Question

If $a+b=0$ & $ab=3$, find $\frac{1}{b^2}+\frac{1}{a^2}$.

• A. $1\frac{1}{9}$
• B. $1\frac{2}{9}$
• C. $2\frac{1}{9}$
• D. $1\frac{1}{7}$

Answer 1

The correct option is A $1\frac{1}{9}$

Given, $a+b=0$ and $ab=3$
$(a+b)=4$
Squaring both sides, we get
$a^2+2ab+b^2=16$
$=>a^2+2\left(3\right)+b^2=16$
$=>a^2+b^2=16-6$
$=>a^2+b^2=10$
Now,
$\frac{1}{b^2}+\frac{1}{a^2}$
$=\frac{a^2+b^2}{a^2{b}^2}$
$=\frac{10}{9}$
$=1\frac{1}{9}$