Complex Conjugate Roots and the Quadratic Formula

Hello, there.

I have the quadratic equation http://www.mathhelpforum.com/math-he...s1_q20_g01.gifhttp://www.mathhelpforum.com/math-he...s1_q20_g01.gif$\displaystyle x^2+2x+7=0$

Using $\displaystyle \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}$, I get $\displaystyle \begin{array}{*{20}c} {x = \frac{{ - 2 \pm \sqrt {-24} }}{{2}}} \\ \end{array}$.

Now, I know the answer to this question is two complex conjugate roots, $\displaystyle x = - 1 \pm i \sqrt{6}$, my question is how? The square root of 24 isn't 6. I'm confused.

I guess I'm just asking for the steps to get there.

Thank you in advance.

Re: Complex Conjugate Roots and the Quadratic Formula

sqrt(-1)=i

and

sqrt(24)=2sqrt(6)

so sqrt(-24)=i2sqrt(6)

now you try.

Re: Complex Conjugate Roots and the Quadratic Formula

I'm still not sure I see where the $\displaystyle \sqrt{6}$ comes from..

This is on the tip of my tongue from when I took Algebra, but I'm not sure I remember how to factor radicals.

Edit: Nevermind I get it. $\displaystyle \sqrt{ab} = \sqrt{a} \sqrt{b}$ so $\displaystyle \sqrt{24} = \sqrt{4} \sqrt{6} = 2 \sqrt{6}$