Find the number of real solutions of the equation $2^x+x^2=1$ My attempt is as follows:-
$$2^x+x^2=1$$
$$\left(1+x\cdot log(2)+\frac{x^2\cdot (log(2))^2}{2!}+\frac{x^3\cdot (log(2))^3}{3!}+\dots\right)+x^2=1$$
$$x\cdot log(2)\left(1+x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}\right)=0$$
$$x=0, \left(1+x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}\right)=0$$
$$x^2+\frac{x\cdot log(2)}{2!}+\frac{x^2\cdot (log(2))^2}{3!}+\dots=-1$$
Now I started to think when this can be negative, it can only be when x is negative.
I also thought about $$S=x^2+\sum_{i=1}^{\infty}\left(\frac{x^i\cdot (log(2))^i}{i+1}\right)$$. But I was not finding the way to transform it into telescopic series.
I am stuck here and not able to proceed further.

Consider the function $$f(x)=2^x+x^2-1.$$
Its first and second derivatives are $$f'(x)=\ln2\cdot2^x+2x$$
and $$f''(x)=(\ln2)^2\cdot2^x+2.$$
Observe:


*

*Second derivative is strictly positive

*First derivative has one root at $x_{min}=\dfrac1{\ln2}W\bigg(\dfrac{(\ln 2)^2}2\bigg)\approx-0.28454$

*$f(x_{min})\approx-0.098034<0$
Therefore, $f(x)$ has two zeroes, and there are two real solutions to your equation.
