QJEEM14Mad5b9: JEE Main 2014 Mathematics - Class 11 JEEM14Mad5b9 Solution: Let and be the roots of equation If in A.P. and then the value…

Question

Verified Solution • 2025–26 Session moderate

Let

\alpha

and

\beta

be the roots of equation

p{x^2} + qx + r = 0,

p \ne 0.

If

p,\,q,\,r

in A.P. and

{1 \over \alpha } + {1 \over \beta } = 4,

then the value of

\left| {\alpha - \beta } \right|

is :

Accepted Answer

Let

p,q,r

are in

AP

\Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Given

{1 \over \alpha } + {1 \over \beta } = 4 \Rightarrow {{\alpha + \beta } \over {\alpha \beta }} = 4

We have

\alpha + \beta = - q/p

and

\alpha \beta = {r \over p}

\Rightarrow {{ - {q \over p}} \over {{r \over p}}} = 4 \Rightarrow q = - 4r\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(i),

we have

2\left( { - 4r} \right) = p + r \Rightarrow p = - 9r

q = - 4r

Now

\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta }

= \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}}

= {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}

= {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}

= {{2\sqrt {13} } \over 9}

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