**Question:**

If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c$ and $d$ are the roots of $x^{2}-$

$12 x+q=0$, where $a, b, c, d$ from a GP, prove that $(q+p):(q-p)=17: 15$

**Solution:**

Given data is,

$x^{2}-3 x+p=0 \rightarrow(1)$

a and b are roots of (1)

So, $(x+a)(x+b)=0$

$x^{2}-(a+b) x+a b=0$

So, $a+b=3$ and $a b=p \rightarrow(2)$

Given data is,

$x^{2}-12 x+q=0 \rightarrow(3)$

c and d are roots of (1)

So, $(x+c)(x+d)=0$

$x^{2}-(c+d) x+c d=0$

So, $c+d=12$ and $c d=q \rightarrow(4)$

a, b, c, d are in GP.(Given data)

Similarly A, AR, AR $^{2}, A R^{3}$ also forms a GP, with common ratio $R$.

From (2)

a + b = 3

A + AR = 3

$\frac{3}{A}=1+R \rightarrow(5)$

From (4),

c + d = 12

$\mathrm{AR}^{2}+\mathrm{AR}^{3}=12$

$\mathrm{AR}^{2}(1+\mathrm{R})=12 \rightarrow(6)$

Substituting value of $(1+R)$ in $(6)$.

R = 2

Now, substitute value of R in (5) to get value of A,

A = 1

Now, the GP required is $A, A R, A R^{2}$, and $A R^{3}$

1, 2, 4, 8…is the required GP.

So,

a = 1, b = 2, c = 4, d = 8

From (2) and (4),

ab = p and cd = q

So, p = 2, and q = 32.

$\frac{q+p}{q-p}=\frac{c d+a b}{c d-a b}=\frac{34}{30}=\frac{17}{15}$

So, $(q+p):(q-p)=17: 15$