Question:
If x + y = 4 and xy = 2, find the value of x2 + y2
Solution:
We have:
$(x+y)^{2}=x^{2}+2 x y+y^{2}$
$\Rightarrow x^{2}+y^{2}=(x+y)^{2}-2 x y$
$\Rightarrow x^{2}+y^{2}=4^{2}-2 \times 2 \quad(\because x+y=4$ and $x y=2)$
$\Rightarrow x^{2}+y^{2}=16-4$
$\Rightarrow x^{2}+y^{2}=12$
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