168 CHAPTER 3 Inequalities
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 polynomialf(x) =ax^2 +bx+chaving zerosx 1 , x 2 ,
we define the “new” quantity
∆ =a^2 det
^1 x^1
1 x 2
2
=a^2 (x 2 −x 1 )^2.At first blush, it doesn’t appear that ∆ has anything to do with the
discriminant D. However, once we have designated the zeros off as
beingx 1 andx 2 , then theFactor Theoremdictates that
f(x) =a(x−x 1 )(x−x 2 ).Since also f(x) = ax^2 +bx+cwe conclude by expanding the above
that
b=−a(x 1 +x 2 ), and c=ax 1 x 2.Now watch this:∆ = a^2 (x 2 −x 1 )^2
= a^2 (x^21 +x^22 − 2 x 1 x 2 )
= a^2 [(x 1 +x 2 )^2 − 4 x 1 x 2 ]
= [−a(x 1 +x 2 )]^2 − 4 a(ax 1 x 2 )
= b^2 − 4 ac=D.In other words, ∆ andDare the same:
D= ∆.Therefore,Dand ∆ will satisfy the same trichotomy rule. But let’s try
to develop the trichotomy rule directly in terms of ∆ instead ofD.