1. Vieta's Formulae — Roots and Coefficients
For any polynomial equation, there are elegant relationships between its roots and its coefficients. These are called Vieta's formulae. They are the foundation of everything in this topic.
| Degree | Equation (a ≠ 0) | Roots | Sum of Roots (Σα) | Sum of Products in Pairs (Σαβ) | Product (αβγ…) |
|---|---|---|---|---|---|
| 2 | ax² + bx + c = 0 | α, β | α + β = −b/a | αβ = c/a | αβ = c/a |
| 3 | ax³ + bx² + cx + d = 0 | α, β, γ | α+β+γ = −b/a | αβ+βγ+γα = c/a | αβγ = −d/a |
| 4 | ax⁴+bx³+cx²+dx+e = 0 | α,β,γ,δ | Σα = −b/a | Σαβ = c/a | αβγδ = e/a |
The signs alternate starting from the coefficient of the next-highest term. For ax³+bx²+cx+d = 0:
The pattern: −, +, −, + … for coefficients b, c, d, e divided by a.
The Quartic — Extra Relation
For ax⁴+bx³+cx²+dx+e = 0 with roots α, β, γ, δ, there is an additional symmetric function:
2. Symmetric Functions — Key Results
A symmetric function of the roots is any expression that is unchanged when any two roots are swapped. Vieta's formulae give us the elementary symmetric functions — all others can be built from them.
For a Cubic — Building Useful Expressions
Given α + β + γ = p, αβ + βγ + γα = q, αβγ = r, we can derive:
= p² − 2q
= p³ − 3pq + 3r
= q/r
= q² − 2pr
= pq − 3r
= (q²−2pr)/r²
The most used identity in this topic is:
⟹ Σα² = (Σα)² − 2(Σαβ). This works for any number of roots — a fundamental algebraic identity.
3. Newton's Power Sums
For higher powers, Newton's identities give a recursive method. Let S_n = αⁿ + βⁿ + γⁿ (power sum). For a cubic with Σα=p, Σαβ=q, αβγ=r:
4. Substitution — Forming New Equations
Given an equation with roots α, β, γ, we can find an equation whose roots are some function of α, β, γ — such as 2α, 1/α, α², α+1, etc.
If the new roots are y = f(α), express x in terms of y (i.e. find x = g(y)), then substitute x = g(y) into the original equation. The result, simplified, gives the new equation in y.
Common Substitutions — No Hints Given in Exam
| New roots | Substitution (x in terms of y) | Note |
|---|---|---|
| kα (scaled) | x = y/k | No hint given |
| 1/α (reciprocal) | x = 1/y | No hint given |
| α² (square) | x = √y (or x² = y) | No hint given |
| α + k (shift) | x = y − k | No hint given |
| α − k | x = y + k | No hint given |
| (aα + b) | x = (y − b)/a | General linear |
Students often confuse which direction to substitute. Remember: if the new root is y = 1/α, you need α = 1/y, so substitute x = 1/y into the original. Always express x in terms of y, not the other way around.
5. Finding Unknown Coefficients
If some information about the roots is given (e.g. one root, or the sum of squares of the roots), Vieta's formulae let us recover the unknown coefficients of the polynomial.
- Write down Vieta's formulae for the polynomial.
- Substitute the given information to form equations in the unknowns.
- Solve the system — often one substitution reduces to a simple equation.
- Always verify by checking all Vieta's conditions hold.
Substitute x = 1/y into the original equation:
For new equation 5y³ − y² + 3y − 2 = 0:
Expand each term carefully:
Collect terms by power:
The zero coefficients of y³ and y make sense — shifting roots by 1 removes the linear and cubic terms here.
Interactive Root Explorer
Enter the coefficients of a cubic equation. The tool computes Vieta's formulae and key symmetric functions instantly — and draws the polynomial.
New Equation Builder
Enter coefficients of a quartic ax⁴ + bx³ + cx² + dx + e = 0:
Practice Questions
From Vieta's: Σα = −p, Σαβ = −7, αβγ = −q.
Since the problem expects a unique answer, this suggests either a specific root is given or integer roots are expected. Let us verify with integer root γ = 1: Σαβ = αβ + γ(α+β) = −3 + 1·(−3) = −6 ≠ −7. Try γ = −3: −3 + (−3)(−2−(−3)) = −3 + (−3)(1) = −6 ≠ −7. Recheck — question as stated gives γ = −1 ± √5.
(a) New roots 2α: let y = 2x → x = y/2. Substitute:
(b) New roots α+2: let y = x+2 → x = y−2. Substitute:
(c) New roots α²: let y = x² → x = ±√y. Since roots of cubic may be positive or negative, use: substitute x = √y into 2x³ − x² + 3x − 1 = 0:
Formula Reference Sheet
Complete reference for Roots of Polynomial Equations — Cambridge 9231 P1, Section 1.1.
- Write Vieta's formulae first, every time — even before reading what is asked. Label them p, q, r or use Σα notation.
- For Σα², the identity
(Σα)² = Σα² + 2Σαβis the single most used result in this topic — commit it to memory. - For substitution, always express x in terms of y, not y in terms of x. Then substitute and clear fractions by multiplying through.
- For reciprocal roots: the new equation's coefficients are the reverse of the original — a quick check.
- For α² roots: after substituting x = √y, group terms with and without √y, then square once to eliminate the radical.
- Always verify the new equation using Vieta's — the sum of the new roots should equal what you can compute from the old roots.