Using quadratic formula solve the given quadratic equation
$p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0, p \neq 0$

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Solution

We have $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0, p \neq 0$

Comparing this equation with $a x^{2} + b x + c = 0$ we have

$a = p^{2}, b = p^{2} - p^{q}, c = - q^{2}$

$∴ D = b^{2} - 4 a c = {(p^{2} - q^{2})}^{2} - 4 \times p^{2} \times - q^{2} = {(p^{2} + q^{2})}^{2} > 0$

So, the given equation has real roots given by

$α = - \frac{b - \sqrt{D}}{2 a} = - \frac{(p^{2} - q^{2}) + (p^{2} + q^{2})}{2 p^{2}} = \frac{q^{2}}{p^{2}}$
$β = \frac{- b - \sqrt{D}}{2 a} = \frac{(p^{2} - q^{2}) - (p^{2} + q^{2})}{2 p^{2}} = - 1$

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