How do you solve $\frac{x-\sqrt x}{\sqrt x-1} \cdot \frac{\sqrt x+1}{\sqrt x+1}$?

Question

I've been having a bit of trouble understanding the formula

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

Apparently it equals sqrt(x) but I have no clue how to get that. Whenever I multiply it out I get

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

How do I simplify it further?

Thank you so much!

Answer 1

For the sake of this question not remaining in the unanswered queue ...

The second fraction is 1 and the first fraction is \begin{eqnarray*} \frac{\color{red}{\sqrt{x}}(\sqrt{x}-1)}{(\sqrt{x}-1)}. \end{eqnarray*}

Answer 2

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

$\frac{\sqrt x+1}{\sqrt x+1}$ go to 1.

So, $\frac{x-\sqrt x}{\sqrt x-1} \cdot\ 1 = \frac{x-\sqrt x}{\sqrt x-1} $

We can say that:

$\frac{x-\sqrt x}{\sqrt x-1}$ = $\frac{x^{\frac{1}{2}}\cdot x^{\frac{1}{2}}-\sqrt{x}}{\sqrt{x}-1}$

which is basically:

$\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}$

Cancel the like terms and you will end up with $\sqrt{x}$

Answer 3

This is a typical high school problem when teaching rationalizing a denominator. A method which actually demonstrates the concept.

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

The top F.O.I.L.'s out or multiplies as follows. Note how x applies to each term in the right parentheses resulting in the first two terms. Then -root x applies to each term in the right parentheses resulting in the 3rd and 4th terms.

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

Note that when $$-\sqrt x \sqrt x$$ simplifies, it becomes -x. So x - x cancels leaving $$\frac{x*\sqrt x -\sqrt x}{(\sqrt x -1)(\sqrt x +1)}$$

The top factors and the questioner seems to understand the bottom becomes x -1

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

Also, for education purposes, this method applies to a more common and broader range of exercises than factoring square root of x.