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Homework Help: 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)


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


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    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:


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