Question

How do I solve: 3x^2 + 6x + 1=0 using the quadratic formula?

Answer 1

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=3,b=6,c=1\]

Answer 2

I did that and I got -6 +/- √24/6

Answer 3

but the answers are: -3 +/- √6/3??

Answer 4

\[\frac{-6\pm\sqrt{36-12}}{6}\]
\[\frac{-6\pm\sqrt{24}}{6}\]
\[\frac{-6\pm2\sqrt{6}}{6}\]
\[\frac{2(-3\pm\sqrt{6})}{6}\]
\[\frac{-3\pm\sqrt{6}}{3}\]

Answer 5

ohh..

Answer 6

only gimmick is
\[24=4\times 6\] so
\[\sqrt{24}=\sqrt{4}\sqrt{6}=2\sqrt{6}\] and you can cancel with the denominator. make sure to factor before you cancel

Answer 7

last question: why is it -3 +/- √6 / 3 in the end? When before it it's 2 (-3 +/- √6)/6?