Answer:
Option D
Explanation:
If $ax^{2}+bx+c=0$ , has roots $\alpha$ and $\beta$ , then
$\alpha+\beta=\frac{-b}{a}$ and $\alpha\beta=\frac{c}{a}$ . Find the valuesw of $\alpha+\beta$ and $\alpha \beta$ and then put in $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta$ to get required value.
Given $\alpha $ and $\beta$ are roots of
$px^{2}+qx+r=0$ , $p\neq0$
$\therefore$ $\alpha+\beta$ = $\frac{-q}{p}$ , $\alpha \beta$= $\frac{r}{p}$ ......(i)
Since p,q and r are in A.P
$\therefore$ 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 $\alpha\beta=\frac{r}{p}=\frac{r}{-9r}=\frac{1}{-9}$
$\therefore$ $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta=\frac{16}{81}+\frac{4}{9}=\frac{16+36}{81}$
$\Rightarrow$ $(\alpha-\beta)^{2}=\frac{52}{81}\Rightarrow|\alpha-\beta|=\frac{2}{9}\sqrt{31}$
