Author 
Topic 

effort
Senior Member
USA
38 Posts 
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.



royhaas
Moderator
USA
3059 Posts 
Posted  12/01/2011 : 14:00:01

Try graphing 2. 


the_hill1962
Advanced Member
USA
1468 Posts 
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
Moderator
USA
3059 Posts 
Posted  12/06/2011 : 13:28:31

Multiple real roots only affect the shape of the graph. Compare the graphs of y=x^3x with y=x^4x^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^21 with x^41. Each has two real roots but teir shapes are different in [1,1]. 


effort
Senior Member
USA
38 Posts 
Posted  12/06/2011 : 13:51:39

Hill, yes, that is my question. Thank you for understanding my question. 


the_hill1962
Advanced Member
USA
1468 Posts 
Posted  12/06/2011 : 17:01:04

royhaas: The only thing that I am getting from your y=x^3x and y=x^4x^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^21 and y=x^41 is, I think, going farther from what the poster wanted explained. y=x^41 has some imaginary roots. However, I am glad that you offered that example because it made me think about y=(x1). Even further, y=(x1). effort: I think this may help you. You see the graph of y=(x1) 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=(x1). I think this describes what you asking! Look how weird it gets at x=1 and x=1 when compared to the standard y=x1 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=[(x1)^5](x1) and y=(x1)^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=[(x1)^5](x1) 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 xaxis. 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.



royhaas
Moderator
USA
3059 Posts 
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. 



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