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

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Homework Help Overview

The discussion revolves around simplifying the expression involving a square root of a quadratic equation, specifically sq.rt.(-x^2-4x-3). The subject area pertains to algebra and complex numbers, particularly focusing on the implications of negative values under a square root.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss factoring out negative values and completing the square as potential approaches to simplify the expression. There are questions regarding the nature of the square root when involving complex numbers and the conditions under which the expression yields real values.

Discussion Status

Several participants have offered insights into the simplification process, with one suggesting the completion of the square and another exploring further simplification of the resulting expression. There appears to be a productive exchange of ideas, though no consensus has been reached on a final method or interpretation.

Contextual Notes

Participants note the importance of understanding the conditions under which the expression is positive or negative, particularly in relation to the values of x that would affect the outcome of the square root.

TKDKicker89
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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)
 
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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 [tex]\sqrt{-x^2- 4x- 3)}[/tex]?

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
[tex]\sqrt{1-(x+2)^2}[/tex]. 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.
 
[tex]\sqrt{1-(x+2)^2}[/tex] can be simplified more
 
Well, yes, of course, how foolish of me! [tex]\sqrt{1-(x+2)^2}= \sqrt{(1-(x+2))(1+(x+2))}= \sqrt{(1-x)(3+x)}[/tex]
 
or even [tex]\sqrt{(-1-x)(3+x)}[/tex] :wink:
 

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