# Homework Help: Question about rewriting an expression with radicals

1. Jul 5, 2011

### Joe_K

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. Jul 5, 2011

### QuarkCharmer

They rationalized the denominator, probably by multiplying it by it's conjugate (both numerator and denominator)

$$\frac{2}{\sqrt{y}-8}$$
$$\frac{2(\sqrt{y}+8)}{(\sqrt{y}-8)(\sqrt{y}+8)}$$

at least I believe, it's hard to see through all those brackets.

3. Jul 5, 2011

### eumyang

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

If so, then all that was done was to subtract, after finding a common denominator:
$\frac{2}{y^{1/2}} - 8 = \frac{2}{\sqrt{y}} - \frac{8\sqrt{y}}{\sqrt{y}} = ...$

4. Jul 7, 2011

### dreasgrech

I simplified it as follows:

$\frac{\frac{-3}{\sqrt{x}}}{\frac{2}{\sqrt{y}}-8} = \frac{\frac{-3}{\sqrt{x}}}{\frac{2-8\sqrt{y}}{\sqrt{y}}} = (\frac{-3}{\sqrt{x}})(\frac{\sqrt{y}}{2-8\sqrt{y}}) = -(\frac{3\sqrt{y}}{2\sqrt{x}-8\sqrt{xy}}) = -(\frac{3\sqrt{y}}{2\sqrt{x}(1-4\sqrt{y})})$

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