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Question about rewriting an expression with radicals

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data


    (-3/x^(1/2)) / ((2/y^(1/2))-8)

    which, from my book, I see can be rewritten with radicals as:

    (-3/sqrt(x)) / ((2-8sqrt(y))/(sqrt(y)))





    3. The attempt at a solution

    Now, what I don't understand is how the denominator was rewritten. I understand that in the numerator, the 1/( x^(1/2)) becomes sqrt(x), but why does the denominator turn out like it does? Why do I end up with two seperate "sqrt(y)" terms? Could someone perhaps walk me through what was done to rewrite the denominator with radicals as shown above^? I would greatly appreciate it. I know that the rewritten expression that I have is correct because it is from my book, but I just don't understand how they rewrote it.
     
  2. jcsd
  3. Jul 5, 2011 #2
    They rationalized the denominator, probably by multiplying it by it's conjugate (both numerator and denominator)

    [tex]\frac{2}{\sqrt{y}-8}[/tex]
    [tex]\frac{2(\sqrt{y}+8)}{(\sqrt{y}-8)(\sqrt{y}+8)}[/tex]

    at least I believe, it's hard to see through all those brackets.
     
  4. Jul 5, 2011 #3

    eumyang

    User Avatar
    Homework Helper

    Actually, it looks like that the original is this:
    [itex]\frac{\frac{-3}{x^{1/2}}}{\frac{2}{y^{1/2}} - 8}[/itex]
    and the rewritten form is
    [itex]\frac{\frac{-3}{\sqrt{x}}}{\frac{2 - 8\sqrt{y}}{\sqrt{y}}}[/itex]

    If so, then all that was done was to subtract, after finding a common denominator:
    [itex]\frac{2}{y^{1/2}} - 8 = \frac{2}{\sqrt{y}} - \frac{8\sqrt{y}}{\sqrt{y}} = ...[/itex]
     
  5. Jul 7, 2011 #4
    I simplified it as follows:

    [itex]
    \frac{\frac{-3}{\sqrt{x}}}{\frac{2}{\sqrt{y}}-8}
    =
    \frac{\frac{-3}{\sqrt{x}}}{\frac{2-8\sqrt{y}}{\sqrt{y}}}
    =
    (\frac{-3}{\sqrt{x}})(\frac{\sqrt{y}}{2-8\sqrt{y}})
    =
    -(\frac{3\sqrt{y}}{2\sqrt{x}-8\sqrt{xy}})
    =
    -(\frac{3\sqrt{y}}{2\sqrt{x}(1-4\sqrt{y})})
    [/itex]

    I'm currently learning algebra, so please correct me if I'm wrong.
     
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