Posted by **jona** on Saturday, February 14, 2009 at 5:24pm.

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

- math -
**Damon**, Saturday, February 14, 2009 at 5:32pm
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)

- math -
**Damon**, Saturday, February 14, 2009 at 5:36pm
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

- math -
**karly**, Sunday, February 15, 2009 at 11:43pm
what is the value of 6 in 146027?

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