What is the algebraic solution to this equation $x^2+2^x-2=0$
I know the numeric solution using graphing which is $\{-1.258,0.653\}$
NB. It's graph looks like parabola
Also what is the algebraic solution to this equation $x^2-2^x=0$
I know the numeric solution using graphing which is $\{-0.767,2,4\}$
NB. It's graph looks like Cubic Equation
The second equation may be solved using the Lambert W function:
$$x^2=2^x\\-x=2^{x/2}\\ [-x\ln(2)/2]e^{[-x\ln(2)/2]}=\ln(2)/2\\-x\ln(2)/2=W(\ln(2)/2)\\x=-\frac{W(\ln(2)/2)}{\ln(2)/2}\approx-0.767$$
Likewise, the positive solutions come down to $x=2$ and $x=4$.
The first equation cannot be solved in the same manner and requires a numerical method:
$$x^2+2^x-2=0\\x=\sqrt{2-2^x}$$
$$x_{n+1}=\sqrt{2-2^{x_n}}$$
Set $x_0=0.5$, then we have
$x_1=0.76537$
$x_2=0.54790$
$x_3=0.73351$
$\vdots$
$x_{35}=0.65374$
The negative solution may be found using
$$x_{n+1}=-\sqrt{2-2^{x_n}}$$
with $x_0=-1$, we get
$x_1=-1.22474$
$x_2=-1.25384$
$x_3=-1.25725$
