x's and 2's

we have $2(2^x-1)x^2+(2^{x^2}-2)x=2^{x+1}-2$ we can transform it as $2*2^x*x^2-2*x^2+2*x^{x^2}-2*x=2*2^x-2$ now we factorize as following : $2*2^x(x^2-1)-2*(x^2-1)+2*x(2^{x^2-1}-1)$ let $y=x^2-1$ and then we have $2*2^x*y-2*y+2*x(2^y-1)$ if y=0 the equation has two roots x1=1 and x2=2.the equation has one more root x3=0. we have to prove that the equation has no more roots. let assume that the equation has roots for y=/0 and x=/0.

we multiply the equation by $1/(y*x)$ . then we have $(2/x)*2^x-(2/x)+(2/y)(2^y-1)=0$ or $(2/x)*(2^x-1)=-(2/y)*(2^y-1)$ or $(2/x)/(2/y)=-(2^y-1)/(2^x-1)$ or $y/x=-(2^y-1)/(2^x-1)$ or $(2^x-1)/x=-(2^y-1)/y$ where y=x^2 -1

let f(x)=(2^x-1)/x.if we prove that there is no x for the expression $f(x)=-f(x^2 -1)$ to be true then the initial equation has no more than the three above roots.

but f(x)>0 for every x, because : -if x>0 then 2^x-1>0 and $(2^x-1)/x>0$ -if x<0 2^x-1<0 and $(2^x-1)/x>0$

so f(x)>0 for every real x and so we have that f(x^2-1)>0. but f(x)>0 and -(f(x^2-1))<0 and there is no x in order to f(x)=-(f(x^2-1)).

thereof the initial equation has exactly three roots x1=1, x2=-1 and x3=0.

graphs work by making the equation look like 2^(x^2) = 2(x^2+x-2)/x

I took a more naive approach. Rhs = 2(2^x - 1). By making Rhs 0, the 1st part of Lhs can become 2(2^x - 1)(x^2 - 1). The 2nd part of LHS can be represented as 2(2^[x^2-1] - 1)x. The sum of the 2 parts should end up as zero. This happens when x =0 or when x^2 -1 =0. So there are 3 answers. (I first misread the question and tried to answer with actual solutions...)

I think trial and error would be an appropriate . So there are three soutions at x= -1, 0, and 1.

TRY and ERROR Solutions. The number is starting with 0,1, -1, 2,-2 etc..but the largest number are not possible, and then we got the answer. The correct answer as the solutions of that equation are 0,-1, and 1. When we enter "0" to the equation, we got 0 = 0 when we enter "-1" , we got -1 = -1 when we enter "1", we got 2 = 2