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(D) Given that $\alpha$ and $\beta$ are roots of $p x^2 + q x + r = 0$. Since $p, q, r$ are in $AP$,we have $2q = p + r$. From $\frac{1}{\alpha} + \fr

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$\frac{2 \sqrt{13}}{9}$

Vedclass · Answer

(D) Given that $\alpha$ and $\beta$ are roots of $p x^2 + q x + r = 0$. Since $p, q, r$ are in $AP$,we have $2q = p + r$. From $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,we get $\frac{\alpha + \beta}{\alpha \beta} = 4$. Using Vieta's formulas,$\alpha + \beta = -\frac{q}{p}$ and $\alpha \beta = \frac{r}{p}$. Substituting these,$\frac{-q/p}{r/p} = -\frac{q}{r} = 4$,so $q = -4r$. Since $p + r = 2q$,we have $p + r = 2(-4r) = -8r$,which implies $p = -9r$. Now,$|\alpha - \beta| = \frac{\sqrt{D}}{|p|} = \frac{\sqrt{q^2 - 4pr}}{|p|}$. Substituting $q = -4r$ and $p = -9r$: $|\alpha - \beta| = \frac{\sqrt{(-4r)^2 - 4(-9r)(r)}}{|-9r|} = \frac{\sqrt{16r^2 + 36r^2}}{9|r|} = \frac{\sqrt{52r^2}}{9|r|} = \frac{2|r|\sqrt{13}}{9|r|} = \frac{2\sqrt{13}}{9}$. Thus,the correct option is $D$.

Let $\alpha$ and $\beta$ be the roots of the equation $p x^2 + q x + r = 0$,where $p \neq 0$. 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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