Results 1 to 3 of 3

Math Help - Solve x^.5 = 6 - x

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    76

    Solve x^.5 = 6 - x

    What I have done is squared both sides which gives
    x=(6-x)^2 =
    0 = x^2-13x+36
    0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function.

    Is there a better way of doing this??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    Quote Originally Posted by p75213 View Post
    What I have done is squared both sides which gives
    x=(6-x)^2 =
    0 = x^2-13x+36
    0=(x-4)(x-9). x=4,x=9. Of these x=4 fits the original function.

    Is there a better way of doing this??
    \sqrt{x} = 6 - x

    x = (6 - x)^2

    x = x^2 - 12x + 36

    0 = x^2 - 13x + 36

    0 = x^2 - 9x - 4x + 36

    0 = x(x - 9) - 4(x - 9)

    0 = (x - 9)(x - 4)


    So x - 9 = 0 or x - 4 = 0.

    Therefore x = 9 or x = 4.


    By the way, both values of x fit the function...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    76
    I see your point. Thanks for that.
    Last edited by p75213; May 30th 2010 at 10:45 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 11:29 PM
  2. Replies: 1
    Last Post: June 9th 2009, 11:37 PM
  3. how do I solve this?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 22nd 2009, 07:21 PM
  4. How could i solve?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 2nd 2009, 03:18 PM
  5. Solve for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 1st 2009, 01:33 PM

Search Tags


/mathhelpforum @mathhelpforum