Question

If $x-y=3$ and $xy=10$, find $x^2+y^2$.

Answer 1

Given that: $x-y=3$ And $xy$ = $10$
Using the given identity: $(a-b{)}^2$ = $a^2-2ab+b^2$
So, $(x-y{)}^2$ = $x^2-2xy+y^2$
$\Rightarrow$ $(x-y{)}^2$ = $x^2-2\left(10\right)+y^2$
$\Rightarrow$ $(3{)}^2$ = $x^2-20+y^2$
$\Rightarrow$ $9$ = $x^2+y^2-20$
$\Rightarrow$ $9+20$ =$x^2+y^2$
$\Rightarrow$ $x^2+y^2$ = $29$
The correct answer is $x^2+y^2=29$