I have been recently given such a system containing irrational equation. Unfortunately for me, i was unable to solve this.
$$\begin{cases}
\sqrt{x^2+y^2}+\sqrt{x-y}=6 \\
x^2-y^2-x+y=12
\end{cases}
$$
I could only find the (4;0) solution.
I would really like to know if there is some pattern you should follow in order to solve this sort of problems.
Thanking in advance.
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d4k3r
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1According to WolframAlpha, there is another solution, but the exact solution revealed the other solution involves solving a septic equation. Numerically, it is approximately $(4.12,1.54)$. – player3236 Dec 24 '20 at 07:35
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sextic ( sixth degree) equation.. – Narasimham Dec 24 '20 at 07:55
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That's absolutely a lot of algebra work. $$ x^2+y^2=\left( 6-\sqrt{x-y} \right) ^2 \\ x^2+y^2=36-12\sqrt{x-y}+x-y \\ \left\{ \begin{array}{c} x^2+y^2+y-x=36-12\sqrt{x-y} \quad(1)\\ x^2-y^2+y-x=12 \quad(2)\\ \end{array} \right. $$ $(1)-(2)$, we have $$ 2y^2=24-12\sqrt{x-y} \\ y^2=12-6\sqrt{x-y} \\ y^2-12=6\sqrt{x-y} \\ y^4-24y^2+144=36\left( x-y \right) \\ x=\frac{1}{36}\left( y^4-24y^2+36y+144 \right) \\ \frac{y^8}{1296}-\frac{y^6}{27}+\frac{y^5}{18}+\frac{23 y^4}{36}-\frac{4 y^3}{3}-\frac{14 y^2}{3}+8 y=0 $$ We can also get a polynomial for $x$ in a similar way. Generally speaking, for high degree polynomials we can hardly find a nice way to write the roots down. According to Mathematica's computation, there's another real root, numerically $(4.12...,1.54...)$ and cannot be represented using radicals. So I think you have reached the limit: $(4,0)$ is the only root that can be computed by hand.
FFjet
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