JEE MAIN - Mathematics (2014 (Offline) - No. 16)
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 :
$${{\sqrt {34} } \over 9}$$
$${{2\sqrt 13 } \over 9}$$
$${{\sqrt {61} } \over 9}$$
$${{2\sqrt 17 } \over 9}$$
Explanation
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 $$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $$
$$ = \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}} $$
$$ = {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}$$
$$ = {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}$$
$$ = {{2\sqrt {13} } \over 9}$$
$$ \Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
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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