Let \alpha and \beta be the roots of the equation p x^2+q x-r=0, where p
eq 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 :

Question

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

Answer: D. \frac{80}{9} Given : p x^2+q x-r=0 Let p=\frac{a}{r_1}, q=a, r=a r_1 \begin{aligned} & 	ext { and } \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{aligned} \begin{aligned}(\alpha-\beta)^2 & =(\alpha+\beta)^2-4 \alpha \beta \\ & =\left(\frac{-q}{p}ight)^2-4\left(\frac{-r}{p}ight) \\ & =\frac{q^2}{p^2}+\frac{4 r}{p} \\ & =r_1^2+4 r_1^2=5 r_1^2 \\ & =5\left(\frac{4}{3}ight)^2=\frac{80}{9}\end{aligned}

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

Answer

Explanation

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