Question

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

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

Answer 1

Answer: (A)

$\frac{2\sqrt{13}}{9}$

GIven $\alpha ,\beta$ are the roots of the equation $p{x}^2+qx+r=0$, $p\ne 0$

$\Rightarrow \alpha +\beta =-\frac{q}{p}$ and $\Rightarrow \alpha \beta =\frac{r}{p}$

$\Rightarrow {|\alpha -\beta |}^2={\alpha }^2+{\beta }^2-2\alpha \beta$

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

$=\frac{q^2}{p^2}-4\frac{r}{p}$

$={\left(\frac{q}{p}\right)}^2-4\left(\frac{r}{p}\right)$ ...$\left(1\right)$

Also, $\frac{2q}{p}=1+\frac{r}{p}$ ...$\left(2\right)$

Given that, $\frac{1}{\alpha }+\frac{1}{\beta }=4$

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

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

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

Given $p,q,r$ are in A.P.

$\Rightarrow 2q=p+r$

$\Rightarrow \frac{2q}{r}=\frac{p}{r}+1$

$\Rightarrow \frac{p}{r}=\frac{2q}{r}-1$

$\Rightarrow \frac{p}{r}=2(-4)-1$ $...$$(\because \frac{q}{r}=-4)$

$\Rightarrow \frac{p}{r}=-9$

$\Rightarrow \frac{r}{p}=-\frac{1}{9}$

Substitute tha value of $\frac{r}{p}$ in equation $\left(2\right)$, we get,

$\frac{2q}{p}=1-\frac{1}{9}$

$\Rightarrow \frac{q}{p}=\frac{1}{2}(1-\frac{1}{9})$

$\Rightarrow \frac{q}{p}=\frac{1}{2}\left(\frac{9-1}{9}\right)$

$\Rightarrow \frac{q}{p}=\frac{1}{2}\cdot \frac{8}{9}$

$\Rightarrow \frac{q}{p}=\frac{4}{9}$

Now, substitute the values of $\frac{q}{p}$ and $\frac{r}{p}$ in the equation $\left(1\right)$, we get

$\Rightarrow {|\alpha -\beta |}^2={\left(\frac{4}{9}\right)}^2-4(-\frac{1}{9})$

$\Rightarrow {|\alpha -\beta |}^2={\left(\frac{4}{9}\right)}^2+\frac{4}{9}$

$\Rightarrow {|\alpha -\beta |}^2=\frac{4}{9}\left(\frac{13}{9}\right)$

$\Rightarrow {|\alpha -\beta |}^2=\frac{52}{81}$

$\Rightarrow |\alpha -\beta |=\frac{2\sqrt{13}}{9}$