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How do we solve the system of equations? \begin{equation*} \begin{cases} 47a^3+129ab^2-93ba^2-83b^3-154a^2-178b^2+276ab = 0,\\ 2a^2+5b^2=23. \end{cases} \end{equation*}

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If you substitute the second into the first and square to get rid of the square roots, you will have a sixth degree equation in a. As you know two solutions, you can divide it down to fourth degree, which is solvable with difficulty or you can go numeric. –  Ross Millikan Sep 24 '12 at 16:54

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Take the first equation and substitute $b^2=\frac15 (23-2a^2)$ $$\begin{align*} 0 &= 47a^3+129ab^2-93ba^2-83b^3-154a^2-178b^2+276ab \\ &= 47a^3+129a\frac15 (23-2a^2)-93ba^2-83b\frac15 (23-2a^2)-154a^2-178\frac15 (23-2a^2)+276ab\\ &= \frac15\left( -23a^3+2967a-299ba^2-1909b-414a^2-4094+1380ab\right)\end{align*}$$ Hence $\hspace{20pt}23a^3+414a^2-2967a+4094=-(299a^2-1380a+1909)b$, i.e. $$b=-\frac{23a^3+414a^2-2967a+4094}{299a^2-1380a+1909}$$ Substituting back, we get: $$5\left(23a^3+414a^2-2967a+4094\right)^2=\left(23-2a^2\right)\left(299a^2-1380a+1909\right)^2$$ $$181447 a^6-1555260 a^5+4210311 a^4-2898920 a^3-1804419 a^2-285660 a-14283=0$$ Using the two known roots, $a=3$ and $a=-\frac17$, we get: $$(a-3)(7a+1)\left(25921 a^4-148120 a^3+189382 a^2+63480 a+4761\right)$$ You have a 4th degree polynomial, which can be solved using Ferrari formula. Using WolframAlpha, we have $529 (a-3)^3 (7 a+1)^3$, so $a=3$ and $a=-\frac17$ are the only roots.

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Here is an "in principle" method.

You can substitute a value for $a^2$ $(= \frac{23-5b^2}2)$ from the second equation into the first, which leaves you with a linear equation for $a$ (coefficients are powers of $b$). Solve for $a$, substitute into the second equation and clear fractions to obtain a polynomial for $b$.

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This is similar in spirit to Ross Millikan's comment, but avoids handling expressions with square roots. –  Mark Bennet Sep 24 '12 at 17:00
    
I think this still gets you to sixth degree, and then to fourth after removing the known roots. The equation for $a$ will be cubic in $b$ and the squaring is the same. –  Ross Millikan Sep 24 '12 at 17:18
    
@RossMillikan - indeed - your comment was posted as I was finishing mine off and wondering whether it merited being put up as an answer. I have just kept $a$ rather than an expression with a square root (I get a fraction instead). I think the method depends on personal taste (I chose mine because I find it easier to organise on paper) and also whether you spot any nice cancellations as you go through. –  Mark Bennet Sep 24 '12 at 17:29

$a=3,b=1$ satisfies the equation $2a^2+5b^2=23$

Putting these values in the first , it satisfies first also. Thus, one point of intersection of these curves is $(3,1)$

I want to add that:

If you find $1$ root , try to check whether it is a repeated root or not.

Here, you can check it as : consider two equations as two curves and since $(3,1)$ is the point of intersection , check whether at $(3,1)$ these two curves have a common tangent; if that's the case, then $(3,1)$ is the repeated root.

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Another solution is $\left(-\dfrac{1}{7}; -\dfrac{15}{7}\right)$. But I want to know the way we can find this solutions. –  minthao_2011 Sep 24 '12 at 16:34

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