1. ## Find All Solutions to "x" In Radical Fraction

Hi everyone!

I've been struggling with this problem involving radical expression as part of my algebra assessment:

[sqrt(x+1) + sqrt(x-1)]/[sqrt(x+1) - sqrt(x-1)] = 3 Solve for all possible solutions to "x".

So far I've tried to simplify the expression by multiplying the denominator by its conjugate, like so:

1) ([sqrt(x+1) + sqrt(x-1)]/[sqrt(x+1) - sqrt(x-1)]) * ([sqrt(x+1) + sqrt(x-1)]/[sqrt(x+1) + sqrt(x-1)]) = 3
2) (x+1 + 2[sqrt(x+1) * sqrt(x-1)] + x-1)/[x+1 - (x-1)] = 3
3) (2x + 2[sqrt(x+1) * sqrt(x-1)])/2 = 3
4) (x + [sqrt(x+1) * sqrt(x-1)]) = 3
5) ?

This is where I'm stuck. I don't know how to proceed from here to solving for all solutions to "x", so if someone could help me from here I would appreciate it!

2. ## Re: Find All Solutions to "x" In Radical Fraction

$3=\dfrac {\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=$

$\dfrac {\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}=$

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

$\dfrac {2x+2\sqrt{x^2-1}}{2}=3$

$x+\sqrt{x^2-1}=3$

$\sqrt{x^2-1}=3-x$

$x^2-1=9-6x+x^2$

$6x=10$

$x=\dfrac 5 3$

This is the only real solution.

3. ## Re: Find All Solutions to "x" In Radical Fraction

Wow it seems so obvious now! Thanks for the help