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Algebra Question

  1. Feb 16, 2005 #1

    DB

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    When

    [tex]\sqrt{16 + (y-6)^2}=\sqrt{9+(y-5)^2}[/tex]

    is simplified to:

    [tex]16+(y-6)^2=9+(y-5)^2[/tex]

    How come the 16 and 9 arent square rooted?

    Thanks
     
  2. jcsd
  3. Feb 16, 2005 #2
    Do you agree that [itex]\sqrt{a} = \sqrt{b}[/itex] implies a = b?

    If so, what happens when you set a = 16 + (y - 6)^2 and b = 9 + (y - 5)^2?
     
  4. Feb 16, 2005 #3

    DB

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    ahhh, o ya its because they are equal, i got mixed up thanks.
    But [tex]\sqrt{16 + (y-6)^2}[/tex]
    doesn't simplify to [tex]16+(y-6)^2[/tex] right?
     
  5. Feb 16, 2005 #4
    Correct. It would have if the expression was [itex]\sqrt{ (16 + (y - 6)^2)^2 }[/itex] though ;)
     
  6. Feb 16, 2005 #5

    DB

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    ya thanks muzza
     
  7. Feb 16, 2005 #6

    HallsofIvy

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    Staff Emeritus
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    If you square both sides of the equation,
    [tex]\sqrt{16- (y-6)^2}= \sqrt{9- (y-5)^2}[/tex]
    you eliminate the two square roots and you get
    16- (y-6)2= 9- (y-5)2.
     
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