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 roots and zeros

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effort Posted - 12/01/2011 : 11:18:47
Solve each equation. State the number and type of roots.
1. (x^4) + 625 =0 2. (x^5) +4x^ = 0

1. I solved for x and did get 4 answers: 5i, -5i, 5i, -5i

four complex root.

2. I solved for x and did get 5 answers: 0,0,0, 2i, -2i

How do you explain that o is an answer 3 times. And how do you explain that the graph goes through the zero 3 times.
6   L A T E S T    R E P L I E S    (Newest First)
royhaas Posted - 12/07/2011 : 06:34:43
The "flattening" occurs as a result of the behavior of the polynomial in |x|<1. This is the case even if all roots are conjugate imaginary roots. The behavior becomes more pronounced with the degree of the polynomial. For example, look at (x^2+1)^2 vs (x^2+1)^4, both of which have repeated complex roots but no real zeroes.
the_hill1962 Posted - 12/06/2011 : 17:01:04
The only thing that I am getting from your
y=x^3-x and y=x^4-x^2 example is that the latter adds the root of x=-1 and that is why it changes shape and goes up through the x axis again.
Your y=x^2-1 and y=x^4-1 is, I think, going farther from what the poster wanted explained. y=x^4-1 has some imaginary roots.
However, I am glad that you offered that example because it made me think about y=(x-1).
Even further, y=(x-1).
I think this may help you.
You see the graph of y=(x-1) has the "in and out" (referencing my previous post) because it has TWO roots at x=1 and TWO roots at x=-1.
Now, look at the graph of y=(x-1). I think this describes what you asking! Look how weird it gets at x=-1 and x=1
when compared to the standard y=x-1
I does go "through" the axis because it has three (not just two) roots at each.
Hopefully you will experiment on your own more. Here are a couple that I did:
y=[(x-1)^5](x-1) and y=(x-1)^4
The latter shows that my theory holds. Since the roots x=-1 and x=1 are repeated FOUR times (an even number), the curve does not go "through" the axis. That got me to thinking and I wondered what
y=[(x-1)^5](x-1) would look like since it would have FIVE x=-1 and SIX x=1 roots. The graph came out interesting. It does go "through" the x axis at x=-1 and "in and out" at x=1.
The best that I can say to answer your original question is that as the more times a root is repeated somewhere, the more it "flattens out" at that spot where it go through the x-axis. If it is an even number of repeated roots there, it does not go through but the same flattening effect happens.
royhaas: Please let me know if I am not correct in what I have stated or done.

effort Posted - 12/06/2011 : 13:51:39
Hill, yes, that is my question. Thank you for understanding my question.
royhaas Posted - 12/06/2011 : 13:28:31
Multiple real roots only affect the shape of the graph. Compare the graphs of y=x^3-x with y=x^4-x^2 over the interval [-2,2]. The overall shape of a graph also depends on first and second derivatives. Note that the distinct roots in both cases is {-1,0,1}. The degree of the polynomial also contributes heavily to the overall shape, as evidenced by the graphs of x^2-1 with x^4-1. Each has two real roots but teir shapes are different in [-1,1].
the_hill1962 Posted - 12/06/2011 : 11:55:23
This is something that I have also wondered about. I know that a "double root" touches the x axis but doesn't cross. a root that is repeated an odd number of times does cross.
effort: You are asking for an explanation. I don't know if I am correct in the following, but here is how I think about it---
The graph of the curve goes "in" and and instead of just continuing through it, it goes "out" on the same side it came in. Therefore I just think, "that is why there are two roots there".
You have a good question asking for an explanation of a triple root!
I have the same question.
Is there anyone that can offer an "explanation"?
royhaas Posted - 12/01/2011 : 14:00:01
Try graphing 2.

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