If $x + y = 4$ and $xy = 2$, find the value of $x^2 + y^2$.

Given:

$x + y = 4$ and $xy = 2$

To do:

We have to find the value of $x^2 + y^2$.

Solution:

The given expressions are $x + y = 4$ and $xy = 2$. Here, we have to find the value of $x^2 + y^2$. So, by squaring the given expression and using the identity $(a+b)^2=a^2+2ab+b^2$, we can find the required value.

$xy = 2$...........(i)

$(a+b)^2=a^2+2ab+b^2$...........(ii)

Now,

$x + y = 4$

Squaring on both sides, we get,

$(x + y)^2 = 4^2$

$x^2+2 \times x \times y+y^2=16$               [Using (ii)]

$x^2+2xy+y^2=16$

$x^2+2(2)+y^2=16$                          [Using (i)]

$x^2+4+y^2=16$

$x^2+y^2=16-4$                 (Transposing 4 to RHS)

$x^2+y^2=12$

Hence, the value of $x^2+y^2$ is $12$.