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I'm reading Sawyer's Prelude to Mathematics, here:

Book's excerpt

I can't understand what's the meaning and application of "condition" here. Also when he gives the example on the cubic equation, stating that the condition is: $$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$

I can understand that it is $b^2-4ac=0$ (I hope I'm right with this), I just have no idea on where is the order of the variables inside the parentheses coming from.

1: I noticed that the $b^2-4ac$ can be found here:

$$-b\pm \frac{\sqrt{b^2-4ac}}{2a}$$

Which could be found by solving a general form quadratic equation:

$$ax^2+bx+c=0$$

Then I thought about searching it on the solutions for cubic equations with some help of Mathematica, but I got nothing that was similar to:

$$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$

or:

$$a^2d^2-6abcd+4b^3d+4ac^3-3b^2c^2=0$$

With no success. You can see it here:

enter image description here

Red Banana
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  • He writes $ax^2+2bx+c=0$ which completing the square gives $a(x+b/a)^2+c-ab^2/a^2=0$ for this to be a perfect square we need $c-b^2/a=0$ or simply $ac-b^2=0$. I guess you understand this part already. I'll have to think about the other part. – James S. Cook Sep 01 '12 at 19:07
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    I'm reading the condition $(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$ as the condition for a cubic to have a three-peated root. You say "quadratic" I think this is a typo. – James S. Cook Sep 01 '12 at 19:10
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    @James: No. If you look at the example $x^3-x^2$, which has two (not three) repeated roots, you will find that the expression is zero. (And I fixed the "quadratic" typo.) – TonyK Sep 01 '12 at 19:11
  • Read the few words just before the condition. This tells where the constants $a$, $b$, $c$, and $d$ come from. As the few sentences before that describe, the author starts by talking about a quadratic equation then changes to talking about a cubic equation. – Code-Guru Sep 01 '12 at 19:14
  • @JamesS.Cook I think the "typo" actually points out where the OP is confused. He might have missed that the second condition is for a cubic equation rather than a quadratic one. – Code-Guru Sep 01 '12 at 19:15
  • @Code-Guru: You haven't answered the OP's question -- where does that expression come from? It's not completely obvious (not to me or the OP, anyway). – TonyK Sep 01 '12 at 19:17
  • @TonyK The OP asked "I just have no idea on where are the variables inside the paretheses coming from". I don't see any question about where the expression itself comes from. I certainly agree that it isn't obvious. It probably isn't even trivial to derive. – Code-Guru Sep 01 '12 at 19:20
  • Consider $ax^3+3bx^2+3cx+d=0$. We wish to write it as a perfect cube. Dividing by $a$ yields: $$ x^3+(3b/a)x^2+(3c/a)x+d/a = (x-m)^3$$ We need to find $m$. Note $$ x^3+(3b/a)x^2+(3c/a)x+d/a = x^3-3mx^2+3xm^2-m^3$$ Equating coefficients gives: $$ b/a=-m, \ \ c/a=m^2, \ \ d/a=-m^3 $$ and eliminating $m$ would reveal the necessary condtions on $a,b,c,d$ for the triple root. Although, clearly this was not asked :( – James S. Cook Sep 01 '12 at 19:22
  • @TonyK I read the OP as confusion about the jump between the example of quadratic equations to a discussion about applying the same concept to cubic equations. This is just my interpretation of what I read. I may be wrong. – Code-Guru Sep 01 '12 at 19:23
  • @Code-Guru: Yes, the OP is not as clear as I thought on first reading. But if there is an interesting question in there, it has to be: Where does the expression $(bc-ad)^2-...$ come from? – TonyK Sep 01 '12 at 19:27
  • With three roots $x_1, x_2, x_3$, the coefficients of the polynomial are symmetric expressions in these, e.g. $\frac{-3b}a=x_1+x_2+x_3$. Two roots are equal iff $(x_1-x_2)(x_1-x_3)(x_2-x_3)=0$, the square of which turns out to be symmetric in the $x_i$ and by a general theorem such can be expressed using $x_1+x_2+x_3$, $x_1x_2+x_1x_3+x_2x_3$ and $x_1x_2x_3$, i.e. essentially the polynomial coefficients. That's how one obtains that weird expression in the coefficients. (That's also how one might arrive at $b^2-(4)ac=0$ because the left hand is in fact $(x_1-x_2)^2$ (up to a factor of $a^2$)) – Hagen von Eitzen Sep 01 '12 at 19:27
  • @TonyK I agree that this is an interesting question. I'm just not inclined to tackle the derivation =p – Code-Guru Sep 01 '12 at 19:37
  • It has just come to my attention that you were unhappy with the change I made to the title of your question, and I wanted to apologize. I am sorry. – MJD Dec 14 '12 at 05:17
  • @MJD No problem, I just wanted to express the exactly as it was in my mind, this is the the most plausible way to me. – Red Banana Dec 14 '12 at 14:13

4 Answers4

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A polynomial has two equal roots when it has a common root with its (formal) derivative. Apply the euclidean algorithm to establish conditions for a common factor. This makes for an easily computable procedure - rather than remembering the equations.

Mark Bennet
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Actually, the stated equation ensures that at least two of the roots are equal. Let $x_1$, $x_2$ and $x_3$ be three roots of the cubic $a x^3 + 3 b x^2 + 3 c x+d =0$. Discriminant of the equation: $$ D = a^4 (x_1-x_2)^2 (x_2-x_3)^2 (x_1-x_3)^2 $$ can be expressed in terms of coefficients of the cubic: $$ D = -27 \left(a^2 d^2-6 a b c d+4 a c^3+4 b^3 d-3 b^2 c^2\right) $$ $D$ equals zero when at least two of the roots are equal.

Sasha
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  • Correct, I just worked out the triple root case and got something entirely different only to reread the post and find "repeated root" for the cubic. My mistake. – James S. Cook Sep 01 '12 at 19:20
  • @Sasha: Yes, but how do you get from the first expression to the second? I think that this is the OP's question... – TonyK Sep 01 '12 at 19:29
  • Please see the wikipedia article. $D$ is defined as a determinant of a matrix, whose elements are expressed in terms of coefficients of the equation. – Sasha Sep 01 '12 at 19:31
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Let $p(x)=x^3+b x^2+c x+d$ have roots $\alpha,\beta,\gamma$. By Vieta's formulas we have

$$\begin{cases}e_1 & =\alpha+\beta+\gamma & =-3b \\ e_2 & =\alpha\beta+\beta\gamma+\gamma\alpha & =~~\,3c \\ e_3 & =\alpha\beta\gamma & = -d.\end{cases}\tag{$*$}$$

In view of the fundamental theorem of symmetric polynomials, we desire to write the discriminant

$$D(p)=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2$$

as a polynomial $F(e_1,e_2,e_3)$ in the first three elementary symmetric polynomials. I touched on a recursive procedure for doing this in an answer here. Begin by setting $\gamma=0$ to get

$$(\alpha-\beta)^2\beta^2\alpha^2=(\bar{e}_1^2-4\bar{e}_2)\bar{e}_2^2 $$

where $\bar{e}_1=\alpha+\beta$ and $\bar{e}_2=\alpha\beta$. We expand out the difference and recombine to get (thanks, W|A)

$$D(p)-(e_1^2-4e_2)e_2^2$$ $$=6\underbrace{\alpha\beta\gamma}_{\large e_3}\big(\,\underbrace{\alpha\beta(\alpha+\beta)+\beta\gamma(\beta+\gamma)+\alpha\gamma(\alpha+\gamma)}_{g}\,\big)-4\underbrace{\alpha\beta\gamma}_{\large e_3}(\underbrace{\alpha^3+\beta^3+\gamma^3}_{\large p_3})+3\underbrace{\alpha^2\beta^2\gamma^2}_{\large e_3^2}$$

The power sum polynomial $p_3=e_1^3-3e_2e_1+3e_3$ is already computed here thankfully, so we only need to worry about the polynomial $g$ labelled above. Then we compute the new difference $g-e_1e_2=-3e_3$ and we're pretty much good to go at this point: $g=e_1e_2-3e_3$ and so

$$D(p)=(e_1^2-4e_2)e_2^2+6e_3(e_1e_2-3e_3)-4e_3(e_1^3-3e_1e_2+3e_3)+3e_3^2.$$

We could have also used Mathematica's SymmetricReduction command to expediate the process and not do any hand computations. Plugging in $(*)$ we get

$$b^2c^2-4c^3-27d^2+18bcd-4db^3 \tag{$\circ$}$$

We may re-homogenize $(\circ)$ by substituting $b\mapsto b/a,c\mapsto c/a,d\mapsto d/a$ and then multiplying by $a^4$, in which case we get the formula on Wikipedia's discriminant article. If we then divide by $27$ and then apply the substitutions $b\mapsto 3b,c\mapsto3c$ we get the formula presented in the text.

There should it seems be a good theoretical reason why the polynomials

$$ax^3+3bx^2+3cx+d\qquad\&\qquad \begin{vmatrix}a&b\\b&c\end{vmatrix}x^2+\begin{vmatrix}b&d\\a&c\end{vmatrix}x+\begin{vmatrix}b&c\\c&d\end{vmatrix} \tag{$\bullet$}$$

have the same discriminant up to a constant independent of $a,b,c,d$, however it escapes me for now. (And probably such a reason was why the author expressed the condition how they did.)

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anon
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  • Eh, I should amend my last statements. As the other two answers get at, we can relate the discriminant with the resultant $R(p,p,')$ and then use a Sylvester matrix for the latter. I will try to think about this more later and tie it all together. – anon Sep 02 '12 at 03:23
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Any cubic polynomial with roots $r$, $s$, $t$ has the form $$k(x-r)(x-s)(x-t) = k x^3 - k x^2 ( r + s + t ) + k x ( r s + s t + t r ) - k r s t$$ for some $k \ne 0$. If we're taking that polynomial to be $a x^3 + 3 b x^2 + 3 c x + d$, then certainly $a=k$ so we won't bother using "$k$" further; also, by equating the remaining coefficients, $$3 b = - a (r+s+t) \qquad 3 c = a (rs+st+tr) \qquad d = -a rst$$

If the cubic two equal roots ---say $s=t$--- then $$3 b = - a (r+2s) \qquad 3 c = a s(2r+s) \qquad d = -a rs^2$$

Now, consider "the condition" that Sawyer proposes: $$(bc-ad)^2-4(ac-b^2)(bd-c^2)=0$$

If you'd be satisfied knowing that the condition does what Sawyer claims, then you can simply apply it to our $a$, $b$, $c$, $d$ and what what happens. Note

$$\begin{align} bc-ad &= -\frac{1}{9}a^2s(r+2s)(2r+s)+a^2rs^2 = -\frac{2}{9}a^2s\left(r-s\right)^2 \\ ac-b^2 &= \frac{1}{3}a^2s(2r+s)-\frac{1}{9}a^2(r+2s)^2 = -\frac{1}{9}a^2\left(r-s\right)^2 \\ bd-c^2 &= \frac{1}{3} a^2rs^2(r+2s)-\frac{1}{9}a^2s^2(2r+s)^2 = -\frac{1}{9}a^2s^2\left(r-s\right)^2 \end{align}$$ so that $$(bc-ad)^2-4(ac-b^2)(bd-c^2) = \frac{4}{81} a^4 s^2\left(r-s\right)^4-\frac{4}{81} a^4 s^2\left(r-s\right)^4 = 0$$

That is, the condition is a somewhat magical relation that combines the coefficients ---which depend on $r$ and $s$--- in just the right way to make everything vanish. That, at least, explains the purpose of the condition; the derivation of the condition, on the other hand, is something of a mess. The powerful concepts of discriminant and (my favorite tool) resultant give high-level views that nicely blur the details, but if you aren't familiar with those ---or just want to try brute force--- how might you proceed?

Well, you can view the expressions for $a$, $b$, $c$, $d$ as a system of three equations in just two parameters $r$ and $s$. Because the system is over-determined, the coefficients cannot be free agents; they must be related somehow. Eliminating $r$ and $s$ from the system is the way to find the relation.

For starters, note that the last equation tells us that $r = -d/a/s^2$; substituting into the first two equations gives: $$3 b s^2 = d-2as^3 \qquad 3 c s = -2 d + as^3$$

And then all we have to do is eliminate $s$ ... which gets back to the mess I mentioned. I'd normally invoke the method of resultants (with the aid of Mathematica), which automates the required clever algebra, but that's as unmotivating as pulling the discriminant out of the air. For now, I'll leave this as a challenge, and will update this answer once can bring myself to TeX-up the steps.

Okay, here's an elementary (though long and drawn-out) way to eliminate $s$.

Note: The point of this is that it's complicated; it makes you appreciate the machinery of the discriminant and resultant!

Combining the $3bs^2$ equation with twice the $3cs$ equation yields $3bs^2+6cs=-3d$ ... or, better, $$bs^2=-(d+2cs)$$ The best way to think of this equation is that it allows us to re-write second-powers of $s$ in terms of first-powers of $s$, thus reducing an expression's $s$-degree in exchange for increasing its overall size. (In effect, we reduce expressions modulo the polynomial $bs^2+2cs+d$ ... curiously enough, a quadratic with discriminant $4(c^2-bd)$.) So, for instance, returning to the $3cs$ equation, we can multiply-through by $b$ a couple of times, ratcheting the $s$-degree down from $3$ to $1$ by replacing $bs^2$ where it appears ... $$\begin{align} 3cs=-2d+as\cdot s^2 &\implies 3bcs = -2bd + as \cdot b s^2 \\ &\implies 3bcs = -2bd - as (d+2cs) \\ &\implies 3bcs = - 2 b d - a d s - 2 a c \cdot s^2 \\ &\implies 3b^2cs = - 2 b^2 d - a b d s + 2 a c (d+2cs) \\ &\implies s(3b^2c+abd-4ac^2) = 2d(ac-b^2) \end{align}$$ We've just effectively solved for $s$, although I don't want to divide by its possibly-zero multiplier, which I'll abbreviate $m$ so that $$s m = 2 d(ac-b^2)$$ Going back to the $bs^2$ equation and multiplying-through by $m$ a time or two ... $$\begin{align} b s^2 = - d - 2 c \cdot s &\implies b s^2 m = - d m - 4 c d (ac-b^2) = bd(bc-ad) \\ &\implies b \cdot s^2 m^2 = bd(bc-ad)m \\ &\implies 4 b d^2 ( b^2 - a c )^2 = bd(bc-ad)m \\ \end{align}$$

From the last equation, we get $$b^2 d \left( a^2 d^2 - 6 a b c d + 4 b^3 d + 4 a c^3 - 3 b^3 c^2 \right) = 0$$

Ignoring the cases $b=0$ and $d=0$ in general, we find that "the condition" must hold when two roots of a cubic coincide! whew!

(In fact, the cases $b=0$ ---in which $r=-2s$--- and $d=0$ ---in which either $r=0$ or $s=0$--- fall under the umbrella of the condition.)

Blue
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