Question 62177
{{{sqrt(3x+1) - sqrt(x-1)= 2}}}

Add sqrt(x-1) to both sides: {{{sqrt(3x+1) = 2+sqrt(x-1)}}}.

Square both sides: {{{3x+1 = (2+sqrt(x-1))^2}}}.
Or, {{{3x+1 = 4+4sqrt(x-1)+x-1}}}.
Combine like terms: {{{2x-2 = 4sqrt(x-1)}}}.

Square both sides: {{{(2x-2)^2 = (4sqrt(x-1))^2}}}.
Or, {{{4x^2-8x+4 = 16(x-1) = 16(x-1) = 16x - 16}}}.
So, {{{4x^2-24x+20 = 0}}}.
Divide the whole equation by 4: {{{x^2-6x+5 = 0}}}.
Factor the equation: {{{(x-5)(x-1) = 0}}} so {{{x=5}}} or {{{x=1}}}.

Verify: 

When x=5, {{{sqrt(3x+1)-sqrt(x-1) = sqrt(16)-sqrt(4)=4-2}}} which is indeed 2.
When x=1, {{{sqrt(3x+1)-sqrt(x-1) = sqrt(4)-sqrt(0) = 2}}}.