Lesson 6: Quadratic Equations

Quadratic equations are the first equations that can refuse to cooperate: two solutions, one, or none, decided by a single number, the discriminant. The Nine Chapters solved quadratics arising from land measurement; Yang Hui gave systematic methods in the 13th century. In the Gaokao, quadratics are everywhere — disguised as geometry areas, profit functions, projectile heights, and parameter problems where the discriminant carries conditions. This lesson builds the four tools — factorising, completing the square, the formula, and Vieta's relations — and the judgement to choose between them.

The tools

For ax² + bx + c = 0 with a ≠ 0, the discriminant Δ = b² − 4ac decides: Δ > 0 gives two distinct real roots, Δ = 0 gives one repeated root, Δ < 0 gives no real roots (the complex roots arrive later in the curriculum). The quadratic formula x = (−b ± √Δ)/(2a) is proved by completing the square on the general equation — you should be able to reproduce that proof, because the Gaokao tests understanding, not recall. Vieta's formulas: if the roots are x₁, x₂, then x₁ + x₂ = −b/a and x₁x₂ = c/a. These convert symmetric expressions in the roots (x₁² + x₂², 1/x₁ + 1/x₂, (x₁ − x₂)²) into arithmetic on coefficients — without ever finding the roots.

Completing the square deserves emphasis: x² + bx = (x + b/2)² − (b/2)². It proves the formula, it finds vertices of parabolas (Lesson 8), and it solves optimisation problems (below). One technique, three payoffs.

Worked examples

Worked example 1 — factorising. Solve x² − 5x + 6 = 0. Seek two numbers with product 6 and sum −5 (watch signs: both negative): −2 and −3. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3. Verify by Vieta: sum 5 = −(−5)/1 ✓, product 6 ✓.

Worked example 2 — completing the square. Solve x² + 6x + 2 = 0. Move the constant: x² + 6x = −2. Add (6/2)² = 9 to both sides: (x + 3)² = 7. Take square roots — with the ±: x + 3 = ±√7, so x = −3 + √7 or x = −3 − √7. Approximate check: √7 ≈ 2.65, roots ≈ −0.35 and −5.65; product ≈ 1.98 ≈ c/a = 2 ✓ (Vieta again as a checker).

Worked example 3 — the formula. Solve 2x² − 3x − 1 = 0. Δ = (−3)² − 4 × 2 × (−1) = 9 + 8 = 17 > 0 — two real roots. x = (3 ± √17)/4. Approximate: √17 ≈ 4.12, roots ≈ 1.78 and ≈ −0.28. Vieta check: sum = 3/2 = 1.5, and 1.78 + (−0.28) = 1.5 ✓; product = −1/2, and 1.78 × (−0.28) ≈ −0.5 ✓.

Worked example 4 — Vieta without solving. The roots x₁, x₂ of x² − 7x + 3 = 0 satisfy x₁ + x₂ = 7 and x₁x₂ = 3. Compute: x₁² + x₂² = (x₁ + x₂)² − 2x₁x₂ = 49 − 6 = 43. And 1/x₁ + 1/x₂ = (x₁ + x₂)/(x₁x₂) = 7/3. And (x₁ − x₂)² = (x₁ + x₂)² − 4x₁x₂ = 49 − 12 = 37, so |x₁ − x₂| = √37. The roots themselves (≈ 6.54 and ≈ 0.46) are never needed — symmetric functions of roots are coefficient arithmetic in disguise.

Worked example 5 — forming the equation. The product of two consecutive positive integers is 156. Let the smaller be n: n(n + 1) = 156, so n² + n − 156 = 0. Factorise: seek product −156, sum 1: (n + 13)(n − 12) = 0. Positive root: n = 12. The integers are 12 and 13 (12 × 13 = 156 ✓). The negative root n = −13 is a genuine algebraic solution discarded by the word "positive" — reading the problem's conditions is part of solving it.

Challenge problem — Gaokao standard

Problem. A firm's daily profit in yuan from selling x hundred units is P(x) = −2x² + 80x − 600. (1) Find the sales level maximising profit, and the maximum profit. (2) For which sales levels does the firm break even?

Solution. (1) Complete the square: P(x) = −2(x² − 40x) − 600 = −2[(x − 20)² − 400] − 600 = −2(x − 20)² + 800 − 600 = −2(x − 20)² + 200. Since −2(x − 20)² ≤ 0 with equality exactly at x = 20, the maximum profit is ¥200 at x = 20 (i.e. 2,000 units). (2) Break-even: P(x) = 0. Using the completed-square form: (x − 20)² = 100, so x − 20 = ±10, giving x = 10 or x = 30. Between 1,000 and 3,000 units the firm profits; outside, it loses. The completed-square form answered both parts — maximisation and equation — which is why it outranks the formula in applied work.

Common mistakes

• Dividing by x and losing a root. x² = 3x has roots 0 and 3. Dividing by x discards x = 0. Factorise instead: x(x − 3) = 0.
• Dropping the ±. (x + 3)² = 7 gives x + 3 = ±√7 — two roots. Writing only +√7 halves your answer.
• Δ computed with sign errors. For 2x² − 3x − 1, the term −4ac = −4 × 2 × (−1) = +8. Write Δ out in full before substituting.
• Vieta sign slips. Sum of roots is −b/a, not b/a. For x² − 7x + 3 the sum is +7 (Worked example 4).
• Forgetting a ≠ 0 in parameter problems. If a "quadratic" has a parameter on x², the case where that parameter makes it linear must be considered separately — a favourite Gaokao trap (see Lesson 7).

Exam Experience

Multiple choice. The equation 3x² − 2x + 1 = 0 has: A. two distinct real roots B. one repeated real root C. no real roots D. roots 1 and 1/3 Answer: C. Δ = (−2)² − 4 × 3 × 1 = 4 − 12 = −8 < 0. Option D is the trap for students who "spot" factors that do not exist (check: 3 − 2 + 1 = 2 ≠ 0).

Fill in the blank. If x₁, x₂ are the roots of x² − 4x − 1 = 0, then x₁² + x₂² = ____. Answer: 18. Sum 4, product −1: x₁² + x₂² = 16 − 2 × (−1) = 18.

Solution question (10 marks). A rectangular garden has area 96 m², and its length is 4 m more than its width. (1) Find the length and width. (2) The garden is fenced on all four sides at ¥25 per metre. Find the fencing cost. Marking scheme. (1) Let the width be w; length w + 4. Equation: w(w + 4) = 96 — 2 marks. Expand and solve: w² + 4w − 96 = 0 factors as (w + 12)(w − 8) = 0 — 2 marks. Take the positive root w = 8 (rejecting −12 explicitly earns the reasoning mark), so width 8 m, length 12 m — 1 mark. [5 marks] (2) Perimeter: 2 × (8 + 12) = 40 m — 2 marks. Cost: 40 × 25 = ¥1,000 — 2 marks, plus 1 mark for a concluding statement with units. [5 marks] Total: 10 marks. Rejecting the negative root in words is not pedantry — markers award it because it shows the model was understood, not merely computed.

Going deeper — discriminants with parameters, and building equations backwards

Parameter problems are where the discriminant earns its name. Problem: for which values of k does x² + kx + 9 = 0 have two equal (repeated) real roots? The condition is Δ = 0: Δ = k² − 4 × 9 = k² − 36 = 0, so k = 6 or k = −6. (Check k = 6: x² + 6x + 9 = (x + 3)², root −3 twice ✓. k = −6: (x − 3)², root 3 twice ✓.) The full classification follows the same dial: |k| > 6 gives two distinct real roots, |k| = 6 one repeated root, |k| < 6 none. Gaokao phrasing varies — "the equation has real roots" means Δ ≥ 0, i.e. |k| ≥ 6 — and reading the exact quantifier is the assessed skill.

Vieta's formulas also run backwards: given desired roots, build the equation. Roots 2 − √3 and 2 + √3: sum = 4, product = (2)² − (√3)² = 1 (difference of squares), so the monic equation is x² − (sum)x + (product) = x² − 4x + 1 = 0. Verify by the formula: x = [4 ± √(16 − 4)]/2 = 2 ± √3 ✓. This reverse direction constructs problems with designed answers — and, more importantly for you, it powers a standard Gaokao move: if a problem hands you a quadratic whose roots appear inside a symmetric expression, Vieta converts without solving (Worked example 4); if it hands you an expression and asks for the equation, reverse Vieta builds it. One more discipline note: in parameter problems where the x² coefficient contains the parameter — say (k − 1)x² + 2x + 1 = 0 — the case k = 1 makes the equation linear (2x + 1 = 0, one root) and must be judged separately before any discriminant is computed. Forgetting the degenerate case is the single most punished step in this problem family.

Second look — one root given, everything recovered

Problem. One root of x² − px + 12 = 0 is x = 3. Find p and the other root. Vieta's formulas (Worked example 4) do everything without solving. The product of the roots equals c/a = 12, so the other root is 12 ÷ 3 = 4. The sum of the roots equals p (since −b/a = p for this monic equation), so p = 3 + 4 = 7. Verify by substitution: with p = 7 the equation is x² − 7x + 12 = (x − 3)(x − 4) = 0 ✓ — roots 3 and 4, as required. This "recover the equation from partial root information" pattern is the reverse-Vieta skill of the Going deeper section in examination dress: examiners give you one root, or the sum, or a relation like x₁ = 2x₂, and Vieta converts the gift into the whole picture. Never reach for the quadratic formula when a symmetric fact about the roots is already in your hand.