1. [3] Suppose that $x$ and $y$ are positive reals such that
$$
x-y^{2}=3, \quad x^{2}+y^{4}=13 .
$$

Find $x$.
Answer: $\frac{3+\sqrt{17}}{2}$ Squaring both sides of $x-y^{2}=3$ gives $x^{2}+y^{4}-2 x y^{2}=9$. Subtract this equation from twice the second given to get $x^{2}+2 x y^{2}+y^{4}=17 \Longrightarrow x+y^{2}= \pm 17$. Combining this equation with the first given, we see that $x=\frac{3 \pm \sqrt{17}}{2}$. Since $x$ is a positive real, $x$ must be $\frac{3+\sqrt{17}}{2}$.