Solve the equation:
$4 x^{2} - 4 p x + (p^{2} - q^{2}) = 0$

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Solution

For the given quadratic equation,

$a = 4, b = - 4 p$ and $c = p^{2} - q^{2}$

Thus, $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- 4 p) \pm \sqrt{(- 4 p)^{2} - 4 \times 4 \times (p^{2} - q^{2})}}{2 \times 4}$

$= \frac{4 p \pm \sqrt{16 p^{2} - 16 p^{2} + 16 q^{2}}}{8} = \frac{4 p \pm 4 q}{8} = \frac{p \pm q}{2}$

So, the roots of the quadratic equation are : $\frac{p + q}{2}$ and $\frac{p - q}{2}$

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x

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