Quadratic Equations — Exercise 5.1

In each of the following, determine whether the given numbers are roots of the given equations or not:

(i) $x^2 - 5x + 6 = 0; 2, -3$

(ii) $3x^2 - 13x - 10 = 0 ; 5, -\dfrac{2}{3}$

Answer

(i) Given,

$x^2 - 5x + 6 = 0$

Substituting x = 2 in the LHS of the given equation, we get:

$\text{LHS} = 2^2 - 5 \times 2 + 6 \\[0.5em] = 4 - 10 + 6 \\[0.5em] = 10 - 10 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ 2 is a root of the given equation.

Substituting x = -3 in the LHS of the given equation, we get:

$\text{LHS} = (-3)^2 - 5 \times (-3) + 6 \\[0.5em] = 9 + 15 + 6 \\[0.5em] = 30$

≠ $\text{RHS}$

∴ 3 is not a root of the given equation.

Hence, 2 is a root but -3 is not a root of the given equation.

(ii) Given,

$3x^2 - 13x - 10 = 0$

Substituting x = 5 in the LHS of the given equation, we get:

$\text{LHS} = 3 \times (5)^2 - 13 \times 5 - 10 \\[0.5em] = 3 \times 25 - 65 -10 \\[0.5em] = 75 - 75 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ 5 is a root of the equation

Substituting x = $-\dfrac{2}{3}$ in the LHS of the given equation, we get:

$\text{LHS} = 3 \times \Big(-\dfrac{2}{3}\Big)^2 - 13 \times \Big( -\dfrac{2}{3} \Big) - 10 \\[1em] = 3 \times \dfrac{4}{9} + \dfrac{26}{3} - 10 \\[1em] = \dfrac{4}{3} + \dfrac{26}{3} - 10 \\[1em] = \dfrac{30}{3} - 10 \\[1em] = 10 - 10 \\[1em] = 0 \\[1em] = \text{RHS}$

∴ $-\dfrac{2}{3}$ is a root of the equation

Hence, both 5 and $-\dfrac{2}{3}$ are roots of the given equation.

In each of the following, determine whether the given numbers are solutions of the given equations or not:

(i) $x^2 - 3\sqrt{3}x + 6 = 0; \sqrt{3}, \space -2\sqrt{3}$

(ii) $x^2 - \sqrt{2}x - 4 = 0; -\sqrt{2}, \space 2\sqrt{2}$

Answer

(i) Given,

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

Substituting x = $\sqrt{3}$ in the LHS of the given equation, we get:

$\text{LHS} = (\sqrt{3})^2 - 3\sqrt{3} \times \sqrt{3} + 6 \\[0.5em] = 3 - 9 + 6 \\[0.5em] = 9 - 9 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $\sqrt{3}$ is a solution of the given equation

Substituting x = $-2\sqrt{3}$ in the LHS of the given equation, we get:

$\text{LHS} = (-2\sqrt{3})^2 - 3\sqrt{3} \times -2\sqrt{3} + 6 \\[0.5em] = 12 + 18 + 6 \\[0.5em] = 36$

≠ $\text{RHS}$

∴ $-2\sqrt{3}$ is not a solution of the given equation

Hence, $\sqrt{3}$ is a solution of the given equation but $-2\sqrt{3}$ is not.

(ii) Given,

$x^2 - \sqrt{2}x - 4 = 0$

Substituting x = $-\sqrt{2}$ in the LHS of the given equation, we get:

$\text{LHS} = (-\sqrt{2})^2 - \sqrt{2} \times (-\sqrt{2}) - 4 \\[0.5em] = 2 + 2 - 4 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $-\sqrt{2}$ is a solution of the given equation

Substituting x = $2\sqrt{2}$ in the LHS of the given equation, we get:

$\text{LHS} = (2\sqrt{2})^2 - \sqrt{2} \times 2\sqrt{2} - 4 \\[0.5em] = 8 - 4 - 4 \\[0.5em] = 8 - 8 \\[0.5em] = 0 \\[0.5em] = \text{RHS}$

∴ $2\sqrt{2}$ is a solution of the given equation

Hence, $-\sqrt{2}$ and $2\sqrt{2}$ both are solutions of the given equation.

If $-\dfrac{1}{2}$ is the solution of the equation 3x² + 2kx - 3 = 0, find the value of k.

Answer

Since, $-\dfrac{1}{2}$ is a solution of the equation 3x² + 2kx - 3 = 0, x = $-\dfrac{1}{2}$ satisfies the given equation.

Substituting x = $-\dfrac{1}{2}$ and solving for k we get:

$3\Big(-\dfrac{1}{2}\Big)^2 + 2k \Big(-\dfrac{1}{2}\Big) -3 = 0 \\[1em] \Rightarrow 3 \times \dfrac{1}{4} - k - 3 = 0 \\[1em] \Rightarrow -k = 3 - \dfrac{3}{4} \\[1em] \Rightarrow k = \dfrac{3}{4} - 3 \\[1em] \Rightarrow k = \dfrac{3 - 12}{4} \\[1em] \Rightarrow k = -\dfrac{9}{4}$

Hence, the value of k is $-\dfrac{9}{4}$

If $\dfrac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, find the value of k.

Answer

Since, $\dfrac{2}{3}$ is the solution of the equation 7x² + kx - 3 = 0, x = $\dfrac{2}{3}$ satisfies the given equation.

Substituting x = $-\dfrac{2}{3}$ and solving for k we get:

$\Rightarrow 7\Big(\dfrac{2}{3}\Big)^2 + k \times \dfrac{2}{3} -3 = 0 \\[1em] \Rightarrow 7 \times \dfrac{4}{9} + \dfrac{2k}{3} -3 = 0 \\[1em] \Rightarrow \dfrac{28}{9} - 3 + \dfrac{2k}{3} = 0 \\[1em] \Rightarrow \dfrac{28}{9} - \dfrac{27}{9} + \dfrac{6k}{9} = 0 \text{ (Taking L.C.M.) } \\[1em] \Rightarrow \dfrac{28 - 27 + 6k}{9} = 0 \\[1em] \Rightarrow 1 + 6k = 0 \\[1em] \Rightarrow 6k = -1 \\[1em] \Rightarrow k = -\dfrac{1}{6}$

Hence, the value of k is $-\dfrac{1}{6}.$

If $\sqrt{2}$ is a root of the equation $kx^2 + \sqrt{2}x - 4 = 0$, find the value of k.

Answer

Since $\sqrt{2}$ is a root of the equation $kx^2 + \sqrt{2}x - 4 = 0$, $x = \sqrt{2}$ satisfies the given equation.

Substituting $x = \sqrt{2}$ in the given equation:

$k ( \sqrt{2} )^2 + \sqrt{2} \times \sqrt{2} - 4 = 0 \\[0.5em] \Rightarrow 2k + 2 - 4 = 0 \\[0.5em] \Rightarrow 2k - 2 = 0 \\[0.5em] \Rightarrow 2k = 2 \\[0.5em] \Rightarrow k = 1$

Hence, the value of k is 1

If a is the root of the equation x² - (a + b)x + k = 0, find the value of k.

Answer

Since, a is the root of the equation x² - (a + b)x + k = 0 , x = a satisfies the given equation.

Substituting x = a in the given equation:

$\Rightarrow a^2 - (a+b)a + k = 0 \\[0.5em] \Rightarrow a^2 - (a^2 + ab) + k = 0 \\[0.5em] \Rightarrow a^2 - a^2 - ab + k = 0 \\[0.5em] \Rightarrow k - ab = 0 \\[0.5em] \Rightarrow k = ab$

Hence, the value of k is ab.

If $\dfrac{2}{3}$ and -3 are roots of the equation px² + 7x + q = 0, find the values of p and q .

Answer

The given equation is px² + 7x + q = 0.

As $\dfrac{2}{3}$ is a root of the equation, so x = $\dfrac{2}{3}$ satisfies the given equation.

Substituting x = $\dfrac{2}{3}$ in the given equation:

$p\big(\dfrac{2}{3} \big)^2 + 7 \times \dfrac{2}{3} + q = 0 \\[1em] \Rightarrow \dfrac{4p}{9} + \dfrac{14}{3} + q = 0 \\[1em] \Rightarrow q = -\dfrac{4p}{9} - \dfrac{14}{3} \\[1em] \Rightarrow q = -\big(\dfrac{4p + 42}{9}\big) \space \text{...(i)} \\[1em]$

Also -3 is a root of the given equation, so x = -3 satisfies the given equation.

Substituting x = -3 in the given equation:

$p(-3)^2 + 7 \times (-3) + q = 0 \\[0.5em] \Rightarrow 9p - 21 + q = 0 \space \text{...(ii)}$

Putting value of q from eqn (i) into eqn (ii)

$9p - 21 - \Big(\dfrac{4p + 42}{9}\Big) = 0$

Taking 9 as the LCM

$\Rightarrow \dfrac{81p - 189 - 4p - 42}{9} = 0 \\[1em] \Rightarrow 77p - 231 = 0 \\[1em] \Rightarrow 77p = 231 \\[1em] \Rightarrow p = \dfrac{231}{77} \\[1em] \Rightarrow p = 3$

Substituting p = 3 in (i), we get

$q = -\Big( \dfrac{4\times 3 + 42}{9} \Big) \\[1em] \Rightarrow q= -\Big(\dfrac{12 + 42}{9}\Big) \\[1em] \Rightarrow q= -\Big(\dfrac{54}{9}\Big) \\[1em] \Rightarrow q= -6$

Hence, p = 3 and q = -6