HMMT February

Let $y=2-x$, so $x+y=2$ and $x^{4}+y^{4}=34$. We know $$(x+y)^{4}=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}=x^{4}+y^{4}+2 x y(2 x^{2}+2 y^{2}+3 x y) .$$ Moreover, $x^{2}+y^{2}=(x+y)^{2}-2 x y$, so the preceding equation becomes $2^{4}=34+2 x y(2. 2^{2}-x y)$, or $(x y)^{2}-8 x y-9=0$. Hence $x y=9$ or -1 . Solving $x y=9, x+y=2$ produces complex solutions, and solving $x y=-1, x+y=2$ produces $(x, y)=(1+\sqrt{2}, 1-\sqrt{2})$ or $(1-\sqrt{2}, 1+\sqrt{2})$. Thus, $x=1 \pm \sqrt{2}$.