$\displaystyle x^2+e^{-x}=8$

Is there an analytic solution? If not, why so?

$\displaystyle e^xx^2+1=8e^x\Rightarrow{e}^x=\frac{-1}{x^2-8}$ ?

All this tells me is that $\displaystyle |x|>\sqrt8$

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- Jul 29th 2009, 08:15 PMVonNemo19A bit of an exponential pickle.
$\displaystyle x^2+e^{-x}=8$

Is there an analytic solution? If not, why so?

$\displaystyle e^xx^2+1=8e^x\Rightarrow{e}^x=\frac{-1}{x^2-8}$ ?

All this tells me is that $\displaystyle |x|>\sqrt8$ - Jul 29th 2009, 09:24 PMearboth
I didn't check your calculations but from the last equation you know that $\displaystyle -\sqrt{8} < x < \sqrt{8}$:

$\displaystyle e^x$ is positive

Therefore the denominator of the fraction must be**negative**

EDIT:

Probably you have noticed that your last equation can't be used to solve it for x because you have the variable on both sides of the equation.

I "solved" (using Newton's method) the equation:

$\displaystyle x^2+e^{-x}-8=0$

and I got 2 real approximate results: $\displaystyle x_1 \approx 2.817847...~\vee~x_2\approx -1.65826...$