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Why is $\Delta = b^2 - 4ac$ so special that it deserves a name of its own ? I saw it only once in quadratic formula and nowhere else.

Does higher polynomials also have discriminant ? if so then is there a general formula for it in terms of coefficients ?

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    $\begingroup$ en.wikipedia.org/wiki/Discriminant $\endgroup$ – carmichael561 May 30 '17 at 23:44
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    $\begingroup$ The discriminant tells us whether a quadratic polynomial will have two (if the disc. is positive), one (if the disc. is zero), or no real roots (if the disc. is negative). $\endgroup$ – Dave May 30 '17 at 23:52
  • $\begingroup$ @carmichael561 I don't have any background to read it. $\endgroup$ – user8277998 May 31 '17 at 0:18
  • $\begingroup$ @Dave Yes I know that. It follows from the quadratic formula. So it is not anything special of $\Delta$ in on itself. It is just a application of the formula. $\endgroup$ – user8277998 May 31 '17 at 0:20
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It was mentioned already that the discriminant of a polynomial is zero if and only if the polynomial has multiple roots (i.e. it has some factor of $(x-a)^n$ for $n>1$).

The discriminant is defined for higher polynomials (there is a formula on wikipedia in terms of the polynomial and its derivative). Perhaps an easier to explain definition is as follows: given a polynomial $$p(x)=\prod_{i=1}^d(x-\alpha_i)$$ (recall from the fundamental theorem of algebra than any polynomial with real coefficients [and leading coefficient 1] can be written in this form, with $\alpha_i$ complex). Then we say $$\text{disc}(p(x))=\prod_{i<j}(\alpha_i-\alpha_j)^2$$ where the product is taken over all pairs $(i,j)$ with $1\le i<j\le d$. (Simple exercise - verify this definition agrees with the discrimiant in the quadratic formula.)

In abstract algebra (particularly in Galois Theory), we can do a bit more with the discriminant than just check that no roots are the same. Consider a polynomial with rational coefficients, $$p(x)=\sum_{i=0}^d a_ix^i=a_dx^d+\dots+a_1x+a_0$$ for some $a_i\in\mathbb{Q}$. In some cases, we can factor such polynomials into linear factors with coefficients in $\mathbb{Q}$: $x^2-2x+1=(x-1)^2$. In other cases, we cannot: $x^2+1=(x+i)(x-i)$. In this case, the polynomial factors into linear polynomials with coefficients in $\mathbb{C}$. Both $\mathbb{C}$ and $\mathbb{Q}$ are sets called fields - essentially they are sets where you can do multiplication, division (except by $0$), addition, and subtraction. There are many other examples of fields, for instance $\mathbb{R}$. Notice that $\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$. We could add a lot of other fields into this chain of inclusions. There are also fields between $\mathbb{Q}$ and $\mathbb{C}$ that do not include $\mathbb{R}$ or have $\mathbb{R}$ as a subset. One of the fundamental questions of Galois theory is: "What is the smallest field over which we can write a polynomial as a product of linear factors?" This smallest field is called the splitting field. It turns out that the discriminant of a polynomial can be used to compute some interesting properties of splitting fields of polynomials.

As a quick note, the splitting field of $x^2+1$ viewed as a polynomial with rational coefficients is the set $\{a+bi:a,b\in\mathbb{Q}\}\subset\mathbb{C}$. This is usually denoted by $\mathbb{Q}(i)$ (or $\mathbb{Q}[i]$ if we are not emphasizing the fact that it is a field). Also, notice that $\mathbb{R}\not\subset\mathbb{Q}(i)$ and $\mathbb{Q}(i)\not\subset\mathbb{R}$.

I am not sure what your background is, so I will leave it at this for now, but feel free to ask if you want me to expand on more things. Also, I am by no means an expert. Hopefully someone with more experience could talk more about what discriminants are useful for.

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    $\begingroup$ Thank you. "the discriminant of a polynomial is zero if and only if the polynomial has multiple roots", But $x^2 - 5x + 6$ has multiple roots but $\Delta$ is not zero ? I have high school background in maths. Also can you tell me when does one learn about Galois Theory ? $\endgroup$ – user8277998 May 31 '17 at 0:50
  • $\begingroup$ We say $a$ is a multiple root of $p(x)$ if we can write $$p(x)=(x-a)^2g(x)$$ for some other polynomial $g(x)$. I tried to explain this in my answer but perhaps wasn't clear. In this case, we can write $$x^2-5x+6=(x-2)(x-3),$$ so the polynomial has two different roots, but neither of them is a multiple root. (Yes, this is confusing terminology. Unfortunately it is standard.) I am in my third year of an undergraduate math degree and I am just learning it now (it is generally part of an abstract algebra course I think). Some schools may offer it earlier or later in the math major. $\endgroup$ – helloworld112358 May 31 '17 at 0:55
  • $\begingroup$ Let me know if I can clarify anything else. I may have used notation that you wouldn't have seen in your classes yet $\endgroup$ – helloworld112358 May 31 '17 at 0:57
  • $\begingroup$ No, your answer is great. I understand the notation you wrote. Just I would like to know about what does $i$ mean in $\Bbb Q(i)$ ? is it imaginary constant $i^2 = -1$ ? $\endgroup$ – user8277998 May 31 '17 at 1:03
  • $\begingroup$ Yes, exactly. Sorry for using $i$ both as the imaginary number and as an index in sums/products. That was poor choice of notation. $\endgroup$ – helloworld112358 May 31 '17 at 1:07
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Here's a typical quadratic: $$ p(x) = ax^2+bx+c. $$ We all know this quadratic can be factorised into the form $$ p(x) = a(x-r_1)(x-r_2), $$ where $r_1$ and $r_2$ are two (not necessarily real) numbers that are called the roots: in particular, $p(r_i)=0$. Multiplying this form out gives $$ p(x) = ax^2 -a(r_1+r_2)x + ar_1r_2. $$ This is supposed to be the same polynomial as our original $p$, so the coefficients must be the same. Hence we find $$ r_1+r_2 = -\frac{b}{a} \qquad r_1r_2 = \frac{c}{a}; $$ these are called Vieta's formulae.

Bearing this in mind, what is this function $\Delta_p = b^2-4ac$ in terms of the roots? $$ \Delta_p = b^2-4ac = a^2(r_1+r_2)^2 - 4a^2r_1r_2 = a^2(r_1^2+2r_1r_2+r_2^2 - 4 r_1r_2) = a^2(r_1-r_2)^2. $$ So the discriminant of the quadratic $p(x)$ is the square of the difference of its roots: it is positive if $p(x)$ two distinct real roots, zero if the roots are equal (we say that $p$ has a double root), and negative if the difference is a multiple of $i$, which occurs when $r_1$ and $r_2$ are complex.

But quadratics are easy. Can we find a similar thing that tells us when the cubic $$ q(x) = ax^3+bx^2+cx+d = a(x-r_1)(x-r_2)(x-r_3) $$ has less than three distinct roots (so, for example, $r_1=r_3$ and it factorises to $a(x-r_1)^2(x-r_2)$)?

The answer is yes: we can define $$\Delta_q = a^6 (r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2,$$ which is obviously zero if (and indeed, only if) two of the $r_i$s are the same: each bracket is the square of the difference of two of the roots, and the expression contains each possible pair. But, you say, this is useless: if we knew the roots, we'd know if they were different! And you'd be correct. So the real question is whether this quantity $\Delta_q$ can be expressed in terms information we do have about $q$, namely the coefficients? The answer turns out to be yes, because

  • $\Delta_q$ is a symmetric polynomial function of the roots (swapping the subscripts on the $r$s does not change its value),
  • we have corresponding Vieta formulae for the coefficients of a cubic in terms of its roots,
  • and there is a theorem that says that any symmetric polynomial can be written in terms of the polynomials that appear in Vieta's formulae.

The fairly gruesome expression we find is $$ \Delta_q = b^2 c^2 - 4 a c^3 - 4 b^3 d + 18 a b c d - 27 a^2 d^2. $$ It is rather nicer if we look at $q(y-b/(3a))/a$, which has no $y^2$ term: it is of the form $$ y^3+Py+Q, $$ which is known as a depressed cubic. This has discriminant $-4P^2-27Q^3$.

Exactly the same may be done for the quartic and higher: since there is no general formula for the root of a degree-5 or higher polynomial, the discriminant is therefore more useful. Although you can probably imagine how unpleasant the expression in terms of the coefficients becomes for a polynomial of high degree if the cubic case looks like this!


In the case of the cubic, one can determine a bit more about the roots from the value of the discriminant. A cubic with real coefficients always has a real root (it has different sign when $x$ is large and positive from when $x$ is very negative, so it must cross the $x$-axis somewhere).

  • If $\Delta_q = 0$, there are at least two equal $r$s.
  • If $\Delta_q < 0$, the other two roots must be complex—there is only one real root.
  • If $\Delta_q > 0$, there are three distinct real roots.

One can prove these simply by considering all the possible combinations in the factorisation $q(x) =a(x-\alpha)(x^2+2bx+c)$, where $\alpha$ is the real root and the quadratic has $0$, $1$ or $2$ real roots.

Above a cubic, there are too many factors in the discriminant for it to classify the roots particularly: $x^4+1$ has positive discriminant and no real roots, for example.

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  • $\begingroup$ Thank you taking the time to write this. As you said there is no general formula for polynomial above deg 4, so the only way to find roots of such polynomial is by guessing some of them and using division to reduced the polynomial or is there any other smarter method ? $\endgroup$ – user8277998 May 31 '17 at 1:21
  • $\begingroup$ One of the original goals of Galois theory was to determine whether the roots of a given polynomial can be expressed only using $+,-,\times,\div,\sqrt[n]{}$ of rational numbers. This has long since been resolved, and research in Galois theory has moved on to pastures new. This does make it rather difficult to find the bit that actually tells you how to do this! An historical introduction you might like to look at is Tignol's Galois' Theory of Algebraic Equations. $\endgroup$ – Chappers May 31 '17 at 1:27
  • $\begingroup$ Actually I just want to know if there is a better way or not ? I understand that you can't tell the method. I will look into the book. $\endgroup$ – user8277998 May 31 '17 at 1:41
  • $\begingroup$ If the coefficients are integers, there's a thing called the rational root theorem, which gives you very limited possibilities on rational roots. Finding roots is a hard problem: most applications use approximations, which tend to be much easier. $\endgroup$ – Chappers May 31 '17 at 1:57
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Higher polynomials also have discrimants which are much more complicated to calculate.

Discriminant gives you an idea about the roots of the polynomial without having to calculate the roots (and above degree 4, you just cannot algebraically).

You can think it like what is the determinant to a matrix is what discrimant is to a polynomial.

See https://en.wikipedia.org/wiki/Discriminant

Edit: I couldn't find the general formula for discrimant of an n'th degree polynomial on the Wikipedia page but it exists and is known to mathematicians; it is simply a sum of some combinations of coefficients. Edit 2: Someone provided it.

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  • $\begingroup$ Thank you. Are there anything special about Discriminant than tell the nature of roots ? $\endgroup$ – user8277998 May 31 '17 at 0:38
  • $\begingroup$ By the way, is that Wages of sin by Arch Enemy ? in your profile picture. :) $\endgroup$ – user8277998 May 31 '17 at 0:40
  • $\begingroup$ First question, not that I know of. Second one, yes :) dreaming of the day I can get it tattoed.. $\endgroup$ – Ussiane Lepsiq May 31 '17 at 0:42
  • $\begingroup$ You are also a AE fan ? I just love their music :). That aside why can't I find $\Delta$ for polynomial above deg 4 ? is there a proof ? $\endgroup$ – user8277998 May 31 '17 at 0:45
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    $\begingroup$ You can find discrimants for every degree of polynomial; you just cannot algebraically compute all the roots for all (not any, all) the polynomials above degree 4. Some of them are easy such as x^5-1=0 and you can, but not all polynomials above degree 4. Yes, there is a proof, but it requires some abstract algebra knowledge, because it is related to simetry groups of polynomials or something like that. I personally don't think I have enough knowledge to understand. Here it is: en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem $\endgroup$ – Ussiane Lepsiq May 31 '17 at 0:55

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