Question #9cf7b

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Question #9cf7b

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Answer 1 · 2015-02-23T20:17:05

Hello,

Answer.

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

or

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

Explanation

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

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

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

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

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Question #9cf7b

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