Firstly, this is not my homework. I am well past high school (finished graduate school several years ago) but I am mentoring a high schooler and I want to explain how to solve this by hand using just pen and paper.

A more presentable form + solution is here

My own attempts could only simplify it to a fourth degree polynomial in $x$ and equate it to $0$. There's gotta be a better way but it is beyond me and I feel embarrassed!

  • $\begingroup$ Thanks @Amzoti but it should be $ y = (2 - x) $ and substituting it led me to what feels like a dead end: $ 250x^4 - 790x^3 + 756x^2 - 232x + 16 = 0$ $\endgroup$ – agks mehx Jun 2 '14 at 2:53
  • $\begingroup$ for which I suppose you could guess (?) $ 1 $ as a root, and arrive at $ 250 x^3-540 x^2+216 x-16 = 0 $ and I don't know how to proceed from there. Wolframalapha tells me there is a root $ x = 2 / 5 $ but I can't see how I could guess it and go through the pain of dividing to finally arrive at a quadratic equation. $\endgroup$ – agks mehx Jun 2 '14 at 2:57
  • $\begingroup$ Thanks @Amzoti At least this helps me confirm that there is no way out of the long, hard slog. Doesn't seem like a fair high school homework question, especially how to guess the $ x = 2/5 $ root in the cubic equation. $\endgroup$ – agks mehx Jun 2 '14 at 3:01
  • 1
    $\begingroup$ To find that $x=\frac{2}{5}$ is a root of the cubic, you could use the rational roots test: en.wikipedia.org/wiki/Rational_root_theorem. That way, you at least see that there only finite number of possible rational solutions to check. $\endgroup$ – Strants Jun 2 '14 at 3:06
  • $\begingroup$ Thank you @Strants, the rational root theorem helps! $\endgroup$ – agks mehx Jun 2 '14 at 3:08

since $xy>0,x+y=2>0, \implies x>0,y>0$

let $x=a^2,y=b^2,a>0,b>0 \implies 4a^4+b^4=5(2a^2-b^2)ab \iff \\ 4a^4-10a^3b+5ab^3+b^4=0$

by observation, $a=b$ is a solution as $4-10+5+1=0$

so we get $(a-b)(4a^3-6a^2b-6ab^2-b^3)=0$

now to check $4a^3-6a^2b-6ab^2-b^3=0$,a factor $b=-2a$ is existed. $\implies$

$(2a+b)(2a^2-4ab-b^2)=0 \implies 2a^2-4ab-b^2=0 $

with $a^2+b^2=2 $ we get $3a^2-4ab=2 \implies b=\dfrac{3a^2-2}{4a} \implies x+\dfrac{(3x-2)^2}{16x}=2$


  • $\begingroup$ @chenbai how did u get the factor of $b=-2a$? $\endgroup$ – tpb261 Jun 2 '14 at 7:03
  • $\begingroup$ @tpb261, it is observation also. first $b=-ka$ is clear, then $4+6k-6k^2+k^3=0,k=2 $is a solution by quick trying. $\endgroup$ – chenbai Jun 2 '14 at 8:18
  • $\begingroup$ hmm... I somehow seem to feel Cardan's or Ferrari's methods are better than to guess. But then, that's me. $\endgroup$ – tpb261 Jun 2 '14 at 8:30

Substituting $y=2-x$ in the second equation gives $16x^4+(2-x)^4+8x^2(2-x)^2 = 25(3x-2)^2(x(2-x))$ The solution is here. Then use the "by-hand"method of solving bi-quadratic equations. It's clumsy and laborious, but is "by-hand". Put back the values of $x$ and eliminate some of the solutions.

  • $\begingroup$ hiw is it bi-quadratic? in a bi-quadratic there should not be terms involving $ x ^ 3 $ and $ x $ $\endgroup$ – agks mehx Jun 2 '14 at 4:06
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    $\begingroup$ Oh, I didn't know that. Sorry. By bi-quadratic, I meant an equation of degree 4. I should have checked the wiki page a bit more carefully. $\endgroup$ – tpb261 Jun 2 '14 at 4:08

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