Question 1068538
I am having a bit of an issue solving sqrt(3x+1)-sqrt(x-1)=2 or √3x+1-√x-1 = 2

This is what I have done so far:

   √3x+1 = √x-1+2  

=> (√3x+1)2 = (√x-1+2)2

=> 3x+1 = x-1+4+4√x-1

=> -4√x-1 = -3x-1+x-1+4  or is it 4 √x+1 = -3x-1+x-1+4?
-4sqrt(x-1) = -3x-1+x-1+4 works.
-4sqrt(x-1) = -2x+2
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16(x-1) = 4x^2 - 8x + 4
4(x-1) = x^2 - 2x + 1
4x-4 = x^2 - 2x + 1
x^2 - 6x + 5 = 0
(x-1)*(x-5) = 0
x = 1, 5
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=>  4√x-1 = 2x-2  *This is where I get lost.  I have seen this as -2x+2 as well.
It doesn't matter.  Comes out the same after it's squared.
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I am really confused at this point and need to get to the next steps.