# 18.  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.

S seema garhwal

Given: a and b are the roots of  $x^2 -3 x + p = 0$

Then, $a+b=3\, \, \, \, and\, \, \, \, ab=p......................1$

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

$c+d=12\, \, \, \, and\, \, \, \, cd=q......................2$

Given: a, b, c, d form a G.P

Let, $a=x,b=xr,c=xr^2,d=xr^3$

From 1 and 2, we get

$x+xr=3$                      and           $xr^2+xr^3=12$

$\Rightarrow x(1+r)=3$                              $xr^2(1+r)=12$

On dividing them,

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

$\Rightarrow r^2=4$

$\Rightarrow r=\pm 2$

When , r=2 ,

$x=\frac{3}{1+2}=1$

When , r=-2,

$x=\frac{3}{1-2}=-3$

CASE (1) when r=2 and x=1,

$ab=x^2r=2\, \, \, and \, \, \, \, cd=x^2r^5=32$

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

i.e. (q + p) : (q – p) = 17:15.

CASE (2) when r=-2 and x=-3,

$ab=x^2r=-18\, \, \, and \, \, \, \, cd=x^2r^5=-288$

$\therefore \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-305}{-270}=\frac{17}{15}$

i.e. (q + p) : (q – p) = 17:15.

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