JEE-Mains 2014 (4/30)

Algebra Level 3

Let $\alpha$ and $\beta$ be the roots of equation $px^2+qx+r \ , \ p \neq 0$ . If $p,q,r$ are in arithmetic progression and $\frac{1}{\alpha}+\frac{1}{\beta}=4$ , then the value of $|\alpha-\beta|$ is :

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

\frac{2 \sqrt{17} }{9}

\frac{\sqrt{34} }{9}

\frac{\sqrt{61}}{9}

1 solution

Rohit Ner
Jun 15, 2015

$\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =4\\ \Rightarrow \frac { \alpha +\beta }{ \alpha \beta } =4\\ \Rightarrow \frac { { -q }/{ p } }{ { r }/{ p } } =4\Rightarrow q=-4r$
Since $p,q$ & $r$ are in arithematic progression, $2q=p+r \\ -8r=p+r \\ \Rightarrow p=-9r$ .
Difference of roots of the equation is given by: $\frac { \sqrt { { q }^{ 2 }-4pr } }{ \left| p \right| } \\ =\frac { \sqrt { 16{ r }^{ 2 }+36{ r }^{ 2 } } }{ \left| -9r \right| } \\ \Huge \color{#3D99F6}{ =\boxed{\frac { 2\sqrt { 13 } }{ 9 } }}$

JEE-Mains 2014 (4/30)

1 solution

0 pending reports