## anonymous 4 years ago How do I solve: 3x^2 + 6x + 1=0 using the quadratic formula?

1. anonymous

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

2. anonymous

I did that and I got -6 +/- √24/6

3. anonymous

but the answers are: -3 +/- √6/3??

4. anonymous

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

5. anonymous

ohh..

6. anonymous

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

7. anonymous

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