I need help! the square root of 3x +1 - the square root x-1=2

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I need help! the square root of 3x +1 - the square root x-1=2

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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it would help out alot in understanding your question if you would use paratnthesis to indicate what would be "under" the radical. like this: sqrt(3x-1) - sqrt(x) -1 = 0 otherwise, we dont know what your problem really is......
ok
sqrt(3x+1)-sqrt(x-1)=2

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solutions are 1,5 dont know how to get there
\[\sqrt{3x+1}-\sqrt{x-1}=2\quad /()^2\] \[3x+1+(x-1)-2\sqrt{(3x+1)(x-1)}=4\] \[4x-2\sqrt{(3x+1)(x-1)}=4\qquad /-4x\] \[-2\sqrt{(3x+1)(x-1)}=4-4x\qquad /:(-2)\] \[\sqrt{(3x+1)(x-1)}=-2+2x\qquad /()^2\] \[(3x+1)(x-1)=(-2+2x)^2\] \[3x^2-2x-1=4+4x^2-8x\qquad /-3x^2+2x+1\] \[0=x^2-6x+5\] \[0=(x-1)(x-5)\] Solutions: \[x_1=1,\quad x_2=5\quad\checkmark\]
isolate a sqrt :) sqrt(x-1) = 2 -sqrt(3x+1) now square both side
and remember to watch your signs.... i appear to have dropped one on accident
thank you I am trying to comprehend all of this

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