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How to solve this system of equations $$\begin{cases} 2x^2+y^2=1,\\ x^2 + y \sqrt{1-x^2}=1+(1-y)\sqrt{x}. \end{cases}$$ I see $(0,1)$ is a root.

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  • $\begingroup$ $(0,1)$ is the only root, found by substituting the first equation in the second as $x^2=\frac{1-y^2}{2}$ and then graphing. $\endgroup$ – wythagoras May 6 '15 at 7:45
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Solution.

First way

From the first equation, we have $$\begin{cases} 2x^2\leqslant 1,\\ y^2 \leqslant 1 \end{cases} \Leftrightarrow \begin{cases} -\dfrac{1}{\sqrt{2}} \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\ - 1 \leqslant y \leqslant 1. \end{cases}$$ Then, the conditions of $x$ and $y$ are $$\begin{cases} 0 \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\ - 1 \leqslant y \leqslant 1. \end{cases}$$ We have $x^2 + y^2 = 1-x^2$. Therefore $x^2 + y^2 \leqslant 1$. Another way, $$1-x^2 = y \sqrt{1-x^2} -(1-y)\sqrt{x} \leqslant y \sqrt{1-x^2}.$$ Because $$ y \sqrt{1-x^2} \leqslant \dfrac{y^2 + 1 - x^2}{2}.$$ Implies $$1-x^2 \leqslant \dfrac{y^2 + 1 - x^2}{2} \Leftrightarrow x^2 + y^2 \geqslant 1 .$$ From $x^2 + y^2 \leqslant 1$ and $x^2 + y^2 \geqslant 1$, we have $x^2 + y^2 = 1.$ Solve $$\begin{cases} x^2 + y^2 = 1,\\ 2x^2 + y^2 = 1,\\ 0 \leqslant x \leqslant 1,\\ - 1 \leqslant y \leqslant \dfrac{1}{\sqrt{2}} \end{cases} \Leftrightarrow \begin{cases} x = 0,\\ y = 1.\end{cases}$$

Second way. We have $2x^2 + y^2 = 1$, therefore $y=\sqrt{1 - 2x^2}$ or $y=-\sqrt{1 - 2x^2}.$

First case, $y=\sqrt{1 - 2x^2}$, subtitution the second equation, we get $$x^2 + \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2}=1+(1-\sqrt{1 - 2x^2})\sqrt{x}.$$ equavalent to $$1 - x^2 - \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2} + (1-\sqrt{1 - 2x^2})\sqrt{x}=0$$ or $$\sqrt{1-x^2} (\sqrt{1-x^2} - \sqrt{1 - 2x^2}) + (1-\sqrt{1 - 2x^2})\sqrt{x}=0.$$ This is equavalent to $$\dfrac{\sqrt{1-x^2}\cdot(1-x^2-1+2x^2)}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}}+\dfrac{(1-1+2x^2)\sqrt{x}}{1+\sqrt{1 - 2x^2}} = 0$$ or $$x^2\left (\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}\right )=0.$$ It is easy to see that $$\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}> 0$$ Thus $x = 0$.

Second cases. $y=-\sqrt{1 - 2x^2}.$ We can check that $$x^2 + y \sqrt{1-x^2} \leqslant \dfrac{1}{2}$$ and $$1+(1-y)\sqrt{x} >1.$$ In this case, the given system of equations has no solution.

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We solve the second system for $y$:

$$y=\frac{-x^2+\sqrt{x}+1}{\sqrt{x}+\sqrt{1-x^2}}$$

and substitute into the first equation and solve for $x$. It is then seen that $(0,1)$ is the only real-valued solution. Computer analysis finds complex solutions where $x$ is the root of a certain $16$ degree polynomial.

Edit: the precise polynomial is $$ x^{16}+12 x^{15}+30 x^{14}-96 x^{13}-79 x^{12}+360 x^{11}+70 x^{10}-804 x^9-92 x^8+972 x^7+230 x^6-600 x^5-207 x^4+192 x^3+62 x^2-36 x+1 =0$$

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(More a comment.) If we allow the non-principal square root, $$\begin{cases} 2x^2+y^2=1,\\ x^2 + y \sqrt{1-x^2}=1\color{red}{\pm} (1-y)\sqrt{x} \end{cases}$$ the $+$ case has one real solution, but the $-$ case has three real solutions: $(x,y)=(0,1)$ and two which surprisingly are roots of $11$-deg equations. Using, $$x=+\sqrt{\frac{1-y^2}{2}}$$ and with $y$ as two appropriate real roots of, $$\small49 - 301 y + 327 y^2 - 51 y^3 - 1038 y^4 + 1334 y^5 - 1066 y^6 + 850 y^7 - 259 y^8 + 279 y^9 + 3 y^{10} + y^{11}=0$$ namely $y_1 \approx -0.619107$, and $y_2 \approx 0.939251$.

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