Question

If $p,q$ are the roots of the equation $a{x}^2+bx+c=0,$ then the value of $(ap+b{)}^{-2}+(aq+b{)}^{-2}$ is

• A. $\frac{b^2+2ac}{a^2{c}^2}$
• B. $\frac{b^2-2ac}{a^2{c}^2}$
• C. $\frac{b^2-2ac}{ac}$
• D. $\frac{b^2-4ac}{a^2{c}^2}$

Answer 1

The correct option is B $\frac{b^2-2ac}{a^2{c}^2}$

Given: $a{x}^2+bx+c=0,$ and it's roots are $p,q.$

We know, sum of roots $=p+q=-\frac{b}{a}$
And product of roots $=p\times q=\frac{c}{a}$

To find:
$(ap+b{)}^{-2}+(aq+b{)}^{-2}=\frac{1}{(ap+b{)}^2}+\frac{1}{(aq+b{)}^2}=\frac{a^2{q}^2+b^2+2abq+a^2{p}^2+b^2+2abp}{(a^{2}pq+abq+abp+b^2{)}^2}=\frac{a^2(p^2+q^2)+2ab(p+q)+2b^2}{(a^{2}pq+ab(p+q)+b^2{)}^2}=\frac{a^2\left[\right(p+q{)}^2-2pq]+2ab(p+q)+2b^2}{(a^{2}pq+ab(p+q)+b^2{)}^2}=\frac{a^2[(-\frac{b}{a}{)}^2-2\times \frac{c}{a}]+2ab(-\frac{b}{a})+2b^2}{(a^2\frac{c}{a}+ab(-\frac{b}{a})+b^2{)}^2}=\frac{b^2-2ac}{a^2{c}^2}$