Finding roots of a quadratic (Lindsay's question from Facebook)

[Jameson]

Lindsay on Facebook writes:

Solve for the roots of 3x^2+12x +8=0. I got +-12 /48 over 6. (/ was the square root sign) yet the answer is the same but -84 instead of 48. What is it?

I think you mean you got -48 instead of 48, but either way let's go through solving this.

\(\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Looking at $3x^2+12x+8$ we see that $a=3$, $b=12$ and $c=8$. Plugging that into the above quadratic equation yields.

\(\displaystyle x = \frac{-12 \pm \sqrt{144-4(3)(8)}}{6}\)

Note that \(\displaystyle \sqrt{144-4(3)(8)}=\sqrt{144-96}=\sqrt{48}\) so the above line simplifies to the following.

\(\displaystyle x = \frac{-12 \pm \sqrt{48}}{6}\)

\(\displaystyle x = \frac{-12 \pm 4 \sqrt{3}}{6}=-2 \pm \frac{2 \sqrt{3}}{3}\)

Now to answer your original question, the 48 under the square root sign should be positive. You are most likely only dealing with real numbers so you can't take the square root of a negative is a good rule to remember.

If you have any other questions about this I invite you to register and ask here.