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

Question

Let $\alpha$ and $\beta$ be the roots of the equation $px^2+qx+r=0,\ p
e0$. If $p,q,r$ are in A.P. and $\dfrac1\alpha+\dfrac1\beta=4$, then the value of $|\alpha-\beta|$ is:

Accepted Answer

$p,q,r$ in A.P.: $2q=p+r$ ... (i)

$\dfrac1\alpha+\dfrac1\beta=\dfrac{\alpha+\beta}{\alpha\beta}=4$. With $\alpha+\beta=-q/p,\ \alpha\beta=r/p$: $\dfrac{-q}{r}=4\Rightarrow q=-4r$ ... (ii)

From (i): $2(-4r)=p+r\Rightarrow p=-9r$ ... (iii)

$$|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4\alpha\beta}=\sqrt{\frac{q^2}{p^2}-\frac{4r}{p}}=\frac{\sqrt{q^2-4pr}}{|p|}=\frac{\sqrt{16r^2+36r^2}}{9|r|}=\frac{\sqrt{52}}{9}=\frac{2\sqrt{13}}{9}.$$

Solutions · 1