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 $|\alpha -\beta |$ 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

The correct option is D $\frac{2\sqrt{13}}{9}$

Given $p{x}^2+qx+r=0,p\ne 0,\frac{1}{\alpha }+\frac{1}{\beta }=4$

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

$2q=p+r$

Divide by $p$

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

where $-\frac{q}{p}=\alpha +\beta ,\frac{r}{p}=\alpha \beta$

$\Rightarrow -2(\alpha +\beta )=1+\alpha \beta$

$\Rightarrow -2(\frac{1}{\alpha }+\frac{1}{\beta })=\frac{1}{\alpha \beta }+1$

$\Rightarrow \frac{1}{\alpha \beta }=-9$

Equation having roots $\alpha ,\beta$ is $9x^2+4x-1=0$

$\alpha ,\beta =\frac{-4\pm \sqrt{16+36}}{2\times 9}$

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