If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$ where $a, b, c, d$ form a $G.P.$ Prove that $(q+p):(q-p)=17: 15$

It is given that $a$ and $b$ are the roots of $x^{2}-3 x+p=0$

$\therefore a+b=3$ and $a b=p$          .......$(1)$

Also, $c$ and $d$ are the roots of $x^{2}-12 x+q=0$

$\therefore c+d=12$ and $c d=q$           .........$(2)$

It is given that $a, b, c, d$ are in $G.P.$

Let $a=x, b=x r, c=x r^{2}, d=x r^{3}$

From $(1)$ and $(2)$

We obtain $x+x y=3 \Rightarrow x(1+r)=3$

$x r^{2}+x^{3}=12$

$\Rightarrow x r^{2}(1+r)=12$

On dividing, we obtain

$\frac{x r^{2}(1+r)}{x(1+r)}=\frac{12}{3}$

$\Rightarrow r^{2}=4$

$\Rightarrow r=\pm 2$

When $r=2, x=\frac{3}{1+2}=\frac{3}{3}=1$

When $r=-2, x=\frac{3}{1-2}=\frac{3}{-1}=-3$

Case $I:$ When $r=2$ and $x=1, \quad a b=x^{2} r=2, \quad c d=x^{2} r^{5}=32$

$\therefore \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}$

i.e., $(q+p):(q-p)=17: 15$

Case $II:$

When $r=-2, x=-3, a b=x^{2} r=-18, c d=x^{2} r^{5}=-288$

$\therefore \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}$

i.e., $(q+p):(q-p)=17: 15$

Thus, in both the cases, we obtain $(q+p):(q-p)=17: 15$