Question

If $a^2+b^2=13$ and $ab=6$, find $a-b$.

• A. $\pm 1$
• B. $\pm 3$
• C. $\pm 9$
• D. $\pm 5$

Answer 1

The correct option is A $\pm 1$

$(a-b{)}^2=a^2+b^2-2ab$
$=13-2\left(6\right)$
$=13-12$
$=1$
$=>(a-b)=\pm 1$