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(D) Given that $\alpha$ and $\beta$ are the roots of $px^2 + qx + r = 0$,we have $\alpha + \beta = -\frac{q}{p}$ and $\alpha\beta = \frac{r}{p}$.Since

Vedclass · Accepted Answer

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

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(D) Given that $\alpha$ and $\beta$ are the roots of $px^2 + qx + r = 0$,we have $\alpha + \beta = -\frac{q}{p}$ and $\alpha\beta = \frac{r}{p}$.Since $p, q, r$ are in $A.P.$,we have $2q = p + r$.Given $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,which implies $\frac{\alpha + \beta}{\alpha\beta} = 4$,so $\alpha + \beta = 4\alpha\beta$.Substituting the expressions for sum and product of roots: $-\frac{q}{p} = 4(\frac{r}{p}) \Rightarrow q = -4r$.Substitute $q = -4r$ into the $A.P.$ condition $2q = p + r$: $2(-4r) = p + r \Rightarrow -8r = p + r \Rightarrow p = -9r$.Now,$\alpha + \beta = -\frac{q}{p} = -\frac{-4r}{-9r} = -\frac{4}{9}$ and $\alpha\beta = \frac{r}{p} = \frac{r}{-9r} = -\frac{1}{9}$.Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$:$(\alpha - \beta)^2 = (-\frac{4}{9})^2 - 4(-\frac{1}{9}) = \frac{16}{81} + \frac{4}{9} = \frac{16 + 36}{81} = \frac{52}{81}$.Therefore,$|\alpha - \beta| = \sqrt{\frac{52}{81}} = \frac{\sqrt{4 	imes 13}}{9} = \frac{2\sqrt{13}}{9}$.

Let $\alpha$ and $\beta$ be the roots of the equation $px^2 + qx + r = 0$ (where $p \neq 0$). If $p, q, r$ are in $A.P.$ and $\frac{1}{\alpha} + \frac{1}{\beta} = 4$,then the value of $|\alpha - \beta|$ is

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