Results 1 to 2 of 2

Math Help - [SOLVED] need help now

  1. #1
    bubba1
    Guest

    [SOLVED] need help now

    How would you solve(for "x"):
    the square root of 4x-12 = x

    I know that you could square both sides to get rid of the square root on the left side, but I don't know how to finish the problem. (Hint: The small 2's means squared. x2 means x squared.)

    square root of(4x-12)2 = (x)2

    4x-12 = x2
    4x = x2-12
    4x-x2 = -12

    What do you do after the last step?
    Last edited by bubba1; May 24th 2006 at 06:43 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by bubba1
    How would you solve(for "x"):
    the square root of 4x-12 = x

    I know that you could square both sides to get rid of the square root on the left side, but I don't know how to finish the problem. (Hint: The small 2's means squared. x2 means x squared.)

    square root of(4x-12)2 = (x)2

    4x-12 = x2
    4x = x2-12
    4x-x2 = -12

    What do you do after the last step?
    You have,
    \sqrt{4x-12}=x
    Square both sides,
    4x-12=x^2
    Thus,
    x^2-4x+12=0
    Thus, but this problem has no real solutions (only complex). Thus, there is no solution.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum