1. Let $x$ and $y$ be real numbers with $x>y$ such that $x^{2} y^{2}+x^{2}+y^{2}+2 x y=40$ and $x y+x+y=8$. Find the value of $x$.
Answer: $\quad 3+\sqrt{7}$ We have $(x y)^{2}+(x+y)^{2}=40$ and $x y+(x+y)=8$. Squaring the second equation and subtracting the first gives $x y(x+y)=12$ so $x y, x+y$ are the roots of the quadratic $a^{2}-8 a+12=0$. It follows that $\{x y, x+y\}=\{2,6\}$. If $x+y=2$ and $x y=6$, then $x, y$ are the roots of the quadratic $b^{2}-2 b+6=0$, which are non-real, so in fact $x+y=6$ and $x y=2$, and $x, y$ are the roots of the quadratic $b^{2}-6 b+2=0$. Because $x>y$, we take the larger root, which is $\frac{6+\sqrt{28}}{2}=3+\sqrt{7}$.