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Math Help - Solving for X Help?

  1. #1
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    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..?
    Last edited by michellederz; September 17th 2013 at 11:38 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Solving for X Help?

    Quote Originally Posted by michellederz View Post
    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
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  3. #3
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    Re: Solving for X Help?

    Quote Originally Posted by topsquark View Post
    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..?
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