Question

Let $\alpha$ and $\beta$ be the roots of equation $p{x}^2+qx+r=0,$ $p\ne 0.$ If $p,q$ and $r$ are in AP 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

Answer: (D)

Given, $\alpha$ and $\beta$ are roots of $p{x}^2+qx+r=0,p\ne 0$. $\therefore \alpha +\beta =\frac{-q}{p},\alpha \beta =\frac{r}{p}....$(i)
Since, $p,q$ and $r$ are in AP.
$\therefore 2q=p+r......$(ii)
Also, $\frac{1}{\alpha }+\frac{1}{\beta }=4\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=4$
$\Rightarrow \alpha +\beta =4\alpha \beta \Rightarrow \frac{-q}{p}=\frac{4r}{p}$ [from Eq. (i)]
$\Rightarrow q=-4r$
On putting the value of $q$ in Eq. (ii), we get
$\Rightarrow 2(-4r)=p+r\Rightarrow p=-9r$
Now, $\alpha +\beta =\frac{-q}{p}=\frac{4r}{p}=\frac{4r}{-9r}=-\frac{4}{9}$
and $\alpha \beta =\frac{r}{p}=\frac{r}{-9r}=\frac{1}{-9}$
$\therefore (\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =\frac{16}{81}+\frac{4}{9}=\frac{16+36}{81}$
$\Rightarrow (\alpha -\beta {)}^2=\frac{52}{81}$
$\Rightarrow |\alpha -\beta |=\frac{2}{9}\sqrt{13}$