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^26x+2=0
3(x^22x)+2=0
3(x^22x+11)+2=0
3(x1)^2 1 =0
3(x1)^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^26x+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
(x1)^2 = 1/3
x1 = +/ 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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