Question

If $(a+b{)}^2=16$ and $ab=3,$ then $(a^2-b^2)$ is ______.

Answer 1

We have $(a+b{)}^2=16$
$=>a^2+b^2+2ab=16$
Subtracting 4ab from both sides,
$=>a^2+b^2+2ab-4ab=16-4ab=>(a-b{)}^2=16-(4\times 3)=>(a-b{)}^2=4=>(a-b)=2$
Since, $(a+b{)}^2=16$
$=>(a+b)=4$
Now, $a^2-b^2=(a+b)(a-b)$
$=>a^2-b^2=4\times 2=>a^2-b^2=8$