"rationalize" the denominator rather than the numerator

1. May 8, 2005

powp

Hello All

Simplify the following

$$\frac{\sqrt{x} + 1}{\sqrt{x} - 1}$$

and Preform the indicated operations and simplfy

$$\sqrt{x}(\sqrt{x} + 1)(2\sqrt{x}-1)$$

Thanks

P

2. May 8, 2005

Anzas

the first:

$$\frac{\sqrt{x} + 1}{\sqrt{x} - 1}$$

$$\frac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} - 1)}$$

$$\frac{x-1}{x -2 \sqrt{x} +1}$$

the second:

$$\sqrt{x}(\sqrt{x} + 1)(2\sqrt{x}-1)$$

$$\sqrt{x}(2x+ \sqrt{x}-1)$$

$$2x \sqrt{x}+x- \sqrt{x}$$

3. May 8, 2005

HallsofIvy

Staff Emeritus
I would have been inclined to "rationalize" the denominator rather than the numerator:
$$\frac{\sqrt{x}+1}{\sqrt{x}-1}= \frac{(\sqrt{x}+1)(\sqrt{x}+1)}{(\sqrt{x}-1})(\sqrt{x}+1)}= \frac{x+ 2\sqrt{x}+1}{x-1}$$

4. May 8, 2005

powp

thanks thats the answers I got but was not sure

When would you "rationalize" the denominator over the numerator or the other way around.

5. May 8, 2005

snoble

Generally simplified form means always "rationalizing" the denominator. Also you can simplify a little bit more. $$\frac{x+ 2\sqrt{x}+1}{x-1} = \frac{ (1+\sqrt{x})^2}{x-1}$$