Question

If $(a+b)=6$, and $ab=2$, then the value of $\frac{1}{a^2}+\frac{1}{b^2}$ is equal to

Answer 1

Given $(a+b)=6$

Squaring both sides we get

$(a+b{)}^2=36$

$a^2+b^2+2ab=36$

$a^2+b^2+2\times 2=36(\text{as}ab=2)$

$a^2+b^2=36-4$

$a^2+b^2=32$ . . . (1)

The given expression is

$\frac{1}{a^2}+\frac{1}{b^2}$

$=\frac{a^2+b^2}{a^2{b}^2}$

$=\frac{(a+b{)}^2-2ab}{a^2{b}^2}$

$=\frac{(6{)}^2-2\times (2)}{(2{)}^2}$

$=\frac{36-4}{4}$

$=8$