Question

If α, β are roots of x² — 3x + p = 0 and Ф. ω are roots of x² — 6x + q = 0 and α,β,Ф,ω are in increasing geometric progression then the value of (2q+p) / (2q-p) is equal to

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\alpha \: , \: \beta \: , \phi \: , \: \omega \:$

are in increasing geometric progression

Let r ( > 1) be the Common ratio

So

$\beta \: = \alpha \:r \:, \: \phi \: = \alpha {r}^{2} \: , \: \omega \: = \alpha \: {r}^{3}$

So From first quadratic Equation

$\alpha + \alpha \:r = 3 \: \: and \: \: \alpha \times \alpha \:r = p \: \: \: ....(1)$

From second quadratic equation

$\alpha {r}^{2} + \alpha {r}^{3} = 6 \: \: and \: \: \alpha {r}^{2} \times \alpha {r}^{3} = q \: \: \: \: ....(2)$

So

$\displaystyle \: \frac{\alpha {r}^{2} + \alpha {r}^{3}}{\alpha + \alpha {r}^{}} = \frac{6}{3}$

$\implies \: {r}^{2} = 2$

$\implies \: r = \pm \: \sqrt{2}$

As r > 1

So

$r = \sqrt{2}$

So

$\displaystyle \: \frac{p}{q} = \frac{1}{ {r}^{4} } = \frac{1}{4}$

So

$\displaystyle \: \frac{q}{p} = 4$

Hence

$\displaystyle \: \frac{2q + p}{2q - p}$

$\displaystyle \: \frac{2 \times \frac{q}{p} + 1}{2 \times \frac{q}{p} - 1}$

$= \displaystyle \: \frac{8 + 1}{8 - 1}$

$= \displaystyle \: \frac{9}{7}$