If the difference of the roots of $x^{2} - p x + q = 0$ is unity, then prove that $p^{2} - 4 q = 1$ and $p^{2} + 4 q^{2} = {(1 + 2 q)}^{2}$

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Solution

Let roots be $α, β$ then
$(α - β)^{2} = (α + β)^{2} - 4 α β$
$\Rightarrow 1^{2} = {(\frac{- (- P)}{1})}^{2} - 4 (\frac{9}{1})$
$\Rightarrow 1 = P^{2} - 4 q$
hence proved.

$\Rightarrow P^{2} = 4 q + 1$
$\Rightarrow P^{2} + (2 q)^{2} = (2 q)^{2} + 2 (1) (2 q) + (1)^{2}$
$P^{2} + 4 q^{2} = (2 q + 1)^{2}$

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If the roots of the equation $x^{2} + p x + q = 0$ differ by $1$ , then $p^{2} = 1 + 4 q$

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