Question

Let $\alpha and\beta$ be the roots of the equation ${px}^2+qx+r=0$. If p, q and r are in AP and $\frac{1}{\alpha }+\frac{1}{\beta }=4$, then value of $|\alpha -\beta |$ is

Answer 1

Given,

$\frac{1}{\alpha }+\frac{1}{\beta }=4$

$\frac{1}{\alpha }+\frac{1}{\beta }$$=\frac{\alpha +\beta }{\alpha \beta }$

$=-\frac{b}{c}$$=-\frac{q}{r}$

$\Rightarrow -\frac{q}{r}=4$

$\therefore q=-4r$

$p,q,r$ are in A.P.

$\Rightarrow 2q=p+r$

$2(-4r)=p+r$

$\therefore p=-9r$

$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$

$={(-\frac{q}{p})}^2-4\frac{r}{p}$

$=\frac{q^2-4pr}{p^2}$

$=\frac{(-4r{)}^2-4(-9r)r}{(-9r{)}^2}$

$\therefore (\alpha -\beta {)}^2=\frac{52}{81}$

$\Rightarrow (\alpha -\beta )=\pm \frac{2\sqrt{13}}{9}$

$\therefore |\alpha -\beta |=\frac{2\sqrt{13}}{9}$