If the difference between the roots of $x^{2} + ax - b = 0$ is equal to the difference between the roots of $x^{2} - px + q = 0$, then $a^{2} - p^{2}$ in terms of $b$ and $q$ is

Question

Dr. Abubacker · Accepted Answer

(a)
Let $\alpha,\ \beta$ are the roots of the equation $x^{2} + ax - b = 0$
$	herefore\ \alpha + \beta = - a,\ \alpha\beta = - b$
And $\gamma,\delta$ are the roots of the equation $x^{2} - px + q = 0$
$	herefore\ \gamma + \delta = p,\ \gamma\delta = q$
Given, $\alpha - \beta = \gamma - \delta\  \Rightarrow (\alpha - \beta)^{2} = (\alpha - \beta)^{2}$
$\Rightarrow \ \ (\alpha + \beta)^{2} - 4\alpha\beta = (\gamma + \delta)^{2} - 4\gamma\delta$
$\Rightarrow a^{2} + 4b = p^{2} - 4q$
$\Rightarrow a^{2} - p^{2} = - 4(b + q)$

Q. If the difference between the roots of $x^{2} + ax - b = 0$ is equal to the difference between the roots of $x^{2} - px + q = 0$, then $a^{2} - p^{2}$ in terms of $b$ and $q$ is

Explanation

Reviewed by Dr. Abubacker

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