Question

Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x ^2-8 ax +2 a =0$ and $q$ and $s$ are the roots of the equation $x^2+12 b x+6 b=0$, such that $\frac{1}{ p }, \frac{1}{ q }, \frac{1}{ r }, \frac{1}{ s }$ are in A.P., then $a ^{-1}- b ^{-1}$ is equal to ______

Answer 1

$ x^2-8 a x+2 a=0 $
$p+r=8 a $
$p r=2 a$
$ \frac{1}{p}+\frac{1}{r}=4 $
$\frac{2}{q}=4 $
$ q=\frac{1}{2} $
$ p=\frac{1}{5}$
$ x^2+12 b x+6 b=0 $
$ q+s=-12 b$
$ q s=6 b $
$\frac{1}{q}+\frac{1}{s}=-2$
$ \frac{2}{r}=-2$
$ r =-1 $
$ s=\frac{-1}{4}$
Now, $\frac{1}{ a }-\frac{1}{ b }=\frac{2}{ pr }-\frac{6}{ qs }=38$