Let $\alpha$ and $\beta$ be the roots of x² - 3x + p=0 and $\gamma$ and $\delta$ be the roots of x² - 6x + q = 0. If $\alpha, \beta, \gamma, \delta$ form a geometric progression.Then ratio (2q + p) : (2q - p) is:

$\alpha$ and $\beta$ are the roots of x² $-$ 3x + p = 0<br><br>$\therefore$ $\alpha$ + $\beta$ = 3 and $\alpha \beta$ = p<br><br>$\gamma$ and $\delta$ are the roots of x² $-$ 6x + q = 0<br><br>$\therefore$ $\gamma$ + $\delta$ = 6 and $\gamma \delta$ = q<br><br>Given that, $\alpha$, $\beta$, $\gamma$, $\delta$ are in G.P.<br><br>$\therefore$ Let $\alpha$ = a, $\beta$ = ar, $\gamma$ = ar² and $\delta$ = ar³<br><br>As $\alpha$ + $\beta$ = 3<br><br>$\Rightarrow$ a + ar = 3<br><br>$\Rightarrow$ a(1 + r) = 3 ...(1)<br><br>Also $\gamma$ + $\delta$ = 6<br><br>$\Rightarrow$ ar² + ar³ = 6<br><br>$\Rightarrow$ ar² (1 + r) = 6 ... (2)<br><br>Dividing (2) by (1),<br><br>r² = 2<br><br>$\Rightarrow r = \sqrt 2$<br><br>$\therefore$ $a = {3 \over {1 + \sqrt 2 }}$<br><br>So, $\alpha = {3 \over {1 + \sqrt 2 }}$, $\beta = {{3\sqrt 2 } \over {1 + \sqrt 2 }}$, $\gamma = {{3 \times 2} \over {1 + \sqrt 2 }}$, $\delta = {{3(2\sqrt {2)} } \over {1 + \sqrt 2 }}$<br><br>$\therefore$ $p = \alpha \beta = {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}$<br><br>and $q = \gamma \delta = {{36\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}$<br><br>Now, ${{2q + p} \over {2q - p}}$<br><br>$$ = {{{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} + {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}} \over {{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} - {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}}}$$<br><br>$= {{72\sqrt 2 + 9\sqrt 2 } \over {72\sqrt 2 - 9\sqrt 2 }}$<br><br>$= {{81} \over {63}}$<br><br>$= {9 \over 7}$