Chapter 3 · Complex Numbers and Equations · Concept 3
3-18 Determination of Cubic Equation
Every lesson so far gave you an equation and asked for its solutions. This one gives you a solution and asks for the equation — and there is only one move to make.
Guiding question
If you already know one answer, what does that tell you about the question?
Learning outcome. Solve higher-degree equations and build cubic equations from what you are given.
Substitute the solution in
When a solution to a cubic equation is given, put it into the equation. A solution makes the equation zero — that is the only thing the word means — and once it is in, the unknown constants are the only things left.
If the solution is a real number
You get one ordinary equation in the constants. Two constants means you need two given solutions.
If the solution is complex
You get one COMPLEX equation, which splits into two — real parts and imaginary parts separately, from lesson 3-1. So a single complex solution is worth two facts.
And a shortcut the book uses without announcing it.
When all the coefficients are real, complex solutions always come in conjugate pairs. So the moment you are told 2 + i is a solution, you also know 2 − i is one — for free, before doing anything.
Which means a cubic with a complex solution always has exactly one real solution: the complex ones pair up, and three is an odd number, so one has to be left over.
One last thing about the wording. “Find the OTHER solutions” means the ones you were not given. The book says so in its margin: do not put 2 + i back in your answer.
The idea takes one sentence. The arithmetic takes a page — so here it is, one line at a time.
the book's Warm Up
x3 + ax2 + bx + 5 = 0 has 2 + i as a solution
x³ + ax² + bx + 5 = 0 has 2 + i as a solution, with a and b real. Find a and b, and the other solutions.
- 1
Work out the powers of 2 + i before anything else.
(2 + i)2 = 3 + 4i, (2 + i)3 = 2 + 11iSquaring first and multiplying by (2 + i) again is far safer than cubing in one go.
Work with a partner. Before either of you substitutes anything, write down how many unknowns there are and how many facts the question gives. If those two numbers do not match, look again — one of the facts is hiding.
x³ + ax + b = 0 has 1 + √2 i as a solution, with a and b real. Find a and b, and the other solutions.
Why is being told one complex solution enough to pin down two unknown constants, when one real solution only pins down one?
Count the equations each one produces.
Cubing in one go.
Work out the square first, then multiply by the bracket once more. (2 + i)² = 3 + 4i, then (3 + 4i)(2 + i) = 2 + 11i.
Leaving the i in the imaginary pile.
From 4ai, the piece that goes in the pile is 4a. The i stays outside the bracket.
Forgetting to split into two equations.
One complex equation is two real ones. Stopping at (…) + (…)i = 0 leaves the question half done.
Putting the given solution back in your answer.
"The OTHER solutions" means the ones you were not told. The book says so specifically.
Not using the conjugate.
With real coefficients, being told 1 − 3i is a solution means 1 + 3i is too — which hands you a quadratic factor and saves the whole root hunt.
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