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Square roots of positive numbers

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    If a and b are positive real numbers, and [tex]\lambda^{2} = ab[/tex], then [tex]\lambda = \pm \sqrt{ab}[/tex].

    2. Relevant equations

    None.

    3. The attempt at a solution

    This is more of a conceptual question that has always escaped me. I do not understand how the square root of two positive numbers could possibly be negative. Since a and b are positive, how can there be any negatives in the square root of their product? Any guidance on this subject would be very much appreciated.
     
  2. jcsd
  3. Mar 9, 2010 #2

    rock.freak667

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    Homework Helper

    (-1)2 =1

    so squaring -√(ab) will give (ab) regardless of a + or - before the square root.
     
  4. Mar 9, 2010 #3
    I think you might be looking too hard into this. There are no negatives in the square root. the negative is outside it. say both a and b are 5. then lambda² = 25. lambda therefore is equal to positive root(25) or negative root(25). Or in other terms, lambda² = 5² or (-5)²
     
  5. Mar 9, 2010 #4
    A numerical example can illustrate this. Let a = 1 and b = 4 for instance. Then we have [tex]\lambda^{2} = 4[/tex]. We realise however, that because negatives cancel upon multiplication, in fact [tex](-2)^2 = 2^(2) = 4[/tex], and so both -2 and 2 are possible solutions.
     
  6. Mar 9, 2010 #5
    Thank y'all for your help, that makes sense now. :)
     
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