Let alpha and beta be the roots of x^{2}-3 x+ | vedclass

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Question

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

Vedclass · Accepted Answer

$9: 7$

Vedclass · Answer

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

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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