Question

Find the values of $p$ and $q$ for which $x=\frac{2}{3}$ and $x=-3$ are the roots of the equation $p{x}^2+7x+q=0$

• A. $p=-3,q=-6$
• B. $p=-3,q=6$
• C. $p=3,q=-6$
• D. $p=3,q=6$

Answer 1

The correct option is C $p=3,q=-6$

Given quadratic equation is $p{x}^2+7x+q=0$

From the theory of equations, we know that
Sum of roots $=\frac{-b}{a}$
$=\frac{-7}{p}$
$=\frac{2}{3}-3$
Hence, $\frac{7}{p}=3-\frac{2}{3}$
$\Rightarrow \frac{7}{p}=\frac{7}{3}$
$\Rightarrow p=3$
And product of roots is $\frac{2}{3}.(-3)$
$=-2$ $=\frac{q}{p}$
$\Rightarrow q=-2p$

$\Rightarrow q=-2\times 3$
$\Rightarrow q=-6$