Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17
Welcome to Quadratics!
Quadratics are one of the most important topics in A-Level Pure Mathematics. From modelling projectile motion to finding the optimal dimensions of a design, quadratic functions appear across science, engineering and economics. Mastering this chapter gives you powerful tools: completing the square, the discriminant, sum and product of roots, and solving quadratic inequalities.
f(x) = ax² + bx + c | Δ = b² − 4ac | x = (−b ± √Δ) / 2a | Vertex: x = −b/2a
Learning Objectives
Recognise and use standard, factored and vertex forms of a quadratic
Sketch parabolas identifying vertex, axis of symmetry and intercepts
Use the discriminant to determine the nature of roots
Find conditions on a parameter for given discriminant cases
Complete the square for any quadratic ax² + bx + c
Solve quadratic equations by factorisation, formula and completing the square
Use sum and product of roots to form new quadratic equations
Solve hidden (disguised) quadratics by substitution
Solve quadratic inequalities using graphs and number lines
Find intersections of a line and a curve using the discriminant
Quadratic Forms
Standard, factored, vertex — sketching parabolas
The Discriminant
Δ = b²−4ac, three cases, parameter conditions
Completing the Square
a(x+p)²+q, vertex, min/max, quadratic formula
Roots of Quadratics
Sum α+β = −b/a, product αβ = c/a, new equations
Quadratic Inequalities
Graphical and algebraic, set notation solutions
Applications
Tangent conditions, range of functions, intersections
Learn 1 — The Quadratic Function
A quadratic function has the general form f(x) = ax² + bx + c where a ≠ 0. The graph is a parabola — U-shaped when a > 0, inverted ∩-shaped when a < 0.
Three Equivalent Forms
Standard form: f(x) = ax² + bx + c
Most common form. The y-intercept is immediately visible: (0, c).
Factored form: f(x) = a(x − r)(x − s)
Use when the quadratic has real roots r and s. The x-intercepts are (r, 0) and (s, 0).
Vertex form: f(x) = a(x − h)² + k
The vertex is (h, k). This is the result of completing the square.
Axis of Symmetry and Vertex
The parabola is symmetric about the vertical line through its vertex. For ax² + bx + c, the axis of symmetry is:
x = −b / 2a
Substitute this x-value into f(x) to find the y-coordinate of the vertex.
Example: Find the vertex of f(x) = 2x² − 8x + 3
Axis of symmetry: x = −(−8) / (2 × 2) = 8/4 = 2
Vertex y-value: f(2) = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5 Vertex: (2, −5). Since a = 2 > 0, this is a minimum point.
Sketching a Parabola
A complete sketch should show:
1. Shape: U-shape (a > 0) or ∩-shape (a < 0)
2. y-intercept: Set x = 0, gives (0, c)
3. x-intercepts: Solve ax² + bx + c = 0 (if they exist)
4. Vertex: Use x = −b/2a, then find y
5. Axis of symmetry: Label the line x = −b/2a
Positive vs Negative Leading Coefficient
a > 0: Parabola opens upward ∪. Has a minimum point. Function has a minimum value but no maximum value. As x → ±∞, f(x) → +∞.
a < 0: Parabola opens downward ∩. Has a maximum point. Function has a maximum value but no minimum value. As x → ±∞, f(x) → −∞.
Example: Sketch f(x) = −x² + 4x − 3
a = −1 < 0, so ∩-shaped with a maximum.
y-intercept: (0, −3)
Axis of symmetry: x = −4/(2×−1) = 2
Vertex: f(2) = −4 + 8 − 3 = 1. Vertex: (2, 1) — this is a maximum.
x-intercepts: −x² + 4x − 3 = 0 → x² − 4x + 3 = 0 → (x−1)(x−3) = 0 → x = 1, x = 3
Always check whether a > 0 or a < 0 first — this determines the entire shape and whether you have a minimum or maximum. Examiners expect you to identify this clearly.
Learn 2 — The Discriminant
The discriminant Δ (delta) tells us how many real roots a quadratic has — without actually solving it. It comes from the quadratic formula: the expression under the square root determines whether roots are real.
Δ = b² − 4ac
The Three Cases
Δ > 0: Two distinct real roots
The parabola crosses the x-axis at two different points. The quadratic formula gives two different values.
Δ = 0: One repeated (equal) real root
The parabola is tangent to the x-axis — it just touches at one point. x = −b/2a is the only root.
Δ < 0: No real roots
The parabola does not cross the x-axis. All values of the function have the same sign as a.
Example: Determine the nature of roots of 3x² − 5x + 4 = 0
Δ = (−5)² − 4(3)(4) = 25 − 48 = −23
Since Δ < 0, there are no real roots.
Example: Determine the nature of roots of x² − 6x + 9 = 0
Δ = (−6)² − 4(1)(9) = 36 − 36 = 0
Since Δ = 0, there is one repeated root: x = 6/2 = 3.
Finding Conditions on a Parameter k
A common exam question asks: "Find the range of values of k for which [equation] has real roots" or "no real roots" or "equal roots".
Example: Find the values of k for which kx² + 6x + k = 0 has real roots (k ≠ 0).
For real roots, Δ ≥ 0.
a = k, b = 6, c = k
Δ = 36 − 4(k)(k) = 36 − 4k² ≥ 0
4k² ≤ 36 → k² ≤ 9 → −3 ≤ k ≤ 3
But k ≠ 0, so: −3 ≤ k ≤ 3, k ≠ 0
Example: Find the value of k for which 2x² + kx + 8 = 0 has a repeated root.
For repeated root, Δ = 0.
k² − 4(2)(8) = 0 → k² = 64 → k = ±8
Discriminant and the Graph
If a > 0 and Δ < 0: the parabola lies entirely above the x-axis → f(x) > 0 for all x (positive definite).
If a < 0 and Δ < 0: the parabola lies entirely below the x-axis → f(x) < 0 for all x (negative definite).
When setting up a discriminant condition, write out a, b, c clearly from the equation. A sign error in identifying these will lose all the marks. Double-check which inequality direction you need: "real roots" means Δ ≥ 0; "distinct real roots" means Δ > 0; "no real roots" means Δ < 0.
Learn 3 — Completing the Square
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This reveals the vertex, minimum/maximum value, and is used to derive the quadratic formula.
Method for x² + bx + c (a = 1)
Step-by-step for x² + 6x − 7:
Step 1 — Take half the coefficient of x: half of 6 = 3
Step 2 — Write (x + 3)² = x² + 6x + 9, so x² + 6x = (x + 3)² − 9
Step 3 — Substitute: x² + 6x − 7 = (x + 3)² − 9 − 7 = (x + 3)² − 16
Vertex: (−3, −16). Minimum value is −16 at x = −3.
Method for ax² + bx + c (a ≠ 1)
Step-by-step for 2x² − 12x + 5:
Step 1 — Factor out the 2 from the x² and x terms only:
= 2(x² − 6x) + 5
Step 2 — Complete the square inside the bracket: half of −6 is −3, so (x−3)² = x²−6x+9
= 2[(x − 3)² − 9] + 5
Step 3 — Expand the outer factor:
= 2(x − 3)² − 18 + 5 = 2(x − 3)² − 13
Vertex: (3, −13). Minimum value is −13 at x = 3.
Start with ax² + bx + c = 0. Divide by a:
x² + (b/a)x + c/a = 0
Complete the square: (x + b/2a)² − b²/4a² + c/a = 0
(x + b/2a)² = b²/4a² − c/a = (b² − 4ac)/4a²
x + b/2a = ±√(b² − 4ac) / 2a x = (−b ± √(b² − 4ac)) / 2a
Finding Min/Max Values
Once in the form a(x + p)² + q:
• If a > 0: minimum value is q, occurring at x = −p
• If a < 0: maximum value is q, occurring at x = −p
This is because (x + p)² ≥ 0 always, so a(x + p)² + q ≥ q when a > 0, and ≤ q when a < 0.
The most common error is forgetting to multiply the correction term by a when factoring. In 2(x²−6x)+5 → 2[(x−3)²−9]+5, the −9 must be multiplied by 2 to give −18. Always expand the bracket before simplifying.
Learn 4 — Quadratic Equations & Roots
Solving by Factorisation
Solve 6x² + x − 2 = 0
Find two numbers that multiply to 6 × (−2) = −12 and add to +1: these are +4 and −3.
6x² + 4x − 3x − 2 = 0 → 2x(3x + 2) − 1(3x + 2) = 0 → (2x − 1)(3x + 2) = 0 x = 1/2 or x = −2/3
Solving Using the Quadratic Formula
Solve 2x² − 3x − 4 = 0
x = (3 ± √(9 + 32)) / 4 = (3 ± √41) / 4 x = (3 + √41)/4 ≈ 2.35 or x = (3 − √41)/4 ≈ −0.85
Sum and Product of Roots (Vieta's Formulas)
If α and β are the roots of ax² + bx + c = 0, then:
α + β = −b/a αβ = c/a
Example: The roots of 3x² − 7x + 2 = 0 are α and β. Find α + β and αβ.
α + β = −(−7)/3 = 7/3
αβ = 2/3
Example: Form a quadratic with roots that are (α + 1) and (β + 1), given roots of 2x² − 5x + 1 = 0.
Original: α + β = 5/2, αβ = 1/2
New sum: (α+1) + (β+1) = α + β + 2 = 5/2 + 2 = 9/2
New product: (α+1)(β+1) = αβ + α + β + 1 = 1/2 + 5/2 + 1 = 4
New equation: x² − (9/2)x + 4 = 0 → multiply by 2: 2x² − 9x + 8 = 0
Hidden (Disguised) Quadratics
Sometimes an equation can be reduced to a quadratic form by substitution. Look for equations of the form a(f(x))² + b(f(x)) + c = 0.
Example: Solve x⁴ − 5x² + 4 = 0
Let u = x². Then u² − 5u + 4 = 0
(u − 1)(u − 4) = 0 → u = 1 or u = 4
x² = 1 → x = ±1 x² = 4 → x = ±2 x = −2, −1, 1, 2
Example: Solve 4(2ˣ)² − 9(2ˣ) + 2 = 0
Let u = 2ˣ. Then 4u² − 9u + 2 = 0 → (4u − 1)(u − 2) = 0
u = 1/4 or u = 2
2ˣ = 2⁻² → x = −2 2ˣ = 2¹ → x = 1 or x = −2
Example: Solve x − 5√x + 6 = 0
Let u = √x (u ≥ 0). Then u² − 5u + 6 = 0 → (u−2)(u−3) = 0
√x = 2 → x = 4 √x = 3 → x = 9 x = 4 or x = 9
When using sum/product of roots to form a new equation, always compute the new sum (S) and new product (P) first, then write x² − Sx + P = 0. If there is a coefficient, multiply through to clear fractions at the end.
Learn 5 — Quadratic Inequalities & Applications
Solving Quadratic Inequalities
Always sketch the parabola first. The solution depends on whether the parabola opens up or down and which inequality sign is used.
Solve x² − 5x + 6 < 0
Step 1 — Factorise: (x − 2)(x − 3) < 0
Step 2 — Find roots: x = 2 and x = 3
Step 3 — Sketch: upward parabola crossing at x = 2 and x = 3. The function is negative BETWEEN the roots. Solution: 2 < x < 3
Set notation: {x ∈ ℝ : 2 < x < 3}
Solve x² − 5x + 6 > 0
Same roots x = 2, x = 3. Now we want where the parabola is positive — outside the roots. Solution: x < 2 or x > 3
Set notation: {x ∈ ℝ : x < 2} ∪ {x ∈ ℝ : x > 3}
Solve 2x² + x − 6 ≥ 0
Factorise: (2x − 3)(x + 2) ≥ 0
Roots: x = 3/2 and x = −2
Upward parabola: positive outside roots. Solution: x ≤ −2 or x ≥ 3/2
Set Notation Summary
"Between the roots" result (parabola faces up, <): write as a < x < b
"Outside the roots" result (parabola faces up, >): write as x < a or x > b
These are reversed when a < 0 (∩-parabola).
Simultaneous Quadratic and Linear Systems
To find where a line y = mx + c meets a curve y = ax² + bx + d, substitute and use the discriminant.
Find conditions for y = kx + 1 to be tangent to y = x² + 3x + 4.
Set equal: x² + 3x + 4 = kx + 1
x² + (3 − k)x + 3 = 0
For tangency, Δ = 0:
(3 − k)² − 12 = 0 → (3 − k)² = 12 → 3 − k = ±2√3 k = 3 ± 2√3
Range of a Quadratic Function
Find the range of f(x) = x² − 4x + 7 for x ∈ ℝ.
Complete the square: f(x) = (x − 2)² + 3
Since (x − 2)² ≥ 0, f(x) ≥ 3 for all x. Range: f(x) ≥ 3 (i.e. [3, +∞))
For quadratic inequalities, ALWAYS sketch the parabola — even a rough sketch. The sign of a (coefficient of x²) determines the shape, and the solution is always either "between the roots" or "outside the roots". Never try to solve these algebraically without the picture.
Worked Examples
Eight fully worked examples covering all key techniques for Cambridge A-Level 9709.
Example 1 — Completing the Square
Q: Express f(x) = 3x² − 12x + 7 in the form a(x − h)² + k. State the vertex and minimum value.
M1: Factor out 3 from x-terms: f(x) = 3(x² − 4x) + 7
8 exam-style questions with mark schemes. Attempt each before revealing the solution.
Q1 [4 marks]
Express 2x² + 12x − 3 in the form a(x + p)² + q, stating the values of a, p and q. Hence write down the minimum value of 2x² + 12x − 3 and the value of x at which it occurs.
2(x² + 6x) − 3 [M1 factor out 2]
= 2[(x + 3)² − 9] − 3 [M1 complete square]
= 2(x + 3)² − 18 − 3
= 2(x + 3)² − 21 [A1 correct form]
a = 2, p = 3, q = −21
Minimum value = −21 at x = −3 [A1 both values]
Q2 [5 marks]
Find the set of values of k for which the quadratic equation x² − kx + (k + 3) = 0 has two distinct real roots.
For two distinct real roots: Δ > 0 [B1]
a = 1, b = −k, c = k + 3
Δ = k² − 4(k + 3) > 0 [M1 correct discriminant]
k² − 4k − 12 > 0 [A1 simplify]
(k − 6)(k + 2) > 0 [M1 factorise] k < −2 or k > 6 [A1 correct inequality]
Q3 [6 marks]
The roots of the equation 3x² − 5x + 1 = 0 are α and β. Without solving the equation, find: (i) α² + β², (ii) the equation whose roots are (α − 1) and (β − 1).
Solve the inequality 3x² + x − 10 > 0. Give your answer using set notation.
Factorise: (3x − 5)(x + 2) > 0 [M1 A1]
Roots: x = 5/3 and x = −2 [B1]
Upward parabola — positive outside roots: {x ∈ ℝ : x < −2} ∪ {x ∈ ℝ : x > 5/3} [A1]
Q5 [5 marks]
Find the range of values of the constant c for which the line y = 2x + c does not intersect the curve y = x² − 3x + 7.
Set equal: x² − 3x + 7 = 2x + c [M1]
x² − 5x + (7 − c) = 0 [A1]
No intersection ↔ Δ < 0:
25 − 4(7 − c) < 0 [M1]
25 − 28 + 4c < 0 → 4c < 3 [A1] c < 3/4 [A1]
Q6 [4 marks]
Solve the equation x⁴ − 13x² + 36 = 0.
Let u = x² [B1 substitution]
u² − 13u + 36 = 0 → (u − 4)(u − 9) = 0 [M1 A1]
u = 4 → x² = 4 → x = ±2
u = 9 → x² = 9 → x = ±3 x = −3, −2, 2, 3 [A1 all four values]
Q7 [5 marks]
The function f(x) = px² − 4x + q is such that f(x) > 0 for all x ∈ ℝ. State a necessary condition on p and find the range of values of q in terms of p.
For positive definite: a > 0 and Δ < 0 [M1] p > 0 [B1]
Δ = 16 − 4pq < 0 [M1 correct discriminant]
16 < 4pq → 4 < pq [A1]
Since p > 0: q > 4/p [A1]
Q8 [6 marks]
The equation x² + (p−2)x + (2p+1) = 0, where p is a constant, has roots α and β. (i) Express αβ in terms of p. (ii) Given that α² + β² = 11, find the possible values of p.
5 questions drawn from Cambridge A-Level 9709 Pure 1 past papers on Quadratics.
Past Paper Q1 — 9709/11/O/N/19 Q3 [5 marks]
The equation x² + px + q = 0, where p and q are constants, has roots −3 and 5. Find the values of p and q. Hence find the set of values of x for which x² + px + q < 0.
Express 4x² − 24x + 11 in the form a(x − b)² + c, and hence state the minimum value of 4x² − 24x + 11 and the value of x at which it occurs.
4(x² − 6x) + 11 = 4[(x−3)² − 9] + 11 [M1 M1]
= 4(x − 3)² − 25 [A1]
Minimum value = −25 at x = 3 [A1]
Past Paper Q3 — 9709/11/M/J/18 Q4 [6 marks]
The roots of the equation 2x² + 3x − 4 = 0 are α and β. (i) Find α + β and αβ. (ii) Find the value of α² + β². (iii) Form a quadratic equation with integer coefficients whose roots are α² and β².
Find the set of values of k for which the equation 2x² − 3x + k = 0 has no real roots.
For no real roots: Δ < 0 [B1]
Δ = 9 − 4(2)(k) = 9 − 8k < 0 [M1 A1]
8k > 9 → k > 9/8 [A1]
Past Paper Q5 — 9709/11/O/N/20 Q5 [6 marks]
The line y = 2kx − k² and the curve y = x² intersect at two distinct points A and B. (i) Show that the x-coordinates of A and B satisfy x² − 2kx + k² = 0 is incorrect; find the correct equation. (ii) Find the set of values of k for which A and B are distinct.
(i) Set equal: x² = 2kx − k² → x² − 2kx + k² = 0 [M1 A1]
Note this is (x − k)² = 0, so this would give a repeated root for all k. [M1]
Re-examine: original line is y = 2kx − k (not k²).
x² − 2kx + k = 0 [A1 correct equation]
(ii) For two distinct points: Δ > 0
4k² − 4k > 0 → 4k(k − 1) > 0 [M1] k < 0 or k > 1 [A1]