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 bx + 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

∵ Roots of 2 ax 2 – 8 ax + 1 = 0 are $\frac{1}{p}$ and $\frac{1}{r}$ and roots of 6 bx 2 + 12 bx + 1 = 0 are $\frac{1}{q}$ and $\frac{1}{s}$.
Let
$\frac{1}{p}$,$\frac{1}{q}$,$\frac{1}{r}$,$\frac{1}{s}$
as α – 3β, α – β, α + β, α + 3β
So sum of roots 2α – 2β = 4 and 2α + 2β = – 2
Clearly
α=$\frac{1}{2}$ and β=−32
Now product of roots,
$\frac{1}{p}$⋅$\frac{1}{r}$=$\frac{1}{2}$a=−5⇒$\frac{1}{a}$=−10
and
$\frac{1}{q}$⋅$\frac{1}{x}$=$\frac{1}{6a}$=−8⇒$\frac{1}{b}$=−48
So,
$\frac{1}{a}$−$\frac{1}{b}$=38