Solve for $x$ in $0 = e^{x^2} (2x)$ $$0 = e^{x^2} (2x)$$
How do I solve for x?
Attempt: Divide both sides by $2x$
$$0 = e^{x^2}$$
Do I simply continue to divide until $x = 0$?
 A: The product of two "things" is zero if and only if one of the factors is zero:
$$
ab = 0 \iff a = 0 \text{ or } b = 0
$$
so
$$
(2x)e^{x^2} = 0 \iff 2x = 0 \text{ or } e^{x^2} = 0.
$$
Now the exponential function is never zero, so ...
A: Hints:
1 You have to be careful about dividing by zero
2 Assuming $x$ is a real number, consider the graph of the exponential function
A: There are two seperate cases to investigate: $x=0$ and $x\not=0$ respectively. 


*

*Suppose $x\not=0$: then it is OK for you to divide away $2x$ and find $$\exp(x^2)=0,$$ which never happens for real $x$. This means that there are no solutions to the original equation for $x\not=0$. 

*Now suppose that $x=0$. Well then you get $0=0$ which is true. Thus $x=0$ is the only solution. 
A: Did you mean $2xe^{x^2} = 0$ ?
If so, since $e^x > 0$ for all $x$, we have only $x=0$ as a solution.
You can't divide by $x$ unless you specify $x\neq 0$, so you can't divide by $x$ and then say : "Oh ! $x=0$ is a solution !"
A: Suppose x is not zero: then you can divide away 2x and find
exp(xx)=0,
which never happens for real x.
