Let \alpha and \beta be the roots of equation p{x^2} + qx + r = 0, p
e 0. If p,\,q,\,r in A.P. and {1 \over \alpha } + {1 \over \beta } = 4, then the value of \left| {\alpha - \beta } ight| is :

Question

Let \alpha and \beta be the roots of equation p{x^2} + qx + r = 0, p 
e 0. If p,\,q,\,r in A.P. and {1 \over \alpha } + {1 \over \beta } = 4, then the value of \left| {\alpha - \beta } ight| is :

Accepted Answer

Answer: B. {{2\sqrt 13 } \over 9} 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}

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 :

Answer

Explanation

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