Solving for X Help?

Printable View

September 17th 2013, 11:13 AM
michellederz

Solving for X Help?

Solve the following equation for X?

-2xe^-6x+x²e^-6x=0

I posted a forum thread similar to this but this one includes e.. Not sure how to solve for x when e is involved..

In the previous thread I posted, you just took the exponents that created a quadratic and solved for x that way. So would this be how you solve it?

-6x+2-6x=0
-12x+2=0
From here you would use the quadratic equation and get the answers 24 , 0

Would this be correct..?
September 17th 2013, 11:46 AM
topsquark

Re: Solving for X Help?

Quote:

Originally Posted by michellederz

Solve the following equation for X?

-2xe^-6x+x²e^-6x=0

I posted a forum thread similar to this but this one includes e.. Not sure how to solve for x when e is involved..

In the previous thread I posted, you just took the exponents that created a quadratic and solved for x that way. So would this be how you solve it?

-6x+2-6x=0
-12x+2=0
From here you would use the quadratic equation and get the answers 24 , 0

Would this be correct..?

Actually I'd start by factoring:
$-2xe^{-6x} + x^2 e^{-6x} = 0$

$\left ( xe^{-6x} \right ) \left ( -2 + x ) = 0$

What possible zeros do we have? (Hint: there's two.)

-Dan
September 17th 2013, 11:49 AM
michellederz

Re: Solving for X Help?

Quote:

Originally Posted by topsquark

Actually I'd start by factoring:
$-2xe^{-6x} + x^2 e^{-6x} = 0$

$\left ( xe^{-6x} \right ) \left ( -2 + x ) = 0$

What possible zeros do we have? (Hint: there's two.)

-Dan

Well one possible zero is 2 (I think..) but I am not sure how to find the other one form the other equation. Could the other possible zero be 0..?