# Division of Radicals Problem (Stuck)

#### EngineMan

So here is the problem:

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

Now I had thought that the method to solve it was to just multiply the numerator and denominator by the conjugate of the denominator, but I keep getting answers that are different from the solution provided, which is:

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

Here is what I am trying to do and how I am getting stuck:

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

This is where I get stuck or realize I am probably doing something totally wrong.

#### skeeter

MHF Helper
So here is the problem:

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

Now I had thought that the method to solve it was to just multiply the numerator and denominator by the conjugate of the denominator, but I keep getting answers that are different from the solution provided, which is:

$$\displaystyle \frac{(2(\sqrt{x+1})-x-2)}{x}$$
if $x = 1$ ...

$$\displaystyle \frac{1-(\sqrt{1+1})}{1+(\sqrt{1-1})} = 1-\sqrt{2}$$

$$\displaystyle \frac{(2\sqrt{1+1}-1-2)}{1} = 2\sqrt{2}-3$$

those two values are not equal ... recheck both the original fraction and the "solution provided"

#### EngineMan

It's definitely the solution. Here is a picture:

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#### skeeter

MHF Helper
as demonstrated, the given solution is invalid ... what else can I say?

1 person

#### EngineMan

I wasn't sure if I was inputting the numbers wrong or something in checking the problem and solution myself, so that is good to know. Thank you for your help. I am relieved that my inability to obtain the solution is just because the solution is wrong

#### skeeter

MHF Helper
fyi ...

$$\displaystyle \frac{1-\sqrt{x+1}}{1+\sqrt{x \color{red}{+} 1}}= \frac{2\sqrt{x+1}-x-2}{x}$$

2 people

#### EngineMan

What's ironic is that I had been thinking of trying it with the same sign in the square roots, but I kept thinking it would be a waste of time, that the problem was with me and not the solution.

#### skeeter

MHF Helper
What's ironic is that I had been thinking of trying it with the same sign in the square roots, but I kept thinking it would be a waste of time, that the problem was with me and not the solution.
No loss ... you learned something.