Let $\alpha$ and $\beta$ be the roots of the equation $p x^2+q x+r=0, p \neq 0$. If $p, q, r$ are in AP and $\frac{1}{\alpha}+\frac{1}{\beta}=4$, then the value of $|\alpha-\beta|$ is

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Aicharya · Accepted Answer

Since, $\alpha$ and $\beta$ are roots of quadratic equation $p x^2+q x+r=0, p \neq 0$ and $p, q, r$ are in AP such that $\frac{1}{\alpha}+\frac{1}{\beta}=4$ ...(i) Let the $p=q-d$ and $r=q+d$, where $d$ is the common difference of AP $\therefore$ Sum of the roots $\alpha+\beta=\frac{-q}{q-d}$ and product of the roots $\alpha \beta=\frac{q+d}{q-d}$ $\because$ from Eq. (i), we have $\frac{\alpha+\beta}{\alpha \beta}=4 \Rightarrow \frac{-q}{q+d}=4 \Rightarrow 5 q+4 d=0$ ...(ii) Now, $|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta}$ $\begin{aligned} & =\sqrt{\frac{q^2}{(q-d)^2}-4 \frac{q+d}{q-d}}=\sqrt{\frac{q^2-4\left(q^2-d^2\right)}{(q-d)^2}} \ & =\sqrt{\frac{4 d^2-3 q^2}{(q-d)^2}}=\sqrt{\frac{4\left(-\frac{5}{4} q\right)^2-3 q^2}{\left(q+\frac{5}{4} q\right)^2}} \ & =\sqrt{\frac{\frac{25}{4}-3}{\left(\frac{9}{4}\right)^2}}=\frac{2}{9} \sqrt{13} \end{aligned}$ Hence, option (d) is correct.

Let \(\alpha\) and \(\beta\) be the roots of the equation \(p x^2+q x+r=0, p \neq 0\). If \(p, q, r\) are in…

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