- Home
- Standard 11
- Mathematics
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
$A.P.$
$G.P.$
$H.P.$
None of these
Solution
(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}}}$…..(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 $A.P.$
