Chapter 3 · Complex Numbers and Equations · Concept 3
3-20 Cubic Equation with Repeated Roots
The last lesson of the chapter, and it introduces nothing new. It takes the Factor Theorem, synthetic division and the discriminant — and asks one question that needs all three.
Guiding question
In how many different ways can a cubic have a repeated root?
Learning outcome. Solve cubic equations with repeated roots.
Split it first, then two cases
Use the Factor Theorem to write the cubic as
(x − a)(x2 + bx + c) = 0
That product has a repeated root in two different ways:
case 1
x² + bx + c = 0 has x = a as a solution
⇒ substitute x = a into x² + bx + c = 0
The root a then belongs to both brackets, so it appears twice.
case 2
x² + bx + c = 0 has a repeated root of its own
⇒ find where its discriminant D becomes 0
The repeat is entirely inside the quadratic, and a has nothing to do with it.
Both cases count, and each may give more than one value. Case 2 sets a quadratic in a to zero, so it often produces two answers on its own. Collect every value you find — some of these questions have three.
The two cases are not two routes to one answer. They are two genuinely different equations, with different values of a and different-looking factorisations at the end.
the book's Warm Up
x3 − 2x2 + (a − 3)x + a = 0 has a repeated root
first, split it — and (x + 1) is a factor whatever a is
(x + 1)(x2 − 3x + a) = 0
P(−1) = −1 − 2 − (a − 3) + a, and the two a's cancel — so it is 0 no matter what a is.
case 1
the quadratic shares the root x = −1
…
case 2
the quadratic has a repeated root all of its own
…
Find the value of the constant a for which x³ − 2x² + (a − 3)x + a = 0 has a repeated root.
- 1
Hunt for a root that works whatever a is. Try the small numbers.
P(−1) = −1 − 2 − (a − 3) + a = 0The two a terms cancel — −a and +a — so this is zero for EVERY a. That is what makes the question possible at all.
Work with a partner. Split the work: one of you takes case 1 and the other case 2. Then write down the finished factorisation for each and compare them — they should look nothing alike.
x3 + (a − 1)x2 + (a − 3)x − 2a + 3 = 0
Find every value of a that gives a repeated root.
A quadratic like x² + 1 = 0 can have no real solutions. Why does a cubic like x³ + x + 1 = 0 always have at least one?
Count the imaginary ones, and remember they never travel alone.
Finding one case and stopping.
There are always two cases to look at, and they give different values of a. Most lost marks in this lesson are simply missing answers.
Forgetting case 2 can give two values.
D = 0 is often a quadratic in a. Exercises 2 and 3 have three answers each for exactly that reason.
Not hunting for the always-a-factor root.
These questions are built so that one root works for every a — the a terms cancel. Try x = 1 and x = −1 first; one of them will.
Losing the a on the way through the division.
The a travels into the quadratic and usually lands in its constant term. Do the synthetic division carefully with the letter still in place.
Not writing out the finished factorisation.
Seeing (x + 1)²(x − 4) is the proof the case worked. Skipping it is how a wrong value survives to the end.
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