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How to solve the system $$ \{\sqrt{x-y}+\sqrt{x-2} = 2, \sqrt{x^2+y^2-xy(x-y)}+\sqrt{y(x-y)} = \sqrt{8(x-y-1)}\}$$ over the reals? I can derive $x=1/16\,{y}^{2}-1/4\,y+9/4$ from the first equation.

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Can we try $x-2=4\cos^4\phi, x-y=4\sin^4\phi$ from the first? as both radicals are $\ge0$ –  lab bhattacharjee Dec 23 '13 at 8:14
    
@ lab bhattacharjee : And then what? –  Markiyan Hirnyk Dec 23 '13 at 8:34
    
Have you tried evaluating the second expression using mathworld.wolfram.com/Double-AngleFormulas.html and mathworld.wolfram.com/Multiple-AngleFormulas.html and –  lab bhattacharjee Dec 23 '13 at 8:36
    
@ lab bhatttacharjee: Have you tried that? –  Markiyan Hirnyk Dec 23 '13 at 8:45

1 Answer 1

This may not be a good way, but this does give the answers.

Letting $A=x^2+y^2-xy(x-y), B=y(x-y), C=8(x-y-1)$, then we get from the second equation, $$A+B+2\sqrt{AB}=C$$$$\Rightarrow 4AB=(C-A-B)^2$$

Also, from the first equation, we get $$x=\frac 1{16}y^2+\frac 14 y+\frac 94.$$ (note! you are wrong.)

Now, if you substitute it for the last equation we got, then you'll get an equation of $y$. And wolfram-alpha will give you the answers. Don't forget to pay attention for the domain of $y$.

Maybe you don't like this way, but calculation does tell you the answer.

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I find this way interesting. Thank you for the correction. Can you elaborate your answer, writing down the equation in $y$? –  Markiyan Hirnyk Dec 23 '13 at 9:30
    
I'm reluctant to do that:) Why don't you do that yourself since you've already known how to do it? –  mathlove Dec 23 '13 at 9:34
    
@MarkiyanHirnyk : wolframalpha.com/input/… –  mathlove Dec 23 '13 at 9:52
    
This is $$-{y}^{10}+18\,{y}^{9}-145\,{y}^{8}+1968\,{y}^{7}-13648\,{y}^{6}+51072 \,{y}^{5}-661344\,{y}^{4}+1624320\,{y}^{3}-3747840\,{y}^{2}+5776896\,y -1597696=0. $$ And then what? –  Markiyan Hirnyk Dec 23 '13 at 9:55
    
I don't find the usage of WA satisfactory. The numerical solution can be found from the original system. –  Markiyan Hirnyk Dec 23 '13 at 9:57

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