How do i solve that equation algebraically ?



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  • $\begingroup$ Most probably you don't. Equations like this very rarely have a closed form solution. $\endgroup$ – 5xum Jul 4 '16 at 11:38
  • $\begingroup$ In which domain? $\endgroup$ – user84976 Jul 4 '16 at 12:16

IN THE CASE $x\in\mathbb{R}$

Firstly you can factor:

$2^x(x^2+2^x)=(x^2-2^x)(x^2+2^x)$, but $x^2+2^x\neq0$ so we simplify:

$2^x=x^2-2^x$ and we obtain just: $2^{x+1}=x^2$.

So you have to find zeroes of $f(x)=2^{x+1}-x^2$ with $x\in \mathbb{R}$. We have that $x=-1$ is a solution.

$f'(x)=\log(2)2^{x+1}-2x$, which is always positive (clearly it's positive for $x<0$), and it's not difficult to verify for $x>0$ So $f$ is monotonic and $-1$ is the only solution.

  • $\begingroup$ $x=-1$ is a solution. $\endgroup$ – lulu Jul 4 '16 at 12:00
  • 2
    $\begingroup$ $-1$ is an integer. The OP didn't even specify real solutions...there are plenty of complex solutions. $\endgroup$ – lulu Jul 4 '16 at 12:03
  • $\begingroup$ I'm sorry, I assumed x being a natural number $\endgroup$ – user84976 Jul 4 '16 at 12:09
  • 1
    $\begingroup$ So you look for complex solution, maybe you could add that to your question $\endgroup$ – user84976 Jul 4 '16 at 12:10

If we divide both sides of the equation by $4^x$, we can rewrite it as


If we now write $u=x^2/2^x$, we have a quadratic, $u+1=u^2-1$, or


Now $x^2/2^x=-1$ clearly has no solution, since $x^2/2^x$ is always non-negative. So we are left with $x^2/2^x=2$, or


A quick sketch of the parabola $y=x^2$ and the exponential curve $y=2^{x+1}$ shows these intersect at exactly one negative value of $x$, which, by inspection, turns out to be at $x=-1$. A simple argument comparing parabolic to exponential growth (and/or a bit of calculus) shows there is no intersection for positive values of $x$. Thus $x=-1$ is the only solution.


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