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If the sum of the roots of the equation $a{x^2} + bx + c = 0$ be equal to the sum of the reciprocals of their squares, then $b{c^2},\;c{a^2},\;a{b^2}$ will be in
$A.P.$
$G.P.$
$H.P.$
None of these
Solution
(a) Given equation $a{x^2} + bx + c = 0$ and let the roots are $\alpha ,\;\beta ,$ so $\alpha + \beta = – \frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$
Now $\frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} = \frac{{{\alpha ^2} + {\beta ^2}}}{{{\alpha ^2}{\beta ^2}}} = \frac{{\frac{{{b^2}}}{{{a^2}}} – \frac{{2c}}{a}}}{{\frac{{{c^2}}}{{{a^2}}}}} = \frac{{{b^2} – 2ac}}{{{c^2}}}$
Under condition $\alpha + \beta = \frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}}$
$ \Rightarrow – \frac{b}{a} = \frac{{{b^2} – 2ac}}{{{c^2}}} \Rightarrow – b{c^2} = a{b^2} – 2{a^2}c$
Hence$2{a^2}c = a{b^2} + b{c^2} \Rightarrow a{b^2},\;c{a^2},\;b{c^2}$
or $b{c^2},\;c{a^2},\;a{b^2}$ be in $A.P.$