How to find number of solutions to $2^x=x^2$ without Graphing Find number of real solutions to $2^x=x^2$ without plotting graph:
I considered $f(x)=2^x -x^2$ 
$$f'(x)=2^x \ln 2-2x=0$$
we get again a transcendental equation. Any good approach please
 A: Your equation is equivalent to $f(x)=\ln x/x=\ln2/2.$ Now $f'(x)=(1-\ln x)/x^2,$ meaning the function is growing until $x=e,$ and decreasing to $0$ after that. So the function crosses every positive level it reaches (below the maximum) twice, once upwards and once downwards. Of course, it's easy to see that those two solutions are $x=2$ and $x=4.$
EDIT: As @Rory Daulton pointed out, this is incomplete. The correct equation should be $f(x)=\ln |x|/x=\ln2/2,$ and this function is growing from $0$ to $\infty$ between $x=-1$ and $x=0,$ giving a third solution. Thanks, Rory Daulton!
A: $a_{0}2^x+b_{0}$ may have at most one zero. Before that it is negative, after that zero, if it exists, it is positive.
Based on just this remark, $a_{1}2^x+b_{1}x$ may have at most two zeros, since its derivative $a_{0}2^x+b_{0}$ is forcing it to behave similar to a parabola, if the derivative has one zero. Again, based on the essential property of $a_{1}2^x+b_{1}x$, $a_{2}2^x+b_{2}x^2$ may have at most three zeros (since it may have at most two extreme values).
That this is really the case is easily seen from its behavior around $2$.
$2^2-2^2=0$, and it is has a negative slope at $2$ since at $1$, $2^1-1^2>0$ and at $3$, $2^3-3^2<0$
Now
$\lim\limits_{x \to -\infty}2^x-x^2<0$ and $\lim\limits_{x \to \infty}2^x-x^2>0$ and since $2^x-x^2$ is continuous there has to be one zero before and one zero after $2$.
