Quadratic Equations — Exercise 5(B)

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

1. -1 and 7

2. 1 and 7

3. -1 or 7

4. 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.

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

1. 6 or 2

2. -6 or -2

3. 6 and -2

4. -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.

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

1. -2

2. 2

3. ±2

4. 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.

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

1. 4

2. 5 and $\dfrac{1}{5}$

3. 2 or $\dfrac{1}{2}$

4. 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.

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

1. x ≠ 0

2. x = 1

3. x = 5

4. 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.

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.

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

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

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.

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.

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

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.

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.

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.

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.

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.

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

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

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).

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.

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.

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.

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.

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.

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 = $\dfrac{\sqrt6}{3}$.