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

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