Solving by completing the square

stacies-katkrazy

New member
Joined
Sep 9, 2006
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5
I am have some difficulty understanding the concept of solving by completing the square.


example: 3x^2-6x-5=0

I get moving the -5 to the right - but I get confused in the division after that.

Any help would be appreciated.[/tex][/code][/list]
 
Move the 5 to the right:

\(\displaystyle \L\\3x^{2}-6x=5\)

Factor out 3:

\(\displaystyle \L\\3(x^{2}-2x)=5\)

Add 1/2 the coefficient of x to both sides. one-half of 2 is 1

Because of the leading 3, we must add 3*1=3 to both sides:

\(\displaystyle \L\\3(x^{2}-\overbrace{2}^{\text{coefficient of x}}x+1)=8\)

Factor:

\(\displaystyle \L\\3(x-1)^{2}-8\)
 
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