Question

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

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

Answer 1

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

Let p,q,r are in AP
$\Rightarrow 2q=p+r....\left(i\right)$
Given $\frac{1}{\alpha }+\frac{1}{\beta }=4\Rightarrow =\frac{\alpha +\beta }{\alpha \beta }=4$
We have $\alpha +\beta =-\frac{q}{p}$ and $\alpha \beta =\frac{r}{p}$
$\Rightarrow \frac{-\frac{q}{p}}{\frac{r}{p}}$ = 4 $\Rightarrow q=-4r$
From (i), we have
$2(-4r)=p+r\Rightarrow p=-9r$
$q=-4r$
Now $|\alpha -\beta |=\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$
= $\sqrt{(\frac{-q}{p}{)}^2-\frac{4r}{p}}=\frac{\sqrt{q^2-4pr}}{|p|}$
= $\frac{\sqrt{16r^2+36r^2}}{|-9r|}=\frac{2\sqrt{13}}{9}$