Chapter 3 · Complex Numbers and Equations · Concept 2
3-7 Relationship between Roots and Coefficients (1)
The two solutions of a quadratic are hidden inside its three numbers — and you can get at their sum and their product without ever solving anything. Once you have those two, a surprising number of questions fall out.
Guiding question
What can the coefficients tell you about the roots, without finding them?
Learning outcome. Find relationships between the roots of a quadratic and its coefficients, and use them.
Sum and product
If the two solutions of ax2 + bx + c = 0 are called α and β, then:
their sum
α + β = −ba
their product
αβ = ca
And the method for any expression built from α and β is always the same: find α + β and αβ first, then rewrite the expression using only those two. Two rewrites do most of the work:
α2 + β2 = (α + β)2 − 2αβ
α3 + β3 = (α + β)3 − 3αβ(α + β)
Where the first one comes from. (α + β)² is α² + 2αβ + β². That is the α² + β² you wanted, plus an extra 2αβ. So take the 2αβ away again. Nothing is being memorised here — it is one expansion and one subtraction.
The roots of 2x² − 6x − 3 = 0 are (3 ± √15)/2 — decimals that go on forever. Watch the long way struggle and the short way walk.
the book's Warm Up 2 · 2x² − 6x − 3 = 0
the long way · find the roots first
α = (3 + √15) ÷ 2 = 3.436492…
β = (3 − √15) ÷ 2 = -0.436492…
the short way · never find them at all
α + β = −(−6) ÷ 2 = 3
αβ = −3 ÷ 2 = −3/2
2(b) α and β solve 2x² − 6x − 3 = 0. Find α³ + β³.
- 1
Get the two numbers you will actually use. Nothing else happens until these are down.
α + β = −−62 = 3 and αβ = −32 = −32Careful with the sum: it is MINUS b over a, and b is already −6, so the two minuses cancel.
Work with a partner. The equation in question 2 has imaginary roots — its discriminant is 16 − 24 = −8. Notice that every answer still comes out real. Talk about why that is not a coincidence.
1. Sum and product for 4x² + 2x − 7 = 0.
2. α and β solve 2x² − 4x + 3 = 0. Find each of these.
1. Sum and product.
2. α and β solve x² − 2x + 3 = 0.
3. α and β solve 3x² + 6x − 1 = 0.
α and β solve 2x² − 6x + 3 = 0, so α + β = 3 and αβ = 3/2.
The roots of x² − 2x + 3 = 0 are imaginary, yet α² + β², α³ + β³ and αβ all come out as ordinary real numbers. Why?
Look at what the two roots have to be, once one of them is known.
Swapping the two formulas.
The SUM is −b/a and the PRODUCT is c/a. The book names this as the main misconception. The minus lives with the sum.
Forgetting to divide by a.
For 3x² − 7x − 6 = 0 the sum is 7/3, not 7. Both formulas have an a underneath.
Using −2αβ where it should be −4αβ.
(α + β)² − 2αβ gives α² + β². (α + β)² − 4αβ gives (α − β)². Different questions, different number.
Losing a minus in a triple product.
In 3αβ(α + β) with αβ = −1/3 and α + β = −2, the piece is +2, and it is being subtracted. Work it out on its own line before combining.
Panicking when the answer is negative.
α² + β² = −2 is fine when the roots are imaginary. Nothing has gone wrong.
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