Let alpha and beta be the roots of the equati | vedclass

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Question

$ p x^2+q x-r=0 < \beta $

$ p=A, q=A R, r=A R^2$

$ A x^2+A R x-A R^2=0$

$ x^2+R x-R^2=0 < \beta $

$ \because \frac{1}{\alpha}+\frac{

Vedclass · Accepted Answer

$\frac{80}{9}$

Vedclass · Answer

$ p x^2+q x-r=0 < \beta $

$ p=A, q=A R, r=A R^2$

$ A x^2+A R x-A R^2=0$

$ x^2+R x-R^2=0 < \beta $

$ \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} $

$ 	herefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-R}{-R^2}=\frac{3}{4} \Rightarrow R=\frac{4}{3} $

$ (\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=R^2-4\left(-R^2ight)=5\left(\frac{16}{9}ight) $

$ =80 / 9$

Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{px}^2+\mathrm{qx}-$ $r=0$, 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 :

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