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

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)}.
  • #1
TKDKicker89
2
0
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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  • #2
i would start by factoring out the -1 and seeing if i can't factor the polynomial more.
 
  • #3
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.
 
  • #4
[tex]\sqrt{1-(x+2)^2}[/tex] can be simplified more
 
  • #5
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]
 
  • #6
or even [tex]\sqrt{(-1-x)(3+x)}[/tex] :wink:
 

What is a quadratic equation?

A quadratic equation is a polynomial equation of the second degree, meaning it contains a squared term. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

How do you solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula or factor the equation. The quadratic formula is (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the constants from the equation. Factoring involves finding two numbers that when multiplied together equal c and when added together equal b. These numbers can then be used to rewrite the equation as (x + m)(x + n) = 0, where m and n are the two numbers. This method is often used when the quadratic equation is easily factorable.

What is the Discriminant?

The Discriminant is the part of the quadratic formula under the square root sign, b^2 - 4ac. It is used to determine the nature of the solutions to a quadratic equation. If the Discriminant is positive, there are two distinct real solutions. If the Discriminant is zero, there is one real solution. If the Discriminant is negative, there are two complex solutions.

What is the role of the square root in the equation?

The square root is used in the quadratic formula to find the solutions to the equation. It allows for both positive and negative solutions to be found. Without the square root, only one solution would be possible.

How is the quadratic equation used in real life?

The quadratic equation has many real-life applications, such as in physics, engineering, and finance. For example, it can be used to find the maximum height of a projectile or the optimal cost for a product. It is also useful in graphing parabolas, which can represent various real-world phenomena.

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