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

[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)

 

[yourdadonapogostick]

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

 

[HallsofIvy]

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.

 

[yourdadonapogostick]

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

 

[HallsofIvy]

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)}

 

[Delta]

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