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Radical equation

  1. Sep 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Sqrt 5x2 +11= x+5

    State the restrictions for x and solve.






    2. Relevant equations
    5x2+11 ≥ 0
    x2≥-11/5
    (then how can u take the sqrt of a negative number??)


    3. The attempt at a solution

    The answer at the back of the book is 7/2, -1

    I don't know what I'm doing wrong but I keep getting 7 and -2 for x??? That is when I isolated the sqrt to one side of the equation, squared it and factored it.
    Also pls explain how to get the restrictions
     
  2. jcsd
  3. Sep 28, 2013 #2

    tiny-tim

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    HiCoco12! :smile:
    The LHS is a square-root, so it has to be positive

    so the RHS must also be positive. :wink:
    Show us how. :confused:
     
  4. Sep 28, 2013 #3
    Show us how. :confused:[/QUOTE]

    by squaring the sqrt 5x2+11 and the x+5 on the other side of the equal sign to get rid of the sqrt which will give me 5x2+11= x2+10x+25
    Then i brought over the x2+10x+25 to the other side which gives me:
    4x2-10x-14
    Then i factored out a 2:
    2(2x2-5x-7)
    and i factored that to give me (x-7) (x+2)
    x= 7,-2...
     
  5. Sep 28, 2013 #4

    eumyang

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    [itex]2x^2 - 5x - 7[/itex] does not factor into [itex](x - 7)(x + 2)[/itex]!
    [itex](x - 7)(x + 2) = x^2 - 5x + 14[/itex]!
     
  6. Sep 29, 2013 #5

    tiny-tim

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    the clue's in the 2x2 ! :biggrin:
     
  7. Sep 29, 2013 #6
    Ohh now I get it.. Oops missed that.. However how do I find the restrictions??
     
  8. Sep 29, 2013 #7

    tiny-tim

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    The LHS is a square-root, so it has to be positive

    so the RHS must also be positive. :wink:
     
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