6.1. Interpreting the Coefficients of a Polynomial 195
(b) Prove that, if x, y, z are real, then a2 >3b and that z (and, by
symmetry, x and y as well) must lie in the closed interval [u, v],
where
u= (l/3) (- a -2&G)
21 = (l/3) (-a + 2JKG)
Note that p’(u) = p’(v) = o2 - 3b.
(c) With the hypotheses of (b), prove that p(u) 5 0 5 p(v) and
deduce that
]2a3 - 9ab + 27~1 5 2(a2 - 3b)3/2
and thence that
a2b2 - 4a3c + 18abc - 4b3 - 27c2 2 0.
(d) Suppose that a2b2 - 4a3c + 18abc - 4b3 - 27c2 > 0. Show that
a2 - 3b 2 0 and that the cubic p(t) has three real zeros.
[Cf. Problems 1.4.4, 5.4.15, 6.2.5.1
- Solve the equation t4 - t3 - 7t2 + 23t - 20 = 0, given that the product
of two of the roots is -5. - Consider the polynomial equation
x4+px3+qx2+rx+s=0.
(a) Prove that the product of two of its roots is equal to the product
of the other two iff r2 = p2s.
(b) Prove that, if the sum of two of its roots is equal to the sum of
the other two, then p” + 8r = 4pq.
(c) Suppose that p3 + 8r = 4pq. Must the sum of two of its roots
equal the sum of the other two?
- In Exercise 1.4.11, the quartic equation
t4 + pt2 + qt + r = 0... (1)
was solved by factoring the left side as (t2 + ut + v)(t2 - ut + w),
where u satisfies the equation
t6 + 2pt4 + (p2 - 4r)t2 - q2 = 0.... (2)
(a) Argue that the roots of (2) can be given as f2a, f2b, f2c, where
a, b, c are selected so that 8abc = -q.