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how do you determine complex solutions?

x^7 + 7x^6 + 13x^5 - 6x^3 + 18x^2 + 10x - 7 = 0

2007-11-04 05:16:32 · 4 answers · asked by ? 2 in Science & Mathematics Mathematics

I need them cuz they are for the interschool test
...wow that was more complicated than I thought

2007-11-04 06:21:30 · update #1

Determine all the complex solutions of x7 + 7x6 + 13x5 - 6x3 + 18x2 + 10x - 7 = 0.

^^^^^^ original problem.

2007-11-04 06:55:13 · update #2

4 answers

You can employ Descartes' Rule of Sign to determine if there are any complex roots.

This will help you to determine the number of possible positive, negative and complex roots.

To find the maximum possible number of positive roots, look at the number of sign changes in f(x): 3 sign changes

To find the maximum number of negative roots, look at f(-x): -x^7 + 7x^6 -13X^5 + 6x^3 + 18^2 - 10x - 7.
There are 4 sign changes here.

Because its possible that you have complex roots due to the quadratic formula returning complex roots you have to look at those roots and count down in pairs (since complex roots always come in pairs).

Thus here are the possibilities:

3 or 1 positive roots
4, 2, or 0 negative roots

So now you need to take stock of the possible combinations:

3 positive, 4 negative, 0 complex
1 positive, 4 negative, 2 complex
1 positive, 2 negative, 4 complex
3 positive, 0 negative, 4 complex
1 positive, 0 negative, 6 complex

Remember, the total number of roots has to be 7.

As far as what those roots are - the rational roots test tells you that your possible rational roots are +/- 7 and +/- 1. You can plug those in and see if any of the produce 0. Otherwise you can try Newton's Method - but most 7th degree polynomials are probably very difficult to solve by conventional means. I plugged this into a function grapher - I saw that it crossed the x axis 5 times: 4 times to the left of the y axis and once to the right of the y axis.

This means that you have 1 positive root, 4 negative roots and 2 complex roots.

That should be a good start.

2007-11-04 05:56:17 · answer #1 · answered by Rock R 3 · 0 0

Real roots
-3.624666
-2.414214
-1.765495
-1
0.41423562
complex roots
0.695080808927+-0.781490191137
Using a scientific calkculator

2007-11-04 14:43:18 · answer #2 · answered by santmann2002 7 · 0 1

do you get that angellica?

2007-11-04 14:03:18 · answer #3 · answered by Math☻Nerd 4 · 0 0

why do you need to do these, jel?

2007-11-04 14:15:24 · answer #4 · answered by Josey 3 · 0 1

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