Question

If the values of $a+b$ and $ab$ are $12$ and $32$ respectively, find the values of $a^2+b^2$ and $(a-b{)}^2$

Answer 1

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

$144=a^2+b^2+2\left(32\right)$

$a^2+b^2=80$

$(a-b{)}^2=(a+b{)}^2-4ab$

$a-b=\sqrt{144-128}$

$a-b=\pm 4$