# Complex Conjugate Roots and the Quadratic Formula

• Feb 29th 2012, 12:56 PM
FatalSylence
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 $x^2+2x+7=0$

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

Now, I know the answer to this question is two complex conjugate roots, $x = - 1 \pm i \sqrt{6}$, 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.

• Feb 29th 2012, 01:19 PM
saravananbs
Re: Complex Conjugate Roots and the Quadratic Formula
sqrt(-1)=i
and
sqrt(24)=2sqrt(6)

so sqrt(-24)=i2sqrt(6)

now you try.
• Feb 29th 2012, 01:46 PM
FatalSylence
Re: Complex Conjugate Roots and the Quadratic Formula
I'm still not sure I see where the $\sqrt{6}$ 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. $\sqrt{ab} = \sqrt{a} \sqrt{b}$ so $\sqrt{24} = \sqrt{4} \sqrt{6} = 2 \sqrt{6}$