If in the equation a{x^2} + bx + c = 0, the s | vedclass

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Question

(a) $\alpha + \beta = \frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} = $ $\frac{{{\alpha ^2} + {\beta ^2}}}{{{{(\alpha \,\beta )}^2}}}$

$ = \frac

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$A.P.$

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(a) $\alpha + \beta = \frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} = $ $\frac{{{\alpha ^2} + {\beta ^2}}}{{{{(\alpha \,\beta )}^2}}}$

$ = \frac{{{{(\alpha + \beta )}^2} - 2\alpha \beta }}{{{{(\alpha \,\beta )}^2}}}$&hellip;..(i)

$\alpha + \beta = - b/a$and $\alpha \beta = c/a$

Putting these value in (i)

==> $\left( {\frac{{ - b}}{a}} \right)\,\left( {\frac{{{c^2}}}{{{a^2}}}} \right) = \frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a}$

or $ - b{c^2} = a{b^2} - 2c{a^2}$or $2c\,{a^2} = a{b^2} + b{c^2}$

Dividing by abc we get, $\frac{{2a}}{b} = \frac{b}{c} + \frac{c}{a}$

==> $\frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in $A.P.$

If in the equation $a{x^2} + bx + c = 0,$ the sum of roots is equal to sum of square of their reciprocals, then $\frac{c}{a},\frac{a}{b},\frac{b}{c}$ are in

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