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Simple Maths Problem

  1. Jun 25, 2005 #1
    Simple Maths Problem....:)

    How to solve this equation?
    ( 3 ((x+1)^1/2))^1/2 = (3x-5)^1/2
     
  2. jcsd
  3. Jun 25, 2005 #2
    first, square both sides
    then, square both sides again
    then, solve quadratic
    then check answers (discard non-solutions)
     
    Last edited: Jun 25, 2005
  4. Jun 25, 2005 #3
    I cant find the correct answer in this way...the answer should be (13+(105)^1/2)/6 actually...
     
  5. Jun 25, 2005 #4
    you'll get that answer (plus 1 other that must be discarded).
    show your work if you still have trouble
     
  6. Jun 25, 2005 #5
    Can you please show me the work? I still cant find it....thanks...
     
  7. Jun 25, 2005 #6
    I'm not sure what you're not getting... Square both of the sides twice, and come up with the quadratic 9x^2-39x+16. The solution for this will of course contain two answers, but one will be invalid.
     
  8. Jun 25, 2005 #7
    ya...i get this final equation...but i get the value of x= (13 +/- (233)^1/2) / 6
     
  9. Jun 25, 2005 #8
    square first time:
    3 ((x+1)^1/2) = (3x-5)
    square second time:
    9(x+1) = 9x^2 - 30x + 25
    rearrange:
    9x^2 - 39x + 16 = 0
    solve quadratic:
    x = (1/18)*(39 ± (1521 - 576)^(1/2))
    x = (1/18)*(39 ± (945)^(1/2))
    divide num & denom by 3:
    x = (13 ± (105)^1/2)/6
    x = 3.8745 or 0.4588
    check answers in problem eq:
    0.4588 - no good (discard)
    3.8745 - ok
     
  10. Jun 25, 2005 #9
    oh..thanks...just meet some mistakes...
     
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