Let and be the roots of the equation px^2+qx-r=0, where p 0.

Question

Let $\alpha $ and $\beta $ be the roots of the equation $p{x}^{2}+qx-r=0$, where $p
e 0$. If $p,q$ and $r$ be the consecutive terms of a non constant G.P. and $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{3}{4}$, then the value of $(\alpha -\beta {)}^{2}$ is :

Accepted Answer

Given : $p{x}^{2}+qx-r=0$

Let $p=\frac{a}{{r}_{1}},q=a,r=a{r}_{1}$

$\begin{gathered} \frac{1}{\alpha }+\frac{1}{\beta }=\frac{3}{4} \ \Rightarrow \frac{\alpha +\beta }{\alpha \beta }=\frac{3}{4} \ \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \ \Rightarrow \frac{q}{r}=\frac{3}{4} \ \Rightarrow \frac{1}{{r}_{1}}=\frac{3}{4} \ \Rightarrow {r}_{1}=\frac{4}{3} \end{gathered}$

$\begin{gathered} (\alpha -\beta {)}^{2} =(\alpha +\beta {)}^{2}-4\alpha \beta \ \ ={(\frac{-q}{p})}^{2}-4(\frac{-r}{p}) \ \ =\frac{{q}^{2}}{{p}^{2}}+\frac{4r}{p} \ \ ={r}_{2}^{1}+4{r}_{2}^{1}=5{r}_{2}^{1} \ \ =5{(\frac{4}{3})}^{2}=\frac{80}{9} \end{gathered}$

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