Expert: Sherry Wallin - 12/27/2010

Question
3x^2-6x+6=0
How do I find them? What are the roots to this equation?

Answer
Sally~

First I would factor out the 3 because the coefficient on the squared term needs to be 1 when you complete the square:

3(x^2-2x+2)=0

What this equality says is that either 3 = 0 or x^-2x+2 = 0. Note: this would need to be factored either by the quadratic formula or by completing the square because it would be near to impossible to factor by sight and apparently you were told to use the method of completing the square to find the solutions to the equation. Solutions, zeros, and roots mean one and the same. Before we go any further you can find out if the roots are real or complex by using the discriminant: b^2-4ac
In this case a = 1, b = -2, and c = 2
b^2-4ac = (-2)^2 -4*1*2 = 4 -8 = -4, so there is a pair of complex roots.

I have a recipe for completing the square and the recipe goes like this:

1) make sure the coefficient on the squared term is 1  
2) move the constant to the right hand side
3) take half the coefficient of the term with a power of one, square it and add it to both sides of the equation
4) the left side is now a perfect square so write it as a perfect square
5) simplify the right hand side
6) take the square root of both sides and solve for the variable (usually x)

Now let's use these steps to find the roots of your equation above:

1  we already did this step
2  x^2 -2x + 2 - 2 = -2  -> simplified is x^2 - 2x = -2
3  x^2 -2x +(-1/2)^2 = -2 + (-1/2)^2  
4  (x-1/2)^2 = -2 + 1/4  -> the perfect square on the LHS is (x-1/2)^2
5  -2 + 1/4 ->  simplified this is (x-1/2)^2 = -7/4
  at this point you have: (x-1/2)^2 = -7/4
6  x -1/2 = +-sqrt(-7/4) = +-sqrt(-7)/2 = +- i*[sqrt(7)/2]
  x -1/2 + 1/2 =  +- i*[sqrt(7)/2]*i + 1/2 -> x = +-i*[sqrt(7) + 1]/2

the individual roots are x = [-i*sqrt(7) + 1]/2, [i*sqrt(7) + 1]/2

Math Prof