Let , be the roots of x^2-3x+p=0 and , be the roots of x^2-6x+q=0.

Question

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

Accepted Answer

Since $\alpha\delta=\beta\gamma$ (G.P.): $\dfrac{\alpha-\beta}{\alpha+\beta}=\dfrac{\gamma-\delta}{\gamma+\delta}$, i.e. $\dfrac{\sqrt{9-4p}}{3}=\dfrac{\sqrt{36-4q}}{6} \Rightarrow 36-16p=36-4q \Rightarrow q=4p.$

$$\frac{2q+p}{2q-p} = \frac{8p+p}{8p-p} = \frac{9}{7}.$$

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