1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Algebra Question

  1. Feb 16, 2005 #1

    DB

    User Avatar

    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

    User Avatar

    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

    User Avatar

    ya thanks muzza
     
  7. Feb 16, 2005 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Algebra Question
  1. Algebra question (Replies: 4)

  2. Algebra question (Replies: 3)

  3. Algebra question (Replies: 2)

  4. Algebra Question (Replies: 8)

  5. Algebra Question (Replies: 1)

Loading...