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346 Answers to Exercises and Solutions to Problems

(b) First solution. Suppose the zeros are a, 6, c, d with a + b = c+ d =

-p/2. Then p2/4 + ab + cd = q and (-p/2)(ab + cd) = -r. For these to be

consistent, we require 2r/p = q - p2/4.

Second solution. Make the change of variables x = - $p - y. Substitution

into the original equation yields

y” + py3 + qy2 + (qp - r - tp3)y + (-p4/16 + p2q/4 - rp/2 + s) = 0.

This equation has exactly the same roots as the original equation and so
its coefficients must be the same. In particular, qp - r - p3/4 = r, which
yields the result.
(c) Yes. The change of variable x = -p/2 - y as in (b) leads to exactly
the same equation for y. Hence the roots a, 6, c, d of the equation in x
are the same as the roots -p/2 - a, -p/2 - b, -p/2 - c, -p/2 - d of the
equation in y.
If, say, a = -p/2 - b and c = -p/2 - d, then a + b = c + d = -p/2. If,
say, a = -p/2 - a, then -p/4 is a root of the given equation. But, in any
case, -p/4 is a zero of the derivative 4x3 + 3px2 + 2qx + r, so -p/4 is a
double root, and again pairs of roots have the same sum.

2.3. (b) If a cubic over R does not have all its zeros real, they must be of
the form r, u + vi, u - vi (r, u, v real, v # 0). The discriminant is then

Kr - u)’ + v2]2(2vi)2 = -4v2[(r - u)” + v212 < 0.
An example of a quartic with nonreal zeros and positive discriminant is
t4 + 4 whose zeros are 1 + i, 1 - i, -1 + i, -1 - i and whose discriminant
is 214.

2.4. Suppose p(t) has a nonreal zero tl. Then its complex conjugate t2 is
also a zero. Let tl = u + vi, t2 = u - vi. Then one of the factors in D is
the square (tl - t2)’ = -4~‘. If t3 is another nonreal zero with complex
conjugate t4, then D contains the product

(t1 - Q2(t2 - t4)2 = It1 - t314.

On the other hand, if t5 is a real zero, then D contains the product

(t1 - t5)2(t2 - t5)2 = It1 - t514.

Thus, one of the square factors involving tl and t2 is negative while the
remaining factors involving either of these zeros can be combined into a
positive product.
Now look at factors of D not involving tl and t2. If there is another
nonreal complex conjugate pair of zeros, the terms involving them can be
combined to give a negative product. This argument can be continued on
to show that the sign of the discriminant is (-l)k where k is the number
of pairs of complex conjugate zeros.