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Q.
Let $\alpha$ and $\beta$ be the roots of equation $px^2+qx+r=0,$ $p\,\ne\,0.$ If $p, q$ and $r$ are in AP and $\frac{1}{\alpha}+\frac{1}{\beta}=4,$ then the value of $| \alpha-\beta|$ is
JEE MainJEE Main 2014Complex Numbers and Quadratic Equations
Solution:
Given, $\alpha$ and $\beta$ are roots of $p x^{2}+q x+r=0, p \neq 0$. $\therefore \alpha+\beta=\frac{-q}{p}, \alpha \beta=\frac{r}{p} ....$(i)
Since, $p, q$ and $r$ are in AP.
$\therefore 2 q=p+r ......$(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{4 r}{p} $ [from Eq. (i)]
$\Rightarrow q=-4 r$
On putting the value of $q$ in Eq. (ii), we get
$\Rightarrow 2(-4 r)=p+r \Rightarrow p=-9 r $
Now, $ \alpha+\beta=\frac{-q}{p}=\frac{4 r}{p}=\frac{4 r}{-9 r}=-\frac{4}{9}$
and $ \alpha \beta=\frac{r}{p}=\frac{r}{-9 r}=\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} $
$\Rightarrow |\alpha-\beta|=\frac{2}{9} \sqrt{13}$