Division of Radicals Problem (Stuck)

Aug 2014
48
3
Upstate NY
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
Jun 2008
16,217
6,765
North Texas
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"
 
Aug 2014
48
3
Upstate NY
It's definitely the solution. Here is a picture:

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skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
as demonstrated, the given solution is invalid ... what else can I say?
 
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Aug 2014
48
3
Upstate NY
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
Jun 2008
16,217
6,765
North Texas
fyi ...

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

typo is the sign in the denominator's radical ... should be +1
 
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Aug 2014
48
3
Upstate NY
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
Jun 2008
16,217
6,765
North Texas
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.