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Math problem - solving for a variable

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data

    sqrt(2x+4) = sqrt(6x+1) - 1

    I Need to solve for x, but cannot seem to get the same answer as the text. (Ans. x= 5/2)

    3. The attempt at a solution

    sqrt(2x+4) - sqrt(6x+1) = -1
    square both sides
    (2x+4) - (6x+1) = 1
    -4x-3=1
    -4x=4
    x=-1 ??

    I know I must be doing something wrong, it has been a while since I have done this sort of problem.

    Thank you for your time.
     
  2. jcsd
  3. Aug 13, 2010 #2

    vela

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    Your mistake is in thinking (a-b)2 = a2-b2, which is what you did when you squared the lefthand side of the equation. You need to multiply it out correctly.
     
  4. Aug 13, 2010 #3
    Im still not there yet:

    My new attempt:

    [​IMG]

    I cant seem to get rid of the sqrt (x)
    i know sqrt(12x) = 2*sqrt(3x)
     
  5. Aug 13, 2010 #4

    vela

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    Now you're making even more mistakes. First,

    [sqrt(2x+4)-sqrt(6x+1)]2 ≠ (2x+4) - (6x+1)

    Second,

    sqrt(a+b) ≠ sqrt(a)+sqrt(b)

    which is what you're doing going from the fifth line to the sixth line. You were, however, correct when you said earlier that

    (sqrt(2x+4))2 = 2x+4



    The problem is actually a bit easier to solve if you square both sides right away:

    (sqrt(2x+4))2 = (sqrt(6x+1)-1)2

    To correctly calculate the righthand side, let a=sqrt(6x+1) and b=1. Then your equation becomes

    (sqrt(2x+4))2 = (a-b)2

    Now FOIL out the righthand side and then substitute back in for a and b.
     
    Last edited: Aug 13, 2010
  6. Aug 13, 2010 #5
    Im sorry, I am unable to see that. All i can see is black and faint white portions. Is it possible for you to take a screen shot and upload to image shack?
     
  7. Aug 13, 2010 #6

    vela

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    I edited my previous post and took out the images. You should be able to read it now.
     
  8. Aug 13, 2010 #7
    I got it, thank you for your help.
    It took a while to come back to me. I appreciate your time, thank you.

    [​IMG]
     
  9. Aug 13, 2010 #8

    Mark44

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    Looks fine.
     
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