State the following statement is True or False
If the roots of the equation $x^{2} + p x + q = 0$ differ by $1$ , then $p^{2} = 1 + 4 q$

True

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False

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Solution

The correct option is A True
The equation is : $x^{2} + p x + q = 0$
The roots of the equation differ by $1$
$∴ x = \frac{- p \pm \sqrt{p^{2} - 4 q}}{2}$
$\frac{- p + \sqrt{p^{2} - 4 q}}{2} - \frac{- p - \sqrt{p^{2} - 4 q}}{2} = 1$
$- p + \sqrt{p^{2} - 4 q} + p + \sqrt{p^{2} - 4 q} = 2$
$2 \sqrt{p^{2} - 4 q} = 2$
$p^{2} - 4 q = 1$
$p^{2} = 1 + 4 q$

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x^{2} - p x + q = 0

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1,

p^{2} + 4 q^{2}

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