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Square roots

  1. Sep 2, 2005 #1
    I'm just wondering about something, look at this:

    [tex]
    \[
    \begin{array}{l}
    \sqrt {3 + x} + \sqrt {2 - x} = 3 \\
    \left( {\sqrt {3 + x} } \right)^2 + \left( {\sqrt {2 - x} } \right)^2 = 3^2 \\
    3 + x + 2 - x = 9 \\
    5 = 9 \\
    \end{array}
    \]
    [/tex]

    Obviously something isn't right here, and I know that if I isolate one of the square roots on the other side, then it all works out. But why doesn't this way work?!
     
  2. jcsd
  3. Sep 2, 2005 #2

    arildno

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    Dearly Missed

    Why should it??
    What you're basically saying in line 2, is that (a+b)^2=a^2+b^2
     
  4. Sep 2, 2005 #3
    Yes, take a look at line two, you have to square the entire left side of the equation.
     
  5. Sep 3, 2005 #4
    TSN79, remember that [tex] ( a + b ) ^ 2 = a^2 + 2ab + b^2 [/tex] :redface:

    Therefore, fixing the second line, you might have:

    [tex] \sqrt {3 + x} + \sqrt {2 - x} = 3 \Rightarrow 2\sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 4 \Rightarrow [/tex]

    [tex] \sqrt {\left( {3 + x} \right)\left( {2 - x} \right)} = 2 \Rightarrow \left( {3 + x} \right)\left( {2 - x} \right) = 4 \Rightarrow [/tex]

    [tex] x^2 + x - 2 = 0 \Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0 \Rightarrow \boxed{x = - 2,1} [/tex]
     
    Last edited: Sep 3, 2005
  6. Sep 3, 2005 #5

    VietDao29

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    That's wrong, bomba923. This line is wrong:
    It should look like this:
    [tex]x ^ 2 + x - 2 = 0 \Rightarrow \left[ \begin{array}{l} x = 1 \\ x = - 2 \end{array} \right.[/tex] :wink:
    Viet Dao,
     
  7. Sep 3, 2005 #6
    Indeed! :redface:, stupid mistakes are the bane of my math education!!!

    Anyway,
    Thanks for pointing that out-->
    Mistake correct, see updated post! :smile:
    -------
     
    Last edited: Sep 3, 2005
  8. Sep 3, 2005 #7

    Curious3141

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    Both of your solutions are admissible.
     
  9. Sep 3, 2005 #8
    Oh yes, that's right! [tex] 1 < 2 [/tex], :cool:
    See updated post!
     
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