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

Answer 1

Answer: (B)

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

$\to$ sum of roots, $\alpha +\beta =-\frac{q}{b}$

Product of roots, $\alpha \beta =\frac{r}{p}$

$\frac{1}{\alpha }+\frac{1}{\beta }=4$

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

$-\frac{q}{b}=4\left(\frac{r}{p}\right)$

$4r=-q$

$\Rightarrow r-q=d=5r\{d=differenceofA.P.\}$

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

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

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

$=\frac{16r^2-4r(r-10r)}{{(r-10r)}^2}$

$=\frac{16r^2+36r^2}{81r^2}=\frac{52}{81}=\frac{13\times 4}{81}$

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