Question

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

Answer 1

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......

Answer 2

ok

Answer 3

sqrt(3x+1)-sqrt(x-1)=2

Answer 4

solutions are 1,5 dont know how to get there

Answer 5

\[\sqrt{3x+1}-\sqrt{x-1}=2\quad /()^2\]
\[3x+1+(x-1)-2\sqrt{(3x+1)(x-1)}=4\]
\[4x-2\sqrt{(3x+1)(x-1)}=4\qquad /-4x\]
\[-2\sqrt{(3x+1)(x-1)}=4-4x\qquad /:(-2)\]
\[\sqrt{(3x+1)(x-1)}=-2+2x\qquad /()^2\]
\[(3x+1)(x-1)=(-2+2x)^2\]
\[3x^2-2x-1=4+4x^2-8x\qquad /-3x^2+2x+1\]
\[0=x^2-6x+5\]
\[0=(x-1)(x-5)\]
Solutions: \[x_1=1,\quad x_2=5\quad\checkmark\]

Answer 6

isolate a sqrt :)

sqrt(x-1) = 2 -sqrt(3x+1) now square both side

Answer 7

and remember to watch your signs.... i appear to have dropped one on accident

Answer 8

thank you I am trying to comprehend all of this