170 CHAPTER 3 Inequalities
With a bit of effort, this determinant can be expanded. It’s easier to
first compute the determinant of the matrix
1 x 1 x^21
1 x 2 x^22
1 x 3 x^23
and then square the result. One has, after a bit of computation, the
highly structured answer
det
1 x 1 x^21
1 x 2 x^22
1 x 3 x^23
= (x 3 −x 2 )(x 3 −x 1 )(x 2 −x 1 ),(this is generalized in the next section) which implies that
∆ =a^4 (x 3 −x 2 )^2 (x 3 −x 1 )^2 (x 2 −x 1 )^2.This is all well and good, but two questions immediately arise:- How does one compute ∆ without knowing the zeros ofP? 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, thenP(x) has one real zero (sayx 1 ) and
a complex-conjugate pair of non-real zeros (x 2 andx 3 ). In this case
(x 2 −x 1 ), (x 3 −x 1 ) would be a complex conjugate pair, forcing 0 <
(x 2 −x 1 )(x 3 −x 1 )∈Rand so certainly that 0<(x 2 −x 1 )^2 (x 3 −x 1 )^2 ∈R.
Furthermore, (x 3 −x 2 ) is purely imaginary and so (x 3 −x 2 )^2 <0, all
forcing ∆<0. Therefore, we see immediately that
∆>0 =⇒P(x) has three distinct real zerosand that