Question

If $\alpha ,\beta$ are the roots of the equation $x^2+x-15=0$ then find $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$

Answer 1

Here $a=1,b=1,c=-15$

Sum of the roots$=\alpha +\beta =\frac{-b}{a}=\frac{-1}{1}=-1$

Product of the roots$=\alpha \beta =\frac{c}{a}=\frac{-15}{1}=-15$

Consider, $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ $=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}$

$=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}$

$=\frac{{(-1)}^2-2\times -15}{{(-15)}^2}$

$=\frac{1+30}{225}$

$=\frac{31}{225}$