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Q. Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0$. If $\alpha$ $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is

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Solution:

$x^{2}-3 x+p=0<_{\beta}^{\alpha}$
$\alpha, \beta, \gamma, \delta$ in G.P.
$\alpha+\alpha=3 \ldots$ (1)
$x^{2}-6 x+q=0<_{\delta}^{\gamma}$
$\alpha r^{2}+\alpha r^{3}=6 \ldots(2)$
$(2) \div(1)$
$r^{2}=2$
So $, \frac{2 q+p}{2 q-p}=\frac{2 r^{5}+r}{2 r^{5}-r}$
$=\frac{2 r^{4}+1}{2 r^{4}-1}=\frac{9}{7}$