## anonymous 5 years ago I need help! the square root of 3x +1 - the square root x-1=2

1. amistre64

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......

2. anonymous

ok

3. anonymous

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

4. anonymous

solutions are 1,5 dont know how to get there

5. anonymous

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

6. amistre64

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

7. amistre64

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

8. anonymous

thank you I am trying to comprehend all of this