Results 1 to 4 of 4

Math Help - solve for x

  1. #1
    Newbie
    Joined
    Apr 2007
    Posts
    2

    solve for x

    Can you show me how to solve:

    x/4^.5 * x/2^.5 = 50 i know the answer is 141 but need help to get there. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Feb 2011
    Posts
    14
    ok here we need to make sense of the left hand side

    (x/4)^0.5 means the square root of that, so we have (x^0.5)/(4^0.5) which equals (x^0.5)/2, if you do the same on the other one you have (x^0.5)/2^0.5.

    sqaure both side to remove the sqaure roots and then you get (x/4) x (x/2) = 2500 ----> x^2 = 20,000 -----> x = 100 x 2^0.5 which approximates to 141
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2007
    Posts
    5
    ok i get (x/4) x (x/2) = 2500 ----> but you lost me here...where does x^2 = 20,000 -----> x = 100 x 2^0.5 come from ???

    I feel like multiplying both sides by 1/4 & 1/2 then take the sqrt of x but it doesn't work.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2007
    Posts
    5
    Quote Originally Posted by reneebu View Post
    ok i get (x/4) x (x/2) = 2500 ----> but you lost me here...where does x^2 = 20,000 -----> x = 100 x 2^0.5 come from ???

    I feel like multiplying both sides by 1/4 & 1/2 then take the sqrt of x but it doesn't work.
    now i feel dumb...its Multiply each side by 8 then sqrt...ouch
    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, 10:29 PM
  2. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  3. how do I solve this?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 22nd 2009, 06:21 PM
  4. How could i solve?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 2nd 2009, 02:18 PM
  5. Solve for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 1st 2009, 12:33 PM

Search Tags


/mathhelpforum @mathhelpforum