Thread: complex roots confusion

Originally Posted by anthonye

So going back to the equation .
How can I tell from just the equation how many roots of each type it will have?

Look at this solution.
The polynomial has degree four and has all real coefficients. That tells us that if there are any complex roots (i.e. the imaginary parts non-zero) they must occur in conjugate pairs. Thus there could be zero, two, or four complex roots. In fact there are two. There are also two irrational roots. See the web link above.

Ok thank you again plato it's a question on a web site and it say that it has four complex roots
thats what was confusing me.

Any sites were I can go and test myself on this matter?

Originally Posted by anthonye

Any sites were I can go and test myself on this matter?

Are you sure that you copied the equation correctly?
You can use WolframAlpha to solve such equations.

Yeah your right that x^2 should be possitive

Hi;
Question state the number of complex roots,the possible number of real roots
of x^3 + 4x^2 + 5x -1 = 0

complex roots = 3 why(complex numbers include real numbers)

real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)

are my statements correct?

Originally Posted by anthonye

Ok getting there now.

This is what I have:

For the equation x^4 - 3x^3 - x^2 - x + 3 = 0

Find the number of complex roots,The possible number of real roots
and the possible number of rational roots.


complex roots = 4 (The reason for my answer is complex numbers are almost any numbers)

Yes, that is true.

real roots = 0,2,4 ( the reason for my answers are 0 the answer may all be complex) and this is where i'm confused

real numbers are on the number line -3,-2,-1,0,1,2,3ect

Well, not just integers, of course, numbers like " ", " ", etc. are real numbers. And you can say a little bit more. By "DeCarte's rule of signs", there can be either 0 or 2 positive real roots (because the sign changes at " " and at "+3"- two sign changes. DesCarte's rule of signs says the number of positive real roots is at most the number of sign changes but may be less by an even number. Now, replacing "x" by "-x" swaps positive and negative roots and which also has 0 or 2 sign changes.

But if a complex number is say 3(imaginary part=0) is 3 complex or
real?

Both. The set of all real numbers is a subset of the set of all complex numbers.

Thanks.

For the question about rational roots, you can use the "rational roots theorem": if the rational number , with m and n integers, of the polynomial equation , all coefficients being integers, then n must evenly divide the "leading coefficient", and m must evenly divide the "constant term", .

Here, so the "leading coefficient" is 1 which is evenly divisible only by 1 and -1. The "constant term" is 3 which is evenly divisible only by 1, -1, 3, and -3. That is, the only possible rational roots are the integers 1, -1, 3, and -3. You can put each into the polynomial to see which, if any, actually satisfy the equation.

Did you miss my last post.

Originally Posted by anthonye

Question state the number of complex roots,the possible number of real roots
of x^3 + 4x^2 + 5x -1 = 0
complex roots = 3 why(complex numbers include real numbers)
real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)
are my statements correct?

I think that you have misunderstood the whole point of this discussion.
Here is a direct quote: " the cubic polynomial has two complex roots"
Now that polynomial has three roots, one real and two complex.
As an accident of history, the language of roots of a polynomial we use the words complex root to mean a complex number with a nonzero imaginary part. Thus , has at least one real root and at most two complex roots.

NOTE: above we are talking about complex roots, not complex numbers as such.

Yes exactly plato two complex and one real thats what I got because complex
roots always come in pairs so that web page is wrong.

Originally Posted by anthonye

Yes exactly plato two complex and one real thats what I got because complex
roots always come in pairs so that web page is wrong.

What web page are you talking about?

Originally Posted by anthonye

Hi;
Question state the number of complex roots,the possible number of real roots
of x^3 + 4x^2 + 5x -1 = 0

complex roots = 3 why(complex numbers include real numbers).

real roots = 1 or 3 why(1 in case two are complex,3 non may be complex)

are my statements correct?

Yes, because non-real solutions to an equation with real coefficients must come in complex conjugate pairs, there must be either 2 or no non-real solutions. That leaves either 1 or 3 real roots. And we can say if there only one real root it must be positive. If all three roots are real, one is positive and the other two negative.

Hi;
Just need my work checking(I'm sure it's all correct now)


For each equation state the number of complex roots,the possible number or real roots
and the possible rational roots.

1) -x^4 = 0 (complex roots 0 or2 or 4)
(real roots 2 or 4)
(rational roots 0)



2) 2x^5 - 4x^4 - 4x^2 + 5 = 0 (complex roots 0 or 2 or 4)
(real roots 1 or 3 or 5)
(rational roots + or - 5/2 or 1/2 or 1 or 5)


3)4x + 8 = 0 (complex roots 0)
(real roots 1)
(rational roots + or - 1/4 or 1/2 or 1 or 2 or or 8)


4)-2x^6 - x^2 + x - 7 = 0 (complex roots 0 or 2 or 4 or 6)
(real roots 2 or 4 or 6)
(rational roots + or - 1 or 1/2 or 7 or 7/2)


5)x^10 + x^8 - x^4 + 3x^2 - x + 1 = 0 (complex roots 0 or 2 or 4 or 6 or 8 or 10)
(real roots 0or 2 or 4 or 6 or 8 or 10)
(rational roots + or - 1)

Please someone Please check my work above

Thank you.

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