SUBIECTUL 1

Determine the rational numbers $x$ and $y$ knowing that:

$16.5-\left(\frac{|x-1|}{\sqrt{2}}-\left|y^{2}-2\right|\right)^{2}=10 \sqrt{2}$

S.G.M. 2014
## SUBJECT 1

Determine the rational numbers $x$ and $y$ knowing that: $16.5 - \left(\frac{|x-1|}{\sqrt{2}} - |y^2 - 2|\right)^2 = 10 \sqrt{2}$.

| The equation is equivalent to: $\left(\frac{|x-1|}{\sqrt{2}} - |y^2 - 2|\right)^2 = 16.5 - 10 \sqrt{2}$ | $1 \text{p}$ |
| :--- | :---: |
| From which $\frac{|x-1|}{\sqrt{2}} - |y^2 - 2| = \pm \sqrt{16.5 - 10 \sqrt{2}}$ | $1 \text{p}$ |
| Or: $\frac{|x-1|}{\sqrt{2}} - |y^2 - 2| = \pm \left(\frac{5}{\sqrt{2}} - 2\right)$ | $1 \text{p}$ |
| The equation $\frac{|x-1|}{\sqrt{2}} - |y^2 - 2| = -\frac{5}{\sqrt{2}} + 2 \Leftrightarrow \frac{|x-1| + 5}{\sqrt{2}} = |y^2 - 2| + 2$ has no rational solutions. | $1 \text{p}$ |
| The equation $\frac{|x-1|}{\sqrt{2}} - |y^2 - 2| = \frac{5}{\sqrt{2}} - 2 \Leftrightarrow \frac{|x-1| - 5}{\sqrt{2}} = |y^2 - 2| - 2$. | $1 \text{p}$ |
| Since $\sqrt{2}$ is irrational, we have: $|x-1| - 5 = 0$ and $|y^2 - 2| - 2 = 0$. From which $x \in \{-4, 6\}$ and $y \in \{-2, 0, 2\}$ | $1 \text{p}$ |
| We obtain $S = \{(-4, -2), (-4, 0), (-4, 2), (6, -2), (6, 0), (6, 2)\}$. | $1 \text{p}$ |