How many real roots does the equation $e^x-x^2=0$ have? How many real roots does the equation  $e^x-x^2=0$ have? 
I can see from wolfram that the curve cuts X-axis only once. How do you go about solving it?
 A: $f(x) = e^x - x^2$
$f'(x) = e^x - 2x$
We will find the critical values of $f'(x)$.
$f''(x) = e^x - 2 = 0 \implies e^x = 2 \implies x = ln(2)$.
Thus, $x = ln(2)$ is the only critical point of $f'(x)$.
$f'''(x) = e^x > 0$ that means this critical point is local minimum of $f'(x)$.
$f'( ln(2) ) = 0.61371xxxx > 0$. That is $f'(x)$ is always positive.
Now, it means $f(x)$ is always increasing. Now choose 2 $x$ values that gives negative and positive results respectively, and conclude that $f(x)$ hhas 1 root.
You can pick $x = -1$ and $x = 1$.
A: Let $f(x) = e^x-x^2$. 
$$\lim_{x \to - \infty} e^x-x^2 =- \infty$$
$$\lim_{x \to + \infty} e^x-x^2 =+ \infty$$
Since $f$ is continuous, by the Intermediate Value Theorem, it has a root, say $a$. Assume for a contradiction that it has another root, say $b$.
The derivative of the function $e^x-2x$ is positive for every $x$.
Now since $f$ is continuous and differentiable, and since $f(a)=f(b)$, by Rolle's theorem, there must exist a point $c \in [a,b]$ such that $f'(c)=0$. This contradicts with the fact that the derivative is always positive.
Hence there is one and only one root.
A: Observe that the real root of $\mathrm e^x=x^2$ is the real root of
$$
-\frac{x}{2}\mathrm e^{-\frac{x}{2}}=\frac{1}{2}
$$
In this way you can use the Lambert-W function to find
$$
-\frac{x}{2}=\mathrm W\left(\frac{1}{2}\right)\Longleftrightarrow x = -2\mathrm W\left(\frac{1}{2}\right)
$$
A: The equation is equivalent to 
$$(e^{x/2}+x)(e^{x/2}-x)=0, $$ which splits in two equations.
The derivatives of the factors are
$$\frac12e^{x/2}+1,\frac12e^{x/2}-1$$ and the second one cancels for $x=\ln(4)$, corresponding to a positive minimum.
So the function is monotonous, ranges from $-\infty$ to $\infty$ and has exactly one root, which Newton will find.
A: In addition to everyone here, I note that not only is the root of the given function found to be $x=-2W\left(\frac12\right)$, but it is more specifically
$$x=-2W_k\left(\frac12\right)$$
Where $k$ denotes the branches of the Lambert W function.  We note simply from graphing that the only real value is given when $k=0$.  All other branches produce complex numbers, so the amount of real roots is one.
A: There's only one real root. The original equation is equivalent to $e^x=x^2$.
For nonnegative values of $x$ function $e^x$ always has bigger value than $x^2$. (Derivative of $e^x$ is $e^x$, derivative of $x^2$ is $2x$. Therefore for nonnegative values $e^x$ grows faster then $x^2$ because of bigger of derivative). And for $x=0$ exponential function already has bigger value.)
If $x$ is negative, then $x^2$ is decreasing and $e^x$ is increasing; also for some value, for example, $-10$, $e^x<x^2$.
(Or simply sketch graphs of both functions.)
Therefore there is one real root. To get approximations of it you can use some numerical method. For example, dividing segment with root in two parts and repeating the process with part which has the root.
