Question

Question about completing the square?
This is my problem:
6m^2 + 5m + -3 = 0

I have no clue where to start. I know how to start a quadratic equation, but completing a square has really confused me.

Answer 1

Hello?

Answer 2

First, divide through by 6 so that the coefficient of the m^2 term is 1.
\[x^2+5x-3=0\]
\[x^2 + \frac{5}{6}x - \frac{1}{2}= 0\]

Now we move the last term to the right by adding 1/2 to each side, and then to complete the square, we take half of the coefficient of the middle term, square it, and add that to both sides:
\[x^2+\frac{5}{6}x + \frac{1}{2}+(\frac{5}{6*2})^2=\frac{1}{2} +(\frac{5}{6*2})^2\]
Having done that, we can write the left side as a square (we just completed it!)
\[(x+\frac{5}{12})^2 = \frac{1}{2}+\frac{25}{144}\] and now the rest is just taking the square root of both sides:

\[x+\frac{5}{12} = \sqrt{\frac{72}{144}+\frac{25}{144}}\]
\[x+\frac{5}{12}=\pm\frac{\sqrt{79}}{12}\]
\[x=\frac{-5\pm\sqrt{79}}{12}\]

Answer 3

Thanks so much for your help! I really appreciate this.