Question

Another one for ya':
2xe^-x - x^2*e^-x = 0
Solve for x?

Answer 1

\[\Large 2xe^{-x} - x^2e^{-x} = 0\]

Answer 2

\[2xe^{-x}-x ^{2}e ^{-x}=0\]

Answer 3

lol

Answer 4

Thats the one yeah!

Answer 5

Start by factoring out e^-x from each term:\[\Large e^{-x}(2x-x^2)=0\]Gives you something like that, yes?

Answer 6

of course! Factoring, doh...

Answer 7

But where do I go from there?

Answer 8

Umm so recall the `Zero Factor Property`:
If, \(\Large (a)(b)=0, \qquad a=0\qquad\text{and/or} \qquad b=0\)

So we have 2 factors, we can set them each equal to zero to solve for x.\[\Large e^{-x}=0, \qquad\qquad 2x-x^2=0\]

Answer 9

For that first one, you need to remember something important about exponentials! :O

Answer 10

I'm a little rusty on those :P