Let and be the roots of the equation px^2+qx-r=0, where p0.

Question

Let $\alpha$ and $\beta$ be the roots of the equation $px^2+qx-r=0$, where $p
eq0$. If $p,q,r$ are consecutive terms of a non-constant G.P. and $\dfrac1\alpha+\dfrac1\beta=\dfrac34$, then the value of $(\alpha-\beta)^2$ is:

Accepted Answer

Since $p,q,r$ are in G.P., $q=pk,\ r=pk^2$ for the common ratio $k$.

$$\alpha+\beta=-\frac{q}{p}=-k, \qquad \alpha\beta=-\frac{r}{p}=-k^2.$$

$$\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{-k}{-k^2}=\frac1k=\frac34 \implies k=\frac43.$$

So $\alpha+\beta=-\dfrac43,\ \alpha\beta=-\dfrac{16}{9}$.

$$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\frac{16}{9}+\frac{64}{9}=\frac{80}{9}.$$

Solutions · 1