 25 or x = -4

⇒ x = $\dfrac{25}{4}$ or x = -4

Since, x ∈ Q

Hence, x = $\Big\lbrace-4, \dfrac{25}{4}\Big\rbrace$.

Solve the following equation by factorization:

3x² + 11x + 10 = 0, when x ∈ I

Answer

Given,

⇒ 3x² + 11x + 10 = 0

⇒ 3x² + 6x + 5x + 10 = 0

⇒ 3x(x + 2) + 5(x + 2) = 0

⇒ (3x + 5)(x + 2) = 0

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

⇒ 3x = -5 or x = -2

⇒ x = $\dfrac{-5}{3}$ or x = -2.

Since, x ∈ I

x = -2

Hence, x = {-2}.

Solve the following equation by factorization:

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

Answer

Given,

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

⇒ x - 3 = 0 or 3x - 1 = 0      [Using Zero-product rule]

⇒ x = 3 or 3x = 1

⇒ x = 3 or x = $\dfrac{1}{3}$

Hence, x = $\Big\lbrace3, \dfrac{1}{3}\Big\rbrace$.

Solve the following equation by factorization:

5x - $\dfrac{35}{x}$ = 18

Answer

Given,

$\Rightarrow 5x - \dfrac{35}{x} = 18 \\[1em] \Rightarrow \dfrac{5x^2 - 35}{x} = 18 \\[1em] \Rightarrow 5x^2 - 35 = 18x \\[1em] \Rightarrow 5x^2 - 18x - 35 = 0 \\[1em] \Rightarrow 5x^2 - 25x + 7x - 35 = 0 \\[1em] \Rightarrow 5x(x - 5) + 7(x - 5) = 0 \\[1em] \Rightarrow (5x + 7)(x - 5) = 0.$

⇒ 5x + 7 = 0 or (x - 5) = 0      [Using Zero-product rule]

⇒ 5x = -7 or x = 5

⇒ x = $\dfrac{-7}{5}$ or x = 5.

Hence, x = $\Big\lbrace5, \dfrac{-7}{5}\Big\rbrace$.

Solve the following equation by factorization:

10x - $\dfrac{1}{x}$ = 3

Answer

Given,

$\Rightarrow 10x - \dfrac{1}{x} = 3 \\[1em] \Rightarrow \dfrac{10x^2 - 1}{x} = 3 \\[1em] \Rightarrow 10x^2 - 1 = 3x \\[1em] \Rightarrow 10x^2 - 3x - 1 = 0 \\[1em] \Rightarrow 10x^2 - 5x + 2x - 1 = 0 \\[1em] \Rightarrow 5x(2x - 1) + 1(2x - 1) = 0 \\[1em] \Rightarrow (5x + 1)(2x - 1) = 0.$

⇒ (5x + 1) = 0 or (2x - 1) = 0      [Using Zero-product rule]

⇒ 5x = -1 or 2x = 1

⇒ x = $-\dfrac{1}{5}$ or x = $\dfrac{1}{2}$.

Hence, x = $\Big\lbrace\dfrac{1}{2}, -\dfrac{1}{5}\Big\rbrace$.

Solve the following equation by factorization:

3a²x² + 8abx + 4b² = 0, a ≠ 0

Answer

Given,

⇒ 3a²x² + 8abx + 4b² = 0

⇒ 3a²x² + 6abx + 2abx + 4b² = 0

⇒ 3ax(ax + 2b) + 2b(ax + 2b) = 0

⇒ (3ax + 2b)(ax + 2b) = 0

⇒ (3ax + 2b) = 0 or (ax + 2b) = 0      [Using Zero-product rule]

⇒ 3ax = -2b or ax = -2b

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

Hence, x = $\Big\lbrace-\dfrac{2b}{a}, -\dfrac{2b}{3a}\Big\rbrace$.

Solve the following equation by factorization:

4x² - 4ax + (a² - b²) = 0, where a, b ∈ R.

Answer

Given,

⇒ 4x² - 4ax + (a² - b²) = 0

⇒ (4x² - 4ax + a²) - b² = 0

⇒ [(2x)² - 2 × a × 2x + (a)²] - b² = 0

⇒ (2x - a)² - b² = 0

⇒ (2x - a + b)(2x - a - b) = 0

⇒ (2x - a + b) = 0 or (2x - a - b) = 0      [Using Zero-product rule]

⇒ 2x = a - b or 2x = a + b

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

Hence, x = $\Big\lbrace \dfrac{a + b}{2}, \dfrac{a - b}{2}\Big\rbrace$.

Solve the following equation by factorization:

5x² - 12x - 9 = 0, when (i) x ∈ I (ii) x ∈ Q

Answer

Given,

⇒ 5x² - 12x - 9 = 0

⇒ 5x² - 15x + 3x - 9 = 0

⇒ 5x(x - 3) + 3(x - 3) = 0

⇒ (5x + 3)(x - 3) = 0

⇒ (5x + 3) = 0 or (x - 3) = 0      [Using Zero-product rule]

⇒ 5x = -3 or x = 3

⇒ x = $\dfrac{-3}{5}$ or x = 3.

(i) Since, x ∈ I

Hence, x = {3}.

(ii) Since, x ∈ Q

Hence, x = $\Big\lbrace3, \dfrac{-3}{5}\Big\rbrace$.

Solve the following equation by factorization:

2x² - 11x + 15 = 0, when (i) x ∈ N (ii) x ∈ I

Answer

Given,

⇒ 2x² - 11x + 15 = 0

⇒ 2x² - 6x - 5x + 15 = 0

⇒ 2x(x - 3) - 5(x - 3) = 0

⇒ (2x - 5)(x - 3) = 0

⇒ (2x - 5) = 0 or (x - 3) = 0      [Using Zero-product rule]

⇒ 2x = 5 or x = 3

⇒ x = $\dfrac{5}{2}$ or x = 3.

(i) Since, x ∈ N

Hence, x = {3}.

(ii) Since, x ∈ I

Hence, x = {3}.

Solve the following equation by factorization:

$\sqrt{3}x^2 + 11x + 6\sqrt{3}$ = 0

Answer

Given,

$\Rightarrow \sqrt{3}x^2 + 11x + 6\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 9x + 2x + 6\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x(x + 3\sqrt{3}) + 2(x + 3\sqrt{3}) = 0 \\[1em] \Rightarrow (\sqrt{3}x + 2)(x + 3\sqrt{3}) = 0 \\[1em] \Rightarrow (\sqrt{3}x + 2) = 0 \text{ or } (x + 3\sqrt{3}) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow \sqrt{3}x = -2 \text{ or } x = -3\sqrt{3} \\[1em] \Rightarrow x = \dfrac{-2}{\sqrt{3}} \text{ or } x = -3\sqrt{3}.$

Hence, $x = \Big\lbrace\dfrac{-2}{\sqrt{3}}, -3\sqrt{3}\Big\rbrace$.

Solve the following equation by factorization:

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

Answer

Given,

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

Hence, $x = \Big\lbrace\dfrac{\sqrt{5}}{2}, \dfrac{-1}{\sqrt{5}}\Big\rbrace$.

Solve the following equation by factorization:

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

Answer

Given,

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

Hence, $x = {1, \sqrt{2}}$.

Solve the following equation by factorization:

$\dfrac{x + 1}{x - 1} = \dfrac{3x - 7}{2x - 5}$

Answer

Given,

$\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{3x - 7}{2x - 5}$

⇒ (x + 1)(2x - 5) = (3x - 7)(x - 1)

⇒ (2x² - 5x + 2x - 5) = (3x² - 3x - 7x + 7)

⇒ (2x² - 3x - 5) = (3x² - 10x + 7)

⇒ (3x² - 10x + 7) - (2x² - 3x - 5) = 0

⇒ 3x² - 10x + 7 - 2x² + 3x + 5 = 0

⇒ x² - 7x + 12 = 0

⇒ x² - 3x - 4x + 12 = 0

⇒ x(x - 3) - 4(x - 3) = 0

⇒ (x - 4)(x - 3) = 0

⇒ (x - 4) = 0 or (x - 3) = 0     [Using Zero-product rule]

⇒ x = 4 or x = 3.

Hence, x = {3, 4}.

Solve the following equation by factorization:

$\dfrac{3x + 1}{7x + 1} = \dfrac{5x + 1}{7x + 5}$

Answer

Given,

$\Rightarrow \dfrac{3x + 1}{7x + 1} = \dfrac{5x + 1}{7x + 5}$

⇒ (3x + 1)(7x + 5) = (5x + 1)(7x + 1)

⇒ (21x² + 15x + 7x + 5) = (35x² + 5x + 7x + 1)

⇒ (21x² + 22x + 5) = (35x² + 12x + 1)

⇒ (35x² + 12x + 1) - (21x² + 22x + 5) = 0

⇒ 35x² + 12x + 1 - 21x² - 22x - 5 = 0

⇒ 14x² - 10x - 4 = 0

⇒ 2(7x² - 5x - 2) = 0

⇒ 7x² - 5x - 2 = 0

⇒ 7x² - 7x + 2x - 2 = 0

⇒ 7x(x - 1) + 2(x - 1) = 0

⇒ (7x + 2)(x - 1) = 0

⇒ (7x + 2) = 0 or (x - 1) = 0     [Using Zero-product rule]

⇒ 7x = -2 or x = 1

⇒ x = $\dfrac{-2}{7}$ or x = 1.

Hence, $x = \Big\lbrace1, \dfrac{-2}{7}\Big\rbrace$.

Solve the following equation by factorization:

$\dfrac{5}{(2x + 1)} + \dfrac{6}{(x + 1)} = 3$

Answer

Given,

$\Rightarrow \dfrac{5}{(2x + 1)} + \dfrac{6}{(x + 1)} = 3 \\[1em] \Rightarrow \dfrac{5(x + 1) + 6(2x + 1)}{(2x + 1)(x + 1)} = 3 \\[1em] \Rightarrow \dfrac{5x + 5 + 12x + 6}{(2x + 1)(x + 1)} = 3 \\[1em] \Rightarrow \dfrac{17x + 11}{(2x^2 + 2x + x + 1)} = 3 \\[1em] \Rightarrow \dfrac{17x + 11}{(2x^2 + 3x + 1)} = 3 \\[1em] \Rightarrow 17x + 11 = 3(2x^2 + 3x + 1) \\[1em] \Rightarrow 17x + 11 = 6x^2 + 9x + 3 \\[1em] \Rightarrow 6x^2 + 9x - 17x + 3 - 11 = 0 \\[1em] \Rightarrow 6x^2 - 8x - 8 = 0 \\[1em] \Rightarrow 2(3x^2 - 4x - 4) = 0 \\[1em] \Rightarrow 3x^2 - 4x - 4 = 0 \\[1em] \Rightarrow 3x^2 - 6x + 2x - 4 = 0 \\[1em] \Rightarrow 3x(x - 2) + 2(x - 2) = 0 \\[1em] \Rightarrow (3x + 2)(x - 2) = 0 \\[1em] \Rightarrow (3x + 2) \text{ or } (x - 2) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 3x = -2 \text{ or } x = 2 \\[1em] \Rightarrow x = \dfrac{-2}{3} \text{ or } x = 2.$

Hence, $x = \Big\lbrace2, \dfrac{-2}{3}\Big\rbrace$.

Solve the following equation by factorization:

$\dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = \dfrac{25}{3}$

Answer

Given,

$\Rightarrow \dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = \dfrac{25}{3} \\[1em] \Rightarrow \dfrac{2x(x - 3) + (2x - 5)(x - 4)}{(x - 4)(x - 3)} = \dfrac{25}{3} \\[1em] \Rightarrow \dfrac{2x^2 - 6x + 2x^2 - 8x - 5x + 20}{x^2 - 3x - 4x + 12} = \dfrac{25}{3} \\[1em] \Rightarrow \dfrac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \dfrac{25}{3} \\[1em] \Rightarrow 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \\[1em] \Rightarrow 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \\[1em] \Rightarrow 25x^2 - 175x + 300 - (12x^2 - 57x + 60) = 0 \\[1em] \Rightarrow 25x^2 - 175x + 300 - 12x^2 + 57x - 60 = 0 \\[1em] \Rightarrow 13x^2 - 118x + 240 = 0 \\[1em] \Rightarrow 13x^2 - 78x - 40x + 240 = 0 \\[1em] \Rightarrow 13x(x - 6) - 40(x - 6) = 0 \\[1em] \Rightarrow (13x - 40)(x - 6) = 0 \\[1em] \Rightarrow (13x - 40) \text{ or } (x - 6) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 13x = 40 \text{ or } x = 6 \\[1em] \Rightarrow x = \dfrac{40}{13} \text{ or } x = 6.$

Hence, $x = \Big\lbrace6, \dfrac{40}{13}\Big\rbrace$.

Solve the following equation by factorization:

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

Answer

Given,

$\Rightarrow \dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = 4\dfrac{1}{4} \\[1em] \Rightarrow \dfrac{x(x + 3) - (1 - x)(x - 2)}{x(x - 2)} = \dfrac{17}{4} \\[1em] \Rightarrow \dfrac{x^2 + 3x - (x - 2 - x^2 + 2x)}{x^2 - 2x} = \dfrac{17}{4} \\[1em] \Rightarrow \dfrac{x^2 + 3x - (3x - 2 - x^2)}{x^2 - 2x} = \dfrac{17}{4} \\[1em] \Rightarrow x^2 + 3x - 3x + 2 + x^2 = \dfrac{17}{4} \times (x^2 - 2x) \\[1em] \Rightarrow 4(2x^2 + 2) = 17 \times (x^2 - 2x) \\[1em] \Rightarrow 8x^2 + 8 = 17x^2 - 34x \\[1em] \Rightarrow 17x^2 - 34x - 8x^2 - 8 = 0 \\[1em] \Rightarrow 9x^2 - 34x - 8 = 0 \\[1em] \Rightarrow 9x^2 - 36x + 2x - 8 = 0 \\[1em] \Rightarrow 9x(x - 4) + 2(x - 4) = 0 \\[1em] \Rightarrow (9x + 2)(x - 4) = 0 \\[1em] \Rightarrow (9x + 2) \text{ or } (x - 4) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 9x = -2 \text{ or } x = 4 \\[1em] \Rightarrow x = \dfrac{-2}{9} \text{ or } x = 4.$

Hence, $x = \Big\lbrace4, \dfrac{-2}{9}\Big\rbrace$.

Solve the following equation by factorization:

$\dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x}$

Answer

Given,

$\Rightarrow \dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{(x - 1) + 2(x - 2)}{(x - 2)(x - 1)} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x - 1 + 2x - 4}{x^2 - x - 2x + 2} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x - 1 + (2x - 4)}{x^2 - 3x + 2} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{3x - 5}{x^2 - 3x + 2} = \dfrac{6}{x} \\[1em] \Rightarrow x(3x - 5) = 6(x^2 - 3x + 2) \\[1em] \Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12 \\[1em] \Rightarrow 6x^2 - 3x^2 - 18x + 5x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow (3x - 4) \text{ or } (x - 3) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 3x = 4 \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{4}{3} \text{ or } x = 3.$

Hence, $x = \Big\lbrace3, \dfrac{4}{3}\Big\rbrace$.

Solve the following equation by factorization:

$2\Big(\dfrac{x}{x + 1}\Big)^2 - 5\Big(\dfrac{x}{x + 1}\Big) + 2 = 0$, x ≠ -1

Answer

Let us consider y = $\dfrac{x}{x + 1}$.

Substituting y = $\dfrac{x}{x + 1}$ in equation $2\Big(\dfrac{x}{x + 1}\Big)^2 - 5\Big(\dfrac{x}{x + 1}\Big) + 2 = 0$, we get :

⇒ 2y² - 5y + 2 = 0

⇒ 2y² - 4y - y + 2 = 0

⇒ 2y(y - 2) - 1(y - 2) = 0

⇒ (2y - 1)(y - 2) = 0

⇒ (2y - 1) = 0 or (y - 2) = 0      [Using Zero-product rule]

⇒ 2y = 1 or y = 2

⇒ y = $\dfrac{1}{2}$ or y = 2.

Now we have,

Case 1 : y = $\dfrac{1}{2}$

$\Rightarrow y = \dfrac{x}{x + 1} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{x}{x + 1} \\[1em] \Rightarrow x + 1 = 2x \\[1em] \Rightarrow 2x - x = 1 \\[1em] \Rightarrow x = 1.$

Case 2 : y = 2

⇒ y = $\dfrac{x}{x + 1}$

⇒ 2 = $\dfrac{x}{x + 1}$

⇒ 2(x + 1) = x

⇒ 2x + 2 = x

⇒ 2x - x = -2

⇒ x = -2.

Hence, x = {-2, 1}.

Solve the following equation by factorization:

5(3x + 1)² + 6(3x + 1) - 8 = 0

Answer

Let us consider y = 3x + 1.

Substituting y = 3x + 1 in equation 5(3x + 1)² + 6(3x + 1) - 8 = 0, we get :

⇒ 5y² + 6y - 8 = 0

⇒ 5y² + 10y - 4y - 8 = 0

⇒ 5y(y + 2) - 4(y + 2) = 0

⇒ (5y - 4)(y + 2) = 0

⇒ (5y - 4) = 0 or (y + 2) = 0      [Using Zero-product rule]

⇒ 5y = 4 or y = -2

⇒ y = $\dfrac{4}{5}$ or y = -2.

Now we have,

Case 1 : y = $\dfrac{4}{5}$

$\Rightarrow y = \dfrac{4}{5} \\[1em] \Rightarrow 3x + 1 = \dfrac{4}{5} \\[1em] \Rightarrow 5(3x + 1) = 4 \\[1em] \Rightarrow 15x + 5 = 4 \\[1em] \Rightarrow 15x = 4 - 5 \\[1em] \Rightarrow 15x = -1 \\[1em] \Rightarrow x = \dfrac{-1}{15}.$

Case 2 : y = -2

⇒ y = -2

⇒ 3x + 1 = -2

⇒ 3x = -2 - 1

⇒ 3x = -3

⇒ x = $\dfrac{-3}{3}$

⇒ x = -1.

Hence, x = $\Big\lbrace-1, \dfrac{-1}{15}\Big\rbrace$.

Solve the following equation by factorization:

$\sqrt{x + 15}$ = (x + 3)

Answer

Given,

⇒ $\sqrt{x + 15}$ = (x + 3)

Squaring both sides, we get :

⇒ (x + 15) = (x + 3)²

⇒ x + 15 = (x)² + (3)² + 2 × x × 3

⇒ x + 15 = x² + 9 + 6x

⇒ x² + 9 + 6x - x - 15 = 0

⇒ x² + 5x - 6 = 0

⇒ x² + 6x - x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x + 6)(x - 1) = 0

⇒ (x + 6) = 0 or (x - 1) = 0      [Using Zero-product rule]

⇒ x = -6 or x = 1.

Substituting x = -6 in the L.H.S. of this equation $\sqrt{x + 15}$ = (x + 3)

$\Rightarrow \sqrt{-6 + 15} \\[1em] \Rightarrow \sqrt{9} \\[1em] \Rightarrow 3$

Substituting x = -6 in the R.H.S. of this equation $\sqrt{x + 15}$ = (x + 3)

⇒ x + 3

⇒ -6 + 3

⇒ -3.

L.H.S ≠ R.H.S .

∴ x = -6 is not valid.

Hence, x = {1}.

Solve the following equation by factorization:

$\sqrt{2x + 9}$ = (13 - x)

Answer

Given,

⇒ $\sqrt{2x + 9}$ = (13 - x)

Squaring both sides we get :

⇒ (2x + 9) = (13 - x)²

⇒ 2x + 9 = (13²) + (x)² - 2 × 13 × x

⇒ 2x + 9 = 169 + x² - 26x

⇒ x² - 26x + 169 - 2x - 9 = 0

⇒ x² - 28x + 160 = 0

⇒ x² - 20x - 8x + 160 = 0

⇒ x(x - 20) - 8(x - 20) = 0

⇒ (x - 20)(x - 8) = 0

⇒ (x - 20) = 0 or (x - 8) = 0      [Using Zero-product rule]

⇒ x = 20 or x = 8

⇒ x = 8.

Substituting x = 20 in the L.H.S. of this equation $\sqrt{2x + 9}$ = (13 - x)

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

Substituting x = 20 in the R.H.S. of this equation $\sqrt{2x + 9}$ = (13 - x)

⇒ 13 - x

⇒ 13 - 20

⇒ -7.

L.H.S ≠ R.H.S .

∴ x = 20 is not valid.

Hence, x = {8}.

Solve the following equation by factorization:

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

Answer

Given,

⇒ $\sqrt{3x^2 - 2}$ = (2x - 1)

Squaring both sides we get :

⇒ (3x² - 2) = (2x - 1)²

⇒ 3x² - 2 = (2x)² + (1)² - 2 × 2x × 1

⇒ 3x² - 2 = 4x² + 1 - 4x

⇒ 4x² + 1 - 4x - 3x² + 2 = 0

⇒ x² - 4x + 3 = 0

⇒ x² - x - 3x + 3 = 0

⇒ x(x - 1) - 3(x - 1) = 0

⇒ (x - 1)(x - 3) = 0

⇒ (x - 1) = 0 or (x - 3) = 0      [Using Zero-product rule]

⇒ x = 1 or x = 3.

Hence, x = {1, 3}.

Solve the following equation by factorization:

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

Answer

Given,

⇒ $\sqrt{3x^2 + x + 5}$ = (x - 3)

Squaring both sides we get :

⇒ (3x² + x + 5) = (x - 3)²

⇒ 3x² + x + 5 = (x²) + (3²) - 2 × x × 3

⇒ 3x² + x + 5 = x² + 9 - 6x

⇒ 3x² + x + 5 - x² - 9 + 6x = 0

⇒ 2x² + 7x - 4 = 0

⇒ 2x² - x + 8x - 4 = 0

⇒ x(2x - 1) + 4(2x - 1) = 0

⇒ (2x - 1)(x + 4) = 0

⇒ (2x - 1) = 0 or (x + 4) = 0      [Using Zero-product rule]

⇒ 2x = 1 or x = -4

⇒ x = $\dfrac{1}{2}$ or x = -4.

Hence, x = $\Big\lbrace-4, \dfrac{1}{2}\Big\rbrace$.

Find the quadratic equation whose solution set is:

(i) {2, -3}

(ii) $\Big\lbrace-3, \dfrac{2}{5}\Big\rbrace$

(iii) $\Big\lbrace\dfrac{2}{5}, -\dfrac{1}{2}\Big\rbrace$

Answer

(i) 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} is x² + x - 6 = 0.

(ii) Since, $\Big\lbrace-3, \dfrac{2}{5}\Big\rbrace$ is solution set.

It means -3 and $\dfrac{2}{5}$ are roots of the equation,

∴ x = -3 or x = $\dfrac{2}{5}$

⇒ x = -3 or 5x = 2

⇒ x + 3 = 0 or 5x - 2 = 0

⇒ (x + 3)(5x - 2) = 0

⇒ (5x² - 2x + 15x - 6) = 0

⇒ 5x² + 13x - 6 = 0.

Hence, quadratic equation with solution set $\Big\lbrace-3, \dfrac{2}{5}\Big\rbrace$ is 5x² + 13x - 6 = 0.

(iii) Since, $\Big\lbrace\dfrac{2}{5}, -\dfrac{1}{2}\Big\rbrace$ is solution set.

It means $\dfrac{2}{5}$ and $-\dfrac{1}{2}$ are roots of the equation,

∴ x = $\dfrac{2}{5}$ or x = $-\dfrac{1}{2}$

⇒ 5x = 2 or 2x = -1

⇒ 5x - 2 = 0 or 2x + 1 = 0

⇒ (5x - 2)(2x + 1) = 0

⇒ (10x² + 5x - 4x - 2) = 0

⇒ 10x² + x - 2 = 0.

Hence, quadratic equation with solution set $\Big\lbrace\dfrac{2}{5}, -\dfrac{1}{2}\Big\rbrace$ is 10x² + x - 2 = 0.

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

Thus, find the other root of the equation.

Answer

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

⇒ (k + 2)(3)² - 3k + 6 = 0

⇒ (k + 2)(9) - 3k + 6 = 0

⇒ 9k + 18 - 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = -24

⇒ k = $\dfrac{-24}{6}$

⇒ k = -4.

Substitute the value of k = -4 in (k + 2)x² - kx + 6 = 0 we get,

⇒ (-4 + 2)(x)² - (-4)x + 6 = 0

⇒ (-2)(x)² - (-4)x + 6 = 0

⇒ -2x² + 4x + 6 = 0

⇒ -2x² - 2x + 6x + 6 = 0

⇒ -2x(x + 1) + 6(x + 1) = 0

⇒ (x + 1)(-2x + 6) = 0

⇒ (x + 1) = 0 or (-2x + 6) = 0      [Using Zero-product rule]

⇒ x = -1 or -2x = -6

⇒ x = -1 or x = $\dfrac{-6}{-2}$

⇒ x = -1 or x = 3.

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