Roots and coefficients of a quadratic
A root of a polynomial is a value of x that makes it equal to zero. On the graph, real roots are where the curve crosses the x-axis.
If ax² + bx + c = 0 has roots α and β, then ax² + bx + c = a(x − α)(x − β). Expanding the right side gives a(x² − (α + β)x + αβ). Compare coefficients:
α + β = −b/a and αβ = c/a.
This works even if the roots are complex numbers.
Cubics and quartics (Vieta's formulas)
For ax³ + bx² + cx + d = 0 with roots α, β, γ:
- Σα = α + β + γ = −b/a
- Σαβ = αβ + βγ + γα = c/a
- αβγ = −d/a
For ax⁴ + bx³ + cx² + dx + e = 0 with roots α, β, γ, δ:
- Σα = −b/a, Σαβ = c/a (6 pairs), Σαβγ = −d/a (4 triples), αβγδ = e/a.
Pattern: the signs go −, +, −, + along the coefficients after the first.
Useful identities for roots (no solving needed):
- α² + β² + γ² = (Σα)² − 2Σαβ
- 1/α + 1/β + 1/γ = Σαβ ÷ αβγ
- α³ + β³ + γ³ = (Σα)³ − 3(Σα)(Σαβ) + 3αβγ
Building an equation from its roots
Go backwards: if you know Σα, Σαβ and αβγ, the cubic is
x³ − (Σα)x² + (Σαβ)x − αβγ = 0
Multiply through to clear fractions if needed. Example: roots 1, 2, 3 give Σα = 6, Σαβ = 11, αβγ = 6, so x³ − 6x² + 11x − 6 = 0.
Transformed roots
Suppose an equation has roots α, β, γ and you want an equation whose roots are changed in the same way, for example 2α, 2β, 2γ.
Substitution method: let the new root be w, write the old root in terms of w and put it into the old equation.
- New roots α + k: w = x + k, so x = w − k. Replace x by (w − k). The graph slides k to the right.
- New roots kα: x = w/k. Replace x by w/k, then clear fractions.
- New roots 1/α: x = 1/w. Replace x by 1/w, then multiply by w³.
- New roots α²: x = √w. Rearrange so only even powers of √w remain, then square.
The other way: compute the new Σ, Σ pairs and product from the old ones. Substitution is usually quicker.
Polynomial identities and how to prove them
An identity is true for every value of the variables (symbol ≡). An equation is true only for some values.
To prove an identity: start with one side (usually the more complicated), expand or factorise step by step, and reach the other side. Do not work on both sides at once as if it were an equation you were solving.
Useful ones:
- (a + b)² = a² + 2ab + b²; a² − b² = (a − b)(a + b)
- a³ + b³ = (a + b)(a² − ab + b²); a³ − b³ = (a − b)(a² + ab + b²)
- (x² + y²)² = (x² − y²)² + (2xy)²: this makes Pythagorean triples. With x = 2, y = 1: 5² = 3² + 4².
Identities let us do mental arithmetic too: 49 × 51 = (50 − 1)(50 + 1) = 2500 − 1 = 2499.
The binomial theorem
Expanding (a + b)ⁿ by hand gets long. The binomial theorem gives the answer directly:
(a + b)ⁿ = Σ C(n, k) aⁿ⁻ᵏ bᵏ, for k = 0 to n,
where C(n, k) = n! ÷ (k!(n − k)!) are the numbers in Pascal's triangle (1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1 …).
The 3D cube shows n = 3: one p³ block, three p²q blocks, three pq² blocks and one q³ block: 1, 3, 3, 1.
Try it: a practical
In the 3D: set roots −1, 2, 2. Predict the sum, the sum of pairs and the product, then check the readout. Set k = −2 and see which way the curve moves.
At home: build a 3 × 3 × 3 cube from 27 sugar cubes or dice. Split it into a 2 × 2 × 2 block, three 2 × 2 × 1 slabs, three 2 × 1 × 1 rods and one single cube. Count: 8 + 12 + 6 + 1 = 27. You have proved (2 + 1)³ by hand.
Key formulas and definitions
- Quadratic: α + β = −b/a, αβ = c/a
- Cubic: Σα = −b/a, Σαβ = c/a, αβγ = −d/a
- Quartic: Σα = −b/a, Σαβ = c/a, Σαβγ = −d/a, αβγδ = e/a
- α² + β² + γ² = (Σα)² − 2Σαβ
- Roots α + k: replace x by (x − k); roots kα: replace x by x/k; roots 1/α: replace x by 1/x
- (a + b)ⁿ = Σ C(n, k) aⁿ⁻ᵏ bᵏ
Worked examples
1. The roots of 2x³ − 6x² + 5x + 4 = 0 are α, β, γ. Find Σα, Σαβ and αβγ.
a = 2, b = −6, c = 5, d = 4. Σα = −b/a = 6/2 = 3. Σαβ = c/a = 5/2. αβγ = −d/a = −4/2 = −2.
2. For the same equation, find α² + β² + γ².
(Σα)² − 2Σαβ = 3² − 2 × 5/2 = 9 − 5 = 4.
3. Find 1/α + 1/β + 1/γ for the same equation.
Σαβ ÷ αβγ = (5/2) ÷ (−2) = −5/4.
4. Find a cubic with roots 1, −2 and 3.
Σα = 2, Σαβ = (1)(−2) + (−2)(3) + (3)(1) = −2 − 6 + 3 = −5, αβγ = −6. Equation: x³ − 2x² − 5x + 6 = 0.
5. x³ − 3x + 1 = 0 has roots α, β, γ. Find an equation with roots α + 1, β + 1, γ + 1.
Let w = x + 1, so x = w − 1. (w − 1)³ − 3(w − 1) + 1 = w³ − 3w² + 3w − 1 − 3w + 3 + 1 = w³ − 3w² + 3 = 0.
6. Same equation: find one with roots 2α, 2β, 2γ.
x = w/2: w³/8 − 3w/2 + 1 = 0. Multiply by 8: w³ − 12w + 8 = 0.
7. Prove that (x² + y²)² ≡ (x² − y²)² + (2xy)².
RHS = x⁴ − 2x²y² + y⁴ + 4x²y² = x⁴ + 2x²y² + y⁴ = (x² + y²)² = LHS. ∎
8. Use the binomial theorem to expand (x + 2)⁴.
Coefficients 1, 4, 6, 4, 1: x⁴ + 4x³(2) + 6x²(4) + 4x(8) + 16 = x⁴ + 8x³ + 24x² + 32x + 16.
Common mistakes
- Forgetting the minus sign: for a cubic the sum of roots is −b/a, not b/a.
- Forgetting to divide by a when the leading coefficient is not 1.
- For transformed roots α + k, replacing x by (x + k) instead of (x − k). The new root is bigger, so you substitute x − k.
- Proving an identity by doing the same thing to both sides as if solving. Start from one side and reach the other.