Quadratic Equations — Exercise 5(A)

4x² - 9 = 0 implies x is equal to :

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

2. $\dfrac{9}{4}$

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

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

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.

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

1. 3

2. 3 or 5

3. 3 or -5

4. 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.

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

1. 3

2. -3

3. 2

4. -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.

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

1. -2

2. 2

3. 4

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.

If x² - 7x = 0; the value of x is :

1. 0 and 7

2. 7

3. 0

4. 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.

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 $-2\sqrt{6}.$

$\dfrac{2}{3}$ 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.