# Solving the Quadratic Equation: sq.rt.(-x^2-4x-3)

• TKDKicker89
In summary, the problem is to simplify the expression \sqrt{-x^2-4x-3} and the approach is to complete the square by adding (4/2)^2 inside the parentheses. The result can be simplified to \sqrt{1-(x+2)^2} and further to \sqrt{(1-x)(3+x)} or \sqrt{(-1-x)(3+x)}.

#### TKDKicker89

I'm having some trouble figuring out how to simplify the following problem.
I know that i= the sq root of -1, and that i^2=-1, but I'm not sure how to approach this problem.
sq.rt.(-x^2-4x-3)

i would start by factoring out the -1 and seeing if i can't factor the polynomial more.

TKDKicker89 said:
I'm having some trouble figuring out how to simplify the following problem.
I know that i= the sq root of -1, and that i^2=-1, but I'm not sure how to approach this problem.
sq.rt.(-x^2-4x-3)

Exactly what is the problem? To simplify $$\sqrt{-x^2- 4x- 3)}$$?

Any time you have something like this, involving a square root,even if it doesn't involve i, think about completing the square.

-x2- 4x- 3= -(x2+ 4x)- 3 and we can see that we need to add (4/2)2= 4 inside the parentheses to complete the square. This is -(x2+ 4x+ 4- 4)- 3= -(x2+ 4x+ 4)+ 1=
-(x+2)2+ 1. The square root can be written as
$$\sqrt{1-(x+2)^2}$$. I don't see much more that can be done and I don't see that it has directly to do with i. Even though the original -x2- 4x- 3 has all "negatives", this can be positive. If x lies between -3 and -1, -x2-4x- 3 will be positive and the square root will be real.

$$\sqrt{1-(x+2)^2}$$ can be simplified more

Well, yes, of course, how foolish of me! $$\sqrt{1-(x+2)^2}= \sqrt{(1-(x+2))(1+(x+2))}= \sqrt{(1-x)(3+x)}$$

or even $$\sqrt{(-1-x)(3+x)}$$