If $x + y = 12$ and $x y = 32$ , find $x^{2} + y^{2}$ .

Open in App

Solution

Given that: $x + y = 12$ and $x y$ = 32
Using the identity, $(x + y)^{2}$ = $x^{2} + 2 x y + y^{2}$
$\Rightarrow$ $(12)^{2}$ = $x^{2} + 64 + y^{2}$
$\Rightarrow$ $144$ = $x^{2} + y^{2} + 64$
$\Rightarrow$ $144 - 64$ = $x^{2} + y^{2}$
$\Rightarrow$ $80$ = $x^{2} + y^{2}$
The answer is $x^{2} + y^{2}$ = $80$

Suggest Corrections

Similar questions

If $x + y = 12$ and $x y = 32$ , find the value of $x^{2} + y^{2}$ .

x + y = 7

x y = 12

x^{2} + y^{2}

If x2+xy+x=12 and y2+xy+y=18 then find the value of x-y is

x^{2} + y^{2} - x y = 3

y - x = 1

\frac{x y}{x^{2} + y^{2}}

Join BYJU'S Learning Program