The the quadratic f(x) = ax2 + bx + c has associated with it the discriminant D = b2 −4ac, which in turn elucidates the nature of the zeros of f(x). In turn, this information gives very helpful information about tangency which in turn can be applied to constrained extrema
problems. This raises at least a couple of questions. The immediate question raised here would be whether higher-degree polynomials also have discriminants. We’ll see that this is the case and will consider the case of cubic polynomials in this section. In the following section we’ll introduce the discriminant for arbitrary polynomials. The notion of the determinant of a matrix will play a central role here.
For the quadratic polynomial f(x) = ax2+bx+chaving zerosx1, x2, we define the “new” quantity
∆ = a2det 1 x1 1 x2 2 = a2(x2 −x1)2.
At first blush, it doesn’t appear that ∆ has anything to do with the discriminant D. However, once we have designated the zeros of f as being x1 and x2, then the Factor Theorem dictates that
f(x) =a(x−x1)(x−x2).
Since also f(x) = ax2 + bx + c we conclude by expanding the above that
b = −a(x1 +x2), and c = ax1x2. Now watch this:
∆ = a2(x2 −x1)2
= a2(x21 +x22 −2x1x2) = a2[(x1 +x2)2 −4x1x2] = [−a(x1 + x2)]2 −4a(ax1x2) = b2 −4ac = D.
In other words, ∆ and D are the same:
D = ∆ .
Therefore, D and ∆ will satisfy the same trichotomy rule. But let’s try to develop the trichotomy rule directly in terms of ∆ instead of D.
SECTION 3.6 Cubic Discriminant 169 Case (i): ∆ > 0. That is to say, (x2 −x1)2 > 0 and so certainly the zeros x1 and x2 are distinct. Furthermore, if they were not real, then they would have to be complex conjugates of one another and this would force (think about it!) (x2−x1)2 < 0 (as x2−x1 would be purely imaginary). Therefore
∆> 0 =⇒ Q has two distinct real zeros.
Case (ii): ∆ = 0. This is clear in that one immediately has that
x1 = x2. That is to say
∆ = 0 =⇒ Q has a double zero.
Case (iii): ∆< 0. Since (x2 −x1)2 < 0 we certainly cannot have both
x1 and x2 real. Therefore, they’re both complex (non-real) as they are complex conjugate. Therefore
∆ < 0 =⇒ Q has two complex (non-real) zeros. That is to say, D and ∆ satisfy the same trichotomy law!
Whereas the definition of D does not suggest a generalization to higher-degree polynomials, the definition of ∆ can be easily generalized. We consider first a natural generalization to the cubic polynomial
P(x) = ax3 +bx2 +cx+d, a, b, c, d ∈ R, a6= 0.
By the Fundamental Theorem of Algebra, we know that (counting multiplicities), P(x) has three zeros; we shall denote them by x1, x2, and x3. They may be real or complex, but we do know that one of these zeros must be real.
We set ∆ = a4det 1 x1 x21 1 x2 x22 1 x3 x23 2 .
With a bit of effort, this determinant can be expanded. It’s easier to first compute the determinant of the matrix
1 x1 x21 1 x2 x22 1 x3 x23
and then square the result. One has, after a bit of computation, the highly structured answer
det 1 x1 x21 1 x2 x22 1 x3 x23 = (x3 −x2)(x3 −x1)(x2 −x1),
(this is generalized in the next section) which implies that ∆ = a4(x3 −x2)2(x3 −x1)2(x2 −x1)2.
This is all well and good, but two questions immediately arise: • How does one compute ∆ without knowing the zeros of P? Also,
and perhaps more importantly, • what is ∆ trying to tell us?
Let’s start with the second bullet point and work out the trichotomy law dictated by ∆.
If P(x) has three distinct real zeros, then it’s obvious that ∆ > 0. If not all of the zeros are real, then P(x) has one real zero (say x1) and a complex-conjugate pair of non-real zeros (x2 and x3). In this case (x2 − x1), (x3 − x1) would be a complex conjugate pair, forcing 0 < (x2−x1)(x3−x1) ∈ Rand so certainly that 0 < (x2−x1)2(x3−x1)2 ∈ R. Furthermore, (x3 −x2) is purely imaginary and so (x3 −x2)2 < 0, all forcing ∆ < 0. Therefore, we see immediately that
∆ > 0 =⇒ P(x) has three distinct real zeros and that
SECTION 3.6 Cubic Discriminant 171
∆ < 0 =⇒ P(x) has one real zero and two non-real complex zeros. This is all rounded out by the obvious statement that
∆ = 0 =⇒P(x) has a multiple zero and all zeros are real.
Of course, none of the above is worth much unless we have a method of computing ∆. The trick is to proceed as in the quadratic case and compute ∆ in terms of the coefficients of P(x). We start with the observation that
P(x) = a(x−x1)(x−x2)(x−x3),
all of which implies that (by expanding)
b = −a(x1 +x2 +x3), c= a(x1x2 +x1x3 +x2x3), d = −ax1x2x3. We set
σ1 = x1 +x2 +x3, σ2 = x1x2 +x1x3 +x2x3, σ3 = x1x2x3,
and call them theelementary symmetric polynomials(inx1, x2, x3). On the other hand, by expanding out ∆, one has that (after quite a bit of very hard work!)
∆ = a4(x3 −x2)2(x3 −x1)2(x2 −x1)2 = a4(−4σ31σ3 + σ21σ 2 2 + 18σ1σ2σ3 −4σ23 −27σ 2 3) = −4b3d+b2c2 + 18abcd−4ac3 −27a2d2
giving a surprisingly complicated homogeneous polynomial in the co- efficient a, b, c, and d. (See Exercise 6 below for a more more direct method for computing ∆.)
We’ll close this section with a representative example. Keep in mind that just as in the case of the quadratic, when the discriminant of a cubic is 0, then the graph of this cubic is tangent to the x-axis at the multiple zero.
Example. Compute the minimum value of the function f(x) = 1
x2 +x, x > 0.
Solution. The minimum value will occur where the line y = c is
tangent to the graph of y = f(x). We may write the equation
f(x) =c in the form of a cubic polynomial in x:
x3 −cx2 + 1 = 0.
As the tangent indicates a multiple zero, we must have ∆ = 0. As
a = 1, b = −c, c = 0, d = 1, we get the equation 4c3 −27 = 0, which implies that the minimum value is given by c = √33
4 (which can be verified by standard calculus techniques).
Now try these:
Exercises.
1. Compute the minimum of the function
h(x) = 1
x +x
2, x > 0.
2. Compute the minimum of 2y −x given that
x3 −x2y + 1 = 0.
3. Compute the maximum value of y +x2 given that
SECTION 3.6 Cubic Discriminant 173 1 2 3 2 4 6 x y
x
3−xy+2=0
y+x =c
2level curves
Equation 1: y=2/x+x² Equation 2: y=c−x² Equation 3: y=2.1−x² Equation 4: y=4.2−x^24. Compute the maximum value of xy, given that
x2 +y = 4.
5. The polynomialS(x1, x2, x3) = x31+x32+x33 is symmetric inx1, x2, x3 and can be expanded in the elementary symmetric polynomials
σ1 = x1 +x2 +x3, σ2 = x1x2 +x1x3 +x2x3, σ3 = x1x2x3. Watch this: x31 +x32 +x33 = (x1 +x2 +x3)3 − 3x21(x2 +x3)−3x22(x1 + x3)−3x23(x1 +x2)−6x1x2x3 = (x1 +x2 +x3)3 −3(x1 +x2 +x3)(x1x2 +x1x3 +x2x3) + 3x1x2x3 = σ13 −2σ1σ2 + 3σ3.
Now try to write the symmetric polynomial x41 + x42 + x43 as a polynomial in σ1, σ2, σ3.
6. Let f(x) = ax3 + bx2 + cx + d, and so the derivative is f0(x) = 3ax2 + 2bx+c. Denote by R(f) the determinant
R(f) = det a b c d 0 0 a b c d 3a 2b c 0 0 0 3a 2b c 0 0 0 3a 2b c .
Show that R(f) = −aD(f), where D = D(f) is the discriminant of f(x). (This is the generalization of the result of Exercise 6 to cubic polynomials.)