# Question #9cf7b

##### 1 Answer
Feb 23, 2015

Hello,

Answer.

• $x = 2 - 2 \sqrt{2}$ and $y = 6 - 4 \sqrt{2}$

or

• $x = 2 + 2 \sqrt{2}$ and $y = 6 + 4 \sqrt{2}$.

Explanation

You have to solve $2 x + 2 = {x}^{2} - 2$, so ${x}^{2} - 2 x - 4 = 0$.

Remark that ${\left(x - 2\right)}^{2} = {x}^{2} - 2 x + 4$, therefore :
${x}^{2} - 2 x - 4 = {\left(x - 2\right)}^{2} - 8$.

So, you have to solve ${\left(x - 2\right)}^{2} = 8$.
Solutions are $x - 2 = \setminus \pm \sqrt{8} = \pm 2 \sqrt{2}$, it means
$x = 2 - 2 \sqrt{2}$ or $x = 2 + 2 \sqrt{2}$.

Finally, calculate corresponding $y$ for each value of $x$ : because $y = 2 x + 2$, it's easy !.