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Problem 35:  Roots of a cubic equation ( ) 1999 Paper III

Consider the cubic equation

x3 px2 + qx r = 0,

where p0 and r0.

If the three roots can be written in the form ak1, a and ak for some constants a and k, show that one root is qp and that q3 rp3 = 0.
If r = q3p3, show that qp is a root and that the product of the other two roots is (qp)2. Deduce that the roots are in geometric progression.
Find a necessary and sufficient condition involving p, q and r for the roots to be in arithmetic progression.


The Fundamental Theorem of Algebra says that a polynomial of degree n can be written as the product of n linear factors, so we can write

x3 px2 + qx r = (x α)(x β)(x γ), ()

where α, β and γ are the roots of the equation x3 px2 + qx r = 0. The basis of this question is the comparison between the left hand side of () and the right hand side, multiplied out, of (). Some of the roots may not be real, but you don’t have to worry about that here.

The ‘necessary and sufficient’ in part (iii) looks a bit forbidding, but if you just repeat the steps of parts (i) and (ii), it is straightforward.

Solution to problem 35

(i) We have

(x ak1)(x a)(x ak) x3 px2 + qx r


x3 a(k1 + 1 + k)x2 + a2(k1 + 1 + k)x a3 = x3 px2 + qx r.

Thus p = a(k1 + 1 + k), q = a2(k1 + 1 + k), and r = a3. Dividing gives qp = a, which is a root, and q3p3 = a3 = r as required.

(ii) Set r = q3p3. Substituting x = qp into the cubic shows that it is a root:

(qp)3 p(qp)2 + q(qp) (qp)3 = 0.

Since qp is one root, and the product of the three roots is q3p3 ( = r in the original equation), the product of the other two roots must be q2p2. The two roots can therefore be written in the form k1(qp) and k(qp) for some number k, which shows that the three roots are in geometric progression.

(iii) The three roots are in arithmetic progression if and only if they are of the form (a d), a and (a + d). (I have followed the lead of part (i) in using this form rather than a, a + d, a + 2d.)

If the roots are in this form, then

x (a d)x ax (a + d) x3 px2 + qx r


x3 3ax2 + (3a2 d2)x a(a2 d2) x3 px2 + qx r.

Thus p = 3a, q = 3a2 d2, r = a(a2 d2). A necessary condition is therefore r = (p3)(q 2p29). Note that one root is p3.

Conversely, if r = (p3)(q 2p29), the equation is

x3 px2 + qx (p3)(q 2p29) = 0.

We can verify that p3 is a root by substitution. Since the roots sum to p and one of them is p3, the others must be of the form p3 d and p3 + d for some d. They are therefore in arithmetic progression.

A necessary and sufficient condition for the roots to be in arithmetic progression is therefore r = (p3)(q 2p29).


As usual, it is a good idea to give a bit of thought to the conditions given, namely p0 and r0.

Clearly, if the roots are in geometric progression, we cannot have a zero root. We therefore need r0. However, we don’t need the condition p0. If the roots are in geometric progression with p = 0, then q = 0, but there is no contradiction: the roots are be2πi3, b and be2πi3 where b3 = r, and these are certainly in geometric progression. So the necessary and sufficient conditions are r0 and p3r = q3.

Neither of these conditions is required if the roots are in arithmetic progression.

Therefore, the condition p0 is only there as a convenience — one thing less for you to worry about. The condition r0 should really have been given only for the first part. It should be said that this sort of question is incredibly difficult to word, which is why the examiner was a bit heavy handed with the conditions.