Using quadratic formula solve the following quadratic equation:

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

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Solution

Given equation is: $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0, p \neq 0$

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

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

$∴$ Discriminant, $D = b^{2} - 4 a c = (p^{2} - q^{2})^{2} - 4 \times p^{2} \times - q^{2}$

$= (p^{2} - q^{2})^{2} + 4 p^{2} q^{2}$

$= (p^{2} + q^{2})^{2}$

$\Rightarrow D > 0$

So, the given equation has real roots and given by,

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

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

$∴$ The roots are $\frac{q^{2}}{p^{2}}, - 1.$

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Using quadratic formula solve the following quadratic equation:

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

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