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
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)
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
what is the value of 6 in 146027?
To solve the equation using completing the square, you've correctly rewritten the equation as:
3(x-1)^2 = 1
Now, to isolate the squared term, you need to divide both sides of the equation by 3:
3(x-1)^2 / 3 = 1 / 3
This simplifies to:
(x-1)^2 = 1/3
Next, you need to square root both sides of the equation to eliminate the squared term:
√((x-1)^2) = √(1/3)
This yields two possible solutions:
x - 1 = √(1/3)
and
x - 1 = -√(1/3)
To solve for x, you need to isolate the variable. Let's start with the first equation:
x - 1 = √(1/3)
Adding 1 to both sides of the equation gives:
x = 1 + √(1/3)
Similarly, for the second equation:
x - 1 = -√(1/3)
Adding 1 to both sides of the equation gives:
x = 1 - √(1/3)
Hence, the two solutions for x are:
x = 1 + √(1/3)
and
x = 1 - √(1/3)