How to solve this sytem of equation: $\begin{cases}(\sqrt{a^2+4}+a)(\sqrt{b^2+1}+b)=1\\27a^6+8b=a^3+2\end{cases}$

I tried to analysis equation (1):

$\sqrt{a^2+4}+a=\sqrt{b^2+1}-b\Leftrightarrow a+b=\sqrt{b^2+1}-\sqrt{a^2+4}$ $\Leftrightarrow 4a^2+16b^2+20ab=9$ ... And then, I don't know how to solve this.

  • 2
    $\begingroup$ Well... you can let $z=a^3$ and then $27z^2 - z + (8b-2) = 0$. Then use quadratic formula and get $a \equiv a(b)$. Then use back substitution into the first equation. $\endgroup$ – Christopher K Dec 13 '13 at 6:12
  • $\begingroup$ I tried to do as you. But I can't continue to do :( $\endgroup$ – abcdxyz Dec 13 '13 at 6:18
  • $\begingroup$ You can also eliminate b from the second equation and plug its expression in the first equation. This will give you a ... MONSTER ! Are they any restriction for the domains ? $\endgroup$ – Claude Leibovici Dec 13 '13 at 6:19
  • $\begingroup$ That way is very complex :( . I will try by maple or wolfram :-ss $\endgroup$ – abcdxyz Dec 13 '13 at 6:23
  • $\begingroup$ Doing what I suggested, I plotted the first equation as a function of "a" and I only found two solutions close to a = 0.893669 and a = -0.736565. $\endgroup$ – Claude Leibovici Dec 13 '13 at 6:27

you can solve the first equation for $b$:$$b=-1/2\,{\frac {2\,{a}^{2}+3+2\,\sqrt {{a}^{2}+4}a}{\sqrt {{a}^{2}+4}+a} } $$ plugging this in the second equation we get $${\frac {27\,{a}^{6}\sqrt {{a}^{2}+4}+27\,{a}^{7}-{a}^{3}\sqrt {{a}^{2} +4}-{a}^{4}-8\,\sqrt {{a}^{2}+4}a-8\,{a}^{2}-2\,\sqrt {{a}^{2}+4}-2\,a -12}{\sqrt {{a}^{2}+4}+a}} =0$$ solving this for $\sqrt{a^2+4}$ and squaring again we get this polynomial:

$$729\,{a}^{12}-54\,{a}^{9}-270\,{a}^{7}-107\,{a}^{6}+10\,{a}^{4}+4\,{a} ^{3}+16\,{a}^{2}+20\,a-32 =0$$ with only two real Solutions. this can be proven by using the Theorem of Sturm.


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