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How to solve this equation in the set of real numbers? $$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$ I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.

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That is an excellent place to start; did you notice that $\sqrt{3-xy-y^2}=0\iff 3-xy-y^2=0$? –  abiessu Sep 22 '13 at 17:02
    
Yes, but how it can help me? –  Alexander Side Sep 22 '13 at 17:12

3 Answers 3

up vote 2 down vote accepted

As you stated, we must have $x^2 + 3y^2 = 7 $ and $3 = xy + y^2$.

This is an ellipse intersecting a hyperbola, so there are at most 4 points of intersection.

Substtuting $ x = \frac{3-y^2}{y}$ into the first equation, we get a quartic equation

$$ (3-y^2) + 3y^4 = 7y^2. $$

This has solutions $y = -\frac{3}{2}, -1, 1, \frac{3}{2}$.

Hence, this gives us the 4 points of intersection

$$ ( -\frac{1}{2}, - \frac{3}{2} ), ( -2, -1), (2, 1), ( \frac{1}{2}, \frac{3}{2}).$$

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Why we have $x^{2} + 3y^{2} = 7$ and $3=xy+y^{2}$? Because we suppose that $(x^{2}+3y^{2}-7)^2=0$ and $3-xy-y^{2}=0$? –  Alexander Side Sep 22 '13 at 17:19
    
@AlexanderSide Yes, and for real numbers $R$, $R^2 = 0 \Leftrightarrow R=0$, and $\sqrt{R} = 0 \Leftrightarrow R=0$. –  Calvin Lin Sep 22 '13 at 17:31
    
@AlexanderSide The original equation is equivalent to saying that "0 is resulted from the summing of (a non-negative quantity) and [another non-negative quantity]". What can you say about these two quantities? –  Mick Sep 30 '13 at 2:38

Hint: $x^{2}+3y^{2}-7+2(3-xy-y^2)=(x-y)^2-1$.

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Сan you give another hint? –  Alexander Side Sep 22 '13 at 17:20
    
From the last expression you have $x-y=\pm 1$. Substitute this into one of the other equations to find $x$ and $y$. –  njguliyev Sep 22 '13 at 17:31

$(x^2 + 3y^2 - 7)^2=0$ and $3-xy-y^2=0$

$x^2 + 3y^2 - 7=0$ and $3-xy-y^2=0$

$x=\pm (7-3y^2)^2$... So you have two cases:

You put $x=(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve

You put $x=-(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve . . .

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