Let a,b be two non-zero real numbers.

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 $\dfrac1p,\dfrac1q,\dfrac1r,\dfrac1s$ are in A.P., then $a^{-1}-b^{-1}$ is equal to

Answer: 38

Accepted Answer

$\dfrac1p,\dfrac1r$ are roots of $2ax^2-8ax+1=0$, and $\dfrac1q,\dfrac1s$ are roots of $6bx^2+12bx+1=0$.

Write the four A.P. terms as $\alpha-3\delta,\alpha-\delta,\alpha+\delta,\alpha+3\delta$ (in some order matching $\frac1p,\frac1q,\frac1r,\frac1s$). Matching sums of the two pairs: $2\alpha-2\delta=4$ (from the first quadratic's root-sum $=8a/(2a)=4$) and $2\alpha+2\delta=-2$ (from the second's, $=-12b/(6b)=-2$), giving $\alpha=\dfrac12,\ \delta=-\dfrac32$.

Product relations: $\dfrac1p\cdot\dfrac1r=\dfrac1{2a}=-5 \implies \dfrac1a=-10$; and $\dfrac1q\cdot\dfrac1s=\dfrac1{6b}=-8 \implies \dfrac1b=-48$.

$$\frac1a-\frac1b=-10-(-48)=38.$$

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