Question

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

• A. $\frac{\sqrt{61}}{9}$
• B. $\frac{2\sqrt{17}}{9}$
• C. $\frac{\sqrt{34}}{9}$
• D. $\frac{2\sqrt{13}}{9}$

Answer 1

The correct option is D $\frac{2\sqrt{13}}{9}$

Given,
$p{x}^2+qx+r=0$
$\Rightarrow \alpha +\beta =-\frac{q}{r},\alpha \cdot \beta =\frac{r}{p}$
Now, $\frac{1}{\alpha }+\frac{1}{\beta }=4$
$\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{q}{p}}{\frac{r}{p}}=-\frac{q}{r}=4$
$\Rightarrow q=-4r...\left(1\right)$
As given, $p,q,r$ are in A.P.
$\therefore 2q=p+r...\left(2\right)$
from equation (1) and (2)
$\Rightarrow 2(-4r)=p+r$
$\Rightarrow p=-9r$
Now, $(\alpha -\beta {)}^2=\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$
$=\sqrt{{(-\frac{q}{p})}^2-4\frac{r}{p}}$
$=\sqrt{\frac{16r^2+4\cdot 9r\cdot r}{(9r{)}^2}}$
$\{\because q=-4r,p=-9r\}$
$=\frac{2\sqrt{13}}{9}$