Question

I'm trying to solve by completing the square but i always get stuck at the end..
3x^2-6x+2=0
3(x^2-2x)+2=0
3(x^2-2x+1-1)+2=0
3(x-1)^2 -1 =0
3(x-1)^2=1
now what do i do?... the answer is supposed to be x=3+-(square root)of3-> the whole thing over 3

Answer 1

3x^2-6x+2=0
well, here is what I do:
3 x^2 - 6 x = -2 get the x^2 and x terms alone on left
x^2 - 2 x = -2/3 divide to get 1x^2
x^2 - 2 x + (2/2)^2 = -2/3 + (2/2)^2 add square if half of coefficient of x to both sides
x^2 - 2 x + 1 = -2/3 + 3/3
(x-1)^2 = 1/3
x-1 = +/- sqrt(1/3)
x = 1 +/- (1/3)sqrt 3
x =(1/3)(3+/- sqrt 3)

Answer 2

what you missed is maybe
sqrt (1/3) = 1/sqrt 3
multiply top and bottom by sqrt 3 to clear denominator of radical
1/sqrt 3 = (1/3)sqrt 3

Answer 3

what is the value of 6 in 146027?