Solve the following

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 :

  1. (1) $3: 1$

  2. (2) $9: 7$

  3. (3) $5: 3$

  4. (4) $33: 31$

Correct Option: , 2


Let $\alpha, \beta, \gamma, \delta$ be in G.P., then $\alpha \delta=\beta \gamma$

$\Rightarrow \frac{\alpha}{\beta}=\frac{\gamma}{\delta} \Rightarrow\left|\frac{\alpha-\beta}{\alpha+\beta}\right|=\left|\frac{\gamma-\delta}{\gamma+\delta}\right|$

$\Rightarrow \frac{\sqrt{9-4 p}}{3}=\frac{\sqrt{36-4 q}}{6}$

$\Rightarrow 36-16 p=36-4 q \Rightarrow q=4 p$

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


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