If x + y = 8 and xy = 12, what is the value of x² + y²?

Use $x^2 + y^2 = (x+y)^2 - 2xy$, which comes from expanding $(x+y)^2 = x^2 + 2xy + y^2$. Substituting $x + y = 8$ and $xy = 12$ gives $64 - 24 = 40$. Check with actual numbers: the pair solves $t^2 - 8t + 12 = 0$, so $x$ and $y$ are 2 and 6, and $4 + 36 = 40$. The most common wrong answer is 64, which results from squaring the sum and forgetting to subtract $2xy$. Another is 52, from subtracting $xy$ only once, and a third is 88, from adding $2xy$ instead of subtracting it.