Question

Obtain the roots of the following quadratic equation by using the general formula the solution:
$p^2{x}^2+(p^2-q^2)x-q^2=0$

Answer 1

Formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Given,

$p^2{x}^2+(p^2-q^2)x-q^2=0$

$x=\frac{-(p^2-q^2)\pm \sqrt{{(p^2-q^2)}^2-4p^2(-q^2)}}{2p^2}$

$=\frac{-(p^2-q^2)\pm (p^2+q^2)}{2p^2}$

$x=\frac{q^2}{p^2},-1$