Find the number of solutions of $ x^2+1=2^x$ I tried solving this  equation $$x^2+1=2^x$$ 
but I was able to get only two roots ,
i.e. $x=0,1$ but the answer given said the equation has 3 roots when I looked at the graph given in the solution it was given as $x=4.25746$ as the other root.
Can anyone please tell me how to get this other root without using a graph plotter ?
 A: It is almost certainly impossible to find an explicit (non-numerical) expression for the third root.
To see that there is a third root, you could notice that
$$
3^2+1 = 10 > 8 = 2^3
$$
so the left-hand side is larger than the right-hand side when $x=3$, and
$$
5^2+1 = 26 < 32 = 2^5
$$
so the right-hand side is larger than the left-hand side when $x=5$. Since both sides of the equation are continuous functions, it follows (essentially by the intermediate value theorem) that the equation must be satisfied for at least one $x \in (3, 5)$.
A: As said by @Micah, it is a desparate fight to find an explicit solution (apart from exceptionaly simple solutions as $0$ and $1$) to such a transcendental equation.
But approximate solutions are easy to find (see below).
Let us first convert the issue into a more manageable one using logarithm function. Let us define: $y=f(x)=ln(x^2+1)$ and $y=g(x)=ln(2^x)=x ln(2)$ (see curves below)
The problem is equivalent to find all $x$ such that $h(x):=f(x)-g(x)=0$
An classical reasoning on the derivative of $h(x)$ helps in separating the roots :
$h'(x)=\dfrac{2x}{1+x^2}-ln(2)$ has the following sign pattern $-,+,-$: in $x_1<x_2<x_3$ all of them on $\mathbb{R_+}$: Thus, $h$ decreases (from $+\infty$ to $h(x_1)<0$) , then increases from $h(x_1)<0$ to $h(x_2)>0$, then decreases from $h(x_2)>0$ to $-\infty$. This implies in a rigorous way that there are exactly 3 roots. 
As said befoe, the first roots $x=0$ and $x=1$ are evident.
For finding the third one, the fixed point iteration $x_{n+1}=h(x_n)$ converges to solution  
x= 4.2574619144479...
in some thirty steps or so...
Nevertheless, Newton method gives, as always, a much more rapid convergence to this third root.

A: About the integer solutions: for any $x\geq 3$, $2^x-1\equiv 7\pmod{8}$, hence $2^x-1$ cannot be a square. If $x<0$, $x^2+1>1$ while $2^x<1$. By direct inspection, the only solutions are given by $x=0$ and $x=1$.
About the real solutions: if there is some real solution besides $0$ or $1$, it has to be positive and fulfill
$$ \log(1+x^2) = x\cdot \log(2), $$
but since $\frac{d^2}{dx^2}\log(1+x^2)=\frac{2(1-x^2)}{(1+x^2)^2}$,  $\;f(x)=\log(1+x^2)$ is convex over $[0,1]$ and concave over $[1,+\infty)$, and the same property is shared by $g(x)=\log(1+x^2)-x\log(2)$.
Since $\lim_{x\to +\infty}g(x)=-\infty$, $g(x)$ has a unique extra root in the interval $(1,+\infty)$, that can be located through Newton's method. Numerically, it is $\approx 4.2574619$.
