Let \[\alpha \] and \[\beta \] be the roots of equation px2+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 :
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Official solution
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 \[\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } \]
\[ = \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}} \]
\[ = {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}\]
\[ = {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}\]
\[ = {{2\sqrt {13} } \over 9}\]