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Here is the question I'm currently looking at:

Show that the discriminant of the equation $y^3+py+q=0$ is $-4p^3-27q^2$.

I've done some research and found this, but we haven't studied Vieta's theorem in my math class just yet. All I know is that if we take a cubic equation of the form $$x^3+ax^2+bx+c=0$$

and we manipulate it to get rid of the $x^2$ term, then we yield $$y^3+py+q=0$$

where $$p=\frac{3b-a^2}{3}$$ and $$q=\frac{27c+2a^3-9ab}{27}$$

Other than that, I don't know how to approach this question.

Rodrigo de Azevedo
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John W. Smith
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  • What is your definition of the discriminant of a polynomial? – Bernard Feb 13 '17 at 19:47
  • Well the discriminant tells us the number of real solutions to a polynomial. So for example, in the simplest case, $D = b^2-4ac$ would be the discriminant and would tell us how many real solutions exist. – John W. Smith Feb 13 '17 at 19:50
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    That's not a definition. It's a property. For a monic polynomial $P$ of degree $n$, its discriminant is $ (-1)^{n(n-1)/2}$ times the resultant of $P$ and $P'$ and characterises polynomials which have a multiple root in $\mathbf C$. – Bernard Feb 13 '17 at 20:04
  • Ahh okay noted. – John W. Smith Feb 13 '17 at 20:06
  • https://brilliant.org/wiki/cubic-discriminant/ –  Dec 21 '18 at 15:37

4 Answers4

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Any $y^3 + py + q$ can be written as $F := (y-x_1)(y-x_2)(y+x_1 + x_2)$ for some $x_1,x_2$ in an algebraic extension. So it suffices to prove the formula just for this one polynomial $F$. Compute the discriminant of $F$ directly from the definition, compute the $x^1$ and $x^0$ coefficients in $F$ (i.e. $p$ and $q$) in terms of $x_1,x_2$, then compute $-4p^3-27q^2$ and compare.

Mark
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  • Should I start by expanding $(y-x_1)(y-x_2)(y+x_1+x_2)$? – John W. Smith Feb 13 '17 at 19:52
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    Yes, that way you find $p,q$ as expressions in terms of $x_1,x_2$. After that, you compute $-4p^3-27q^2$ as an expression in terms of $x_1,x_2$. You compare that expression with the discriminant of $F$ (computed directly from the definition). That completes the proof (provided that they are equal of course). – Mark Feb 13 '17 at 19:59
  • Alright I'll give this a shot. – John W. Smith Feb 13 '17 at 20:00
  • Mark, can you give an explanation or a reference for the possibility to write $F := (y-x_1)(y-x_2)(y+x_1 + x_2)$: I don't see why it is possible. – Jean Marie Feb 13 '17 at 21:46
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    Suppose for example that your polynomial is in $\mathbb{C}[y]$. Any polynomial in this ring factors into linear factors so $F = (y-x_1)(y-x_2)(y-x_3)$. But since the $x^2$ term in $F$ is zero, you find $x_3 = -x_1 - x_2$. If your polynomial is in $K[y]$ for some field $K$, the same applies, just replace $\mathbb{C}$ by the algebraic closure of $K$. – Mark Feb 13 '17 at 21:54
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    @Mark Thank you: it's clear now. – Jean Marie Feb 15 '17 at 23:40
6

The discriminant is a quantity such that its sign decides the number of real roots. When it is zero, we have two roots, of which a double one. This double root corresponds to an extremum which is on the horizontal axis (blue curve).

By canceling the derivative

$$3y^2+p=0,$$

we solve for $y$,

$$y=\pm\sqrt{-\frac p3}.$$

Then plugging in the cubic function,

$$f(y)=\mp\frac p3\sqrt{-\frac p3}\pm p\sqrt{-\frac p3}+q=\pm\frac{2p}3\sqrt{-\frac p3}+q=0.$$

This quantity cancels when

$$4p^3+27q^2=0.$$

enter image description here

Just for fun, we can use the method for the quadratic equation. With a deflated equation

$$y^2+p=0$$ we trivially have $$y=0,\\f(0)=p=0.$$

[The deflated equation is obtained by completing the square, $ax^2+bx+c=a\left(\left(x-\frac b{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right)$.]

  • I really like this explanation. On your $f(y)$ function, the first symbol after the equal sign is that a plus or minus? – John W. Smith Feb 13 '17 at 21:03
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    @Oliver_821: it's a "minus-plus", which is the negative of the "plus-minus" that follows. –  Feb 14 '17 at 07:42
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@Mark There is a nice graphical interpretation of the discriminant. Let us consider

$$x^3+px+q=0$$

as the equation giving the abscissas of intersection point(s) of

$$\begin{cases}y=x^3 \ \ & \text{cubic curve (C)}\\y=-px-q \ \ & \text{straight line } \ (D_{p,q})\end{cases}$$

(see graphics below) According to the values of $p$ and $q$, $(C) \cap (D_{p,q})$ have one, two or three intersection points (which correspond to one, two or three real roots).

For example the green line has one intersection point with (C), whereas the blue line has three. The transition case between 1 root ans three roots is when there is a double root; this means exactly for $(D_{p,q})$ that it must be tangent to $(C)$, as are all the black lines represented on the figure below.

Writing the general equation of the tangent to (C) under the form:

$$y=x_0^3+3x_0^2(x-x_0) \ \ \iff \ \ y=(3x_0^2)x+(-2x_0^3)$$

By identification with the generic equation $y=-px-q$ of $(D_{p,q})$, we obtain

$$\begin{cases}\ \ \ 3x_0^2&=&-p\\-2x_0^3&=&-q\end{cases}$$

Eliminating $x_0$ between these two equations give the equation $4p^3+27q^2=0$, which is a condition of transition between the case "one real solution" and 3 "real solutions" (with a complete analogy with the rôle of condition $\Delta=0$ for a quadratic).

enter image description here

Jean Marie
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3

Let $(y-a)(y-b)(y-c)=0$ our equation,

$a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Thus, we have $y^3-3uy^2+3v^2y-w^3=0$.

In our case $v^2=\frac{p}{3}$, $u=0$ and $w^3=-q$.

Hence, the discriminant it's $$(a-b)^2(a-c)^2(b-c)^2$$ or $$27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$27\left(-4\cdot\frac{p^3}{27}-q^2\right)$$ or $$-4p^3-27q^2$$

Michael Rozenberg
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