Question

Let $a,b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^2-8ax+2a=0$ and $q$ and $s$ are the roots of the equation $x^2+12bx+6b=0$, 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

Answer: 38

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