Question

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

• A. $\frac{\sqrt{52}}{9}$
• B. $\frac{2\sqrt{17}}{9}$
• C. $\frac{\sqrt{34}}{9}$
• D. $\frac{2\sqrt{13}}{9}$

Answer 1

The correct option is A $\frac{\sqrt{52}}{9}$

As we know,

Sum of roots $(\alpha +\beta )=\frac{-q}{p}$

Product of roots $\left(\alpha \beta \right)=\frac{r}{p}$

Given that:

$p,q,r$ are in $AP$ so,

$2q=p+r$ .....(1)

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

$\alpha +\beta =4\alpha \beta$ ......(2)

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

$q=-4r,$

$p=-9r$ (from equation 1)

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

$=16{\alpha }^2{\beta }^2-4\alpha \beta$

$=4\alpha \beta (4\alpha \beta -1)$

$=4\frac{r}{p}(4\frac{r}{p}-1)$

$=\frac{4r}{-9r}(\frac{4r}{-9r}-1)$

$=\frac{-4}{9}\left(\frac{-4-9}{9}\right)$

$=\frac{52}{81}$

$(\alpha -\beta )=\frac{\sqrt{52}}{9}$

$|\alpha -\beta |=\frac{\sqrt{52}}{9}.$