SOLUTION: Hi, I'm being asked to solve for x by completing the square:
-2x^2 + 12x - 6 = 0
-2x^2 + 12x = 6 < I moved the 6 to the other side
x^2 - 6x = -3 < I divide
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Quadratic Equations and Parabolas
-> SOLUTION: Hi, I'm being asked to solve for x by completing the square:
-2x^2 + 12x - 6 = 0
-2x^2 + 12x = 6 < I moved the 6 to the other side
x^2 - 6x = -3 < I divide
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Question 355495: Hi, I'm being asked to solve for x by completing the square:
-2x^2 + 12x - 6 = 0
-2x^2 + 12x = 6 < I moved the 6 to the other side
x^2 - 6x = -3 < I divided everything by -2 to get the x^2 by itself
x^2 - 6x + 9 = 6 < I took half of the coefficient 6 (6/2 = -3), squared
it (9), and added it to both sides of the equation
This is where I get stuck. I'm supposed to write the left side in squared form, but because there is a plus and minus, it wouldn't be just one factor squared. Would I write this as (x-3)(x+3)? If so, how would I take the square root of that?
Thanks for any guidance you can provide! Answer by scott8148(6628) (Show Source):
