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No of roots of a equation

  1. Sep 23, 2012 #1
    A linear equation has 1 root
    A quadratic equation has 2 roots(including two equal roots and two complex roots)
    A cubic equation has 3 roots
    Is this means that the no of roots of a equation in one unknown depends on the degree of the polynomial?Why?Any proof and explanation?
    Thx a lot :)
  2. jcsd
  3. Sep 23, 2012 #2
    Yes, every nonzero n-degree polynomial has exactly n roots (counting multiplicity and complex roots). The result is known as the fundamental theorem of algebra. The proof of this result is not easy at all.

    A proof is given in Spivak's "Calculus" in the chapter "Complex functions".
  4. Sep 23, 2012 #3


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    Yes it means exactly that! It's the fundamental theorem of algebra.
  5. Sep 23, 2012 #4
    I am exactly the same age as you but I can't understand what is fundamental theorem of algebra that I have searched in wiki,it is too difficult for me to understand as I am just a silly A-level student,anyway thx :)I hope I will understand it in the future.
  6. Sep 23, 2012 #5
    LOL micromass isn't 15. :bugeye:
  7. Sep 23, 2012 #6
    It simply says that any nonconstant polynomial has a root. For example, [itex]x^6+x^4+2x^3+1=0[/itex] has a solution, even though we cannot find it explicitely.

    Using that result, we can prove that any n-th degree polynomial (with n>0) has exactly n roots.

    I'm 15 and a half. Almost 16. :biggrin:
  8. Sep 23, 2012 #7
    Pfft. :grumpy:
  9. Sep 23, 2012 #8


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    Although not part of the statement of the 'Fundamental Theorem of Algebra' (which says simply that any polynomial equation has at least one root), but an immediate result is that any polynomial can be written in linear factors: [itex]a(x- x_0)(x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_n)[/itex].

    In order that the polynomial have degree n, there must be n "x"s multiplied together and so n values of [itex]x_i[/itex]. That is what is meant by "a polynomial of degree n has n roots, not necessarily different"- that how we distinguish how many roots there are that are the same number.
  10. Sep 28, 2012 #9
    I forget where i read it but i am certain i once read that the Fundamental theorem of algebra is not a theorem at all. It requires that i^2 = -1 which is a formalization or dressing up so to speak of the given number systems. We have to break an axioms to jump from the real to complex system.
  11. Sep 28, 2012 #10


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    That's BS. FTA is a theorem about polynomials over the complex numbers. The statement i2 = -1 is perfectly valid in ℂ. And no, we don't "break axioms" at all. The complex numbers are constructed using pairs of reals.
  12. Sep 28, 2012 #11

    The mere fact that one complex number is not considered larger or smaller than any other complex is in itself the breaking of the unsaid axiom of "MAGNITUDE" which is certainly true in the set of reals.

    Also, its either the associative or commutative law which is broken when jumping to the complex system. Its been a while since i have been into this stuff so i,m going to look it up.
  13. Sep 28, 2012 #12


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    What do you mean by "unsaid axiom"?. Unless otherwise stated, the axioms of mathematics are ZFC.
    Secondly, you can compare "magnitude": the reals have an absolute value and so do the complex numbers.
    What is true is that complex numbers do not form a totally ordered field.

    The complex numbers form a field: addition and multiplication are both associative and commutative.
    Last edited: Sep 28, 2012
  14. Sep 28, 2012 #13

    It indeed has been a long time since i have been into this stuff.

    Iv'e got a feeling i have been mixing up two entirely unrelated things.

    The axiom dropping actually has to do with the Minkowski (Spelling?) equation for Special Relativity.

    One of the components in R4 is time and an axiom has to be dropped in order for the equation to geometrically explain the time difference for space travelers relative to a stationary individual.

    However i have just remembered where i have read that the FTA is not a theorem at all. It was the Encyclopedia Britannica 1968 under the section "Foundations of Mathematics." It indeed also said that Complex numbers were a formalization and that there was also no proof that negative numbers even exist.

    This even to me seems bizarre but i know what i read and i reread that section
    many times.

    If none of this is true then the encyclopedias we had at home as children were a fraud.

    This is actually freaking me a bit.
  15. Sep 28, 2012 #14
    You are confusing math with physics. In math, ℝ4 is just the set of 4-tuples (x1, x2, x3, x4) where each xi is a real number. And ℝ8678 is the set of 8678-tuples. There's no physical interpretation needed ... that's for the physicists.

    As far as negative numbers not existing, I quite agree. Positive numbers too. I've seen 3 books, 3 apples, and 3 visually challenged mice; but I've never seen the number 3 in isolation. It's just a mathematical abstraction. Abstraction is what's gotten us from cave-dwelling to our present state of civilization. Law, justice, politics, truth, wisdom, works of fiction, reading and writing, all these things are human abstractions. They exist, but not in the physical world. Now we're into philosophy. In math, imaginary numbers exist. In the physical world, even the familiar counting numbers don't have concrete existence.

    Anyway the number i is just a counterclockwise turn of 90 degrees in the plane. Do it twice and it's the same as multiplying your direction by -1. Voila, i2 = -1. The powers of i are just the sequence up, left, down, right repeated over and over. We're a little off-topic here but imaginary numbers are very real ... if by real you allow abstractions. If you don't allow abstractions, even the number 3 isn't real.

    FTA is a theorem of mathematics usually proved in undergrad math major classes in algebra or complex analysis. It was first proved by Gauss in 1799 when he was 21 years old. The rest of you 30-something Ph.D. candidates, get busy!! :-)
  16. Sep 28, 2012 #15

    I think you have cleared the matter up nicely.

    on a side note ...

    Is this solvable

  17. Sep 29, 2012 #16
    What do you mean? Are you trying to integrate it?
  18. Oct 1, 2012 #17
    Yes...integrate it

    I have seen the integral symbol used in other posts but i'm not so computer savvy with all the whistle and bells.
    Oh here it is.

  19. Oct 1, 2012 #18


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    And the integers break the rule basic for the naturals:
    "There exists a LEAST number".

    The rationals break with the fundamental rule of integers that says:
    "There exists a number "a", so that adding or subtracting "a" from an integer (some number of times) will yield any other integer"

    Whar was your point again?
  20. Oct 1, 2012 #19


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    Instead of using the ^ symbol, I recommend using the superscript option of the forum, so it looks like ∫(x2+2)5 dx.

    There are two methods you can use. The first is the binomial formula and expand out the brackets. Then integrate term by term. The only thing to be careful is calculating the coefficients, but a good calculator can take care of that.

    The other option is factorise into ##\int(x^2+2)^5 \, dx = \int (x+i\sqrt2)^5(x-i\sqrt2)^5\,dx## and use integration by parts.
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