Lesson 3: Algebra II | O Level Mathematics

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1. Quadratic Equations

A quadratic equation has the form ax² + bx + c = 0 where a ≠ 0. There are three methods to solve quadratic equations: factorisation, the quadratic formula, and completing the square.

Standard Form: ax² + bx + c = 0. Always rearrange to this form before solving. A quadratic equation has at most two solutions (roots).

Method 1 — Factorisation

Zero Product Rule: If A × B = 0, then A = 0 OR B = 0 (or both). This is the foundation of solving by factorisation.

📐 Worked Example 1 — Solve by Factorisation

Solve: 2x² − 5x − 3 = 0

Find two numbers that multiply to (2 × −3) = −6 and add to −5.
1 × (−6) = −6 and 1 + (−6) = −5 ✓

Split the middle term and factorise by grouping:
2x² + x − 6x − 3 = 0
x(2x + 1) − 3(2x + 1) = 0
(x − 3)(2x + 1) = 0

Apply the zero product rule:
x − 3 = 0 → x = 3
2x + 1 = 0 → x = −½

Method 2 — The Quadratic Formula

Quadratic Formula — Use when factorisation is difficult or impossible

x = [−b ± √(b² − 4ac)] / 2a

For ax² + bx + c = 0. The expression b² − 4ac is called the discriminant (Δ).

Δ > 0 → two distinct real roots | Δ = 0 → one repeated root | Δ < 0 → no real roots

📐 Worked Example 2 — Quadratic Formula

Solve 3x² − 4x − 2 = 0. Give answers correct to 2 decimal places.

Identify a = 3, b = −4, c = −2. Calculate the discriminant:
Δ = (−4)² − 4(3)(−2) = 16 + 24 = 40

Apply the formula:
x = [4 ± √40] / 6 = [4 ± 6.3246] / 6

Two solutions:
x = (4 + 6.3246)/6 = 10.3246/6 = 1.72 (2 d.p.)
x = (4 − 6.3246)/6 = −2.3246/6 = −0.39 (2 d.p.)

Method 3 — Completing the Square

Completing the Square rewrites ax² + bx + c in the form a(x + p)² + q. It is used to:
(1) Solve quadratic equations (2) Find the vertex of a parabola (3) Identify minimum/maximum values

📐 Worked Example 3 — Completing the Square

Solve x² + 6x + 7 = 0 by completing the square.

Move the constant to the right:
x² + 6x = −7

Add (half of 6)² = 9 to BOTH sides:
x² + 6x + 9 = −7 + 9 = 2
(x + 3)² = 2

Take square roots of both sides:
x + 3 = ±√2
x = −3 + √2 ≈ −1.59 or x = −3 − √2 ≈ −4.41

📐 Worked Example 4 — Write in Completed Square Form

Write 2x² − 8x + 5 in the form a(x + p)² + q. State the minimum value and the value of x at which it occurs.

Factor out the coefficient of x² from the first two terms:
2(x² − 4x) + 5

Complete the square inside the bracket: half of (−4) = −2, square = 4.
2(x² − 4x + 4 − 4) + 5
2[(x − 2)² − 4] + 5

Expand the outer bracket:
2(x − 2)² − 8 + 5 = 2(x − 2)² − 3
Minimum value = −3 when x = 2.

Which method to use?
• Factorisation — quick; works when roots are integers or simple fractions.
• Quadratic formula — always works; use when told to give answers to d.p. or when Δ is not a perfect square.
• Completing the square — use when asked for vertex, minimum/maximum, or to write in a(x+p)²+q form.
If the question says "give exact answers" → use formula leaving √ in answer, or complete the square.

2. Variation (Direct, Inverse, and Other)

Type	Statement	Equation	Graph Shape
Direct	y is directly proportional to x	y ∝ x → y = kx	Straight line through origin
Direct (Square)	y is directly proportional to x²	y ∝ x² → y = kx²	Parabola through origin
Direct (Square Root)	y is directly proportional to √x	y ∝ √x → y = k√x	Curve increasing, decreasing rate
Inverse	y is inversely proportional to x	y ∝ 1/x → y = k/x	Hyperbola (xy = constant)
Inverse (Square)	y is inversely proportional to x²	y ∝ 1/x² → y = k/x²	Steeper hyperbola

📐 Worked Example 5 — Variation Problem

y is inversely proportional to x². When x = 3, y = 4. Find y when x = 6. Find x when y = 16.

Write the equation: y = k/x². Find k using x=3, y=4:
4 = k/9 → k = 36
So: y = 36/x²

Find y when x = 6:
y = 36/36 = 1

Find x when y = 16:
16 = 36/x² → x² = 36/16 = 2.25 → x = 1.5

3. Functions

Function: A rule that maps each input value (from the domain) to exactly one output value (in the range). Written as f(x) or g(x). The domain is the set of allowable inputs; the range is the set of corresponding outputs.

Function Notation and Evaluation

📐 Worked Example 6 — Evaluating Functions

f(x) = 3x² − 2x + 1. Find: (a) f(2) (b) f(−1) (c) f(x+1)

(a) Substitute x = 2:
f(2) = 3(4) − 2(2) + 1 = 12 − 4 + 1 = 9

(b) Substitute x = −1:
f(−1) = 3(1) − 2(−1) + 1 = 3 + 2 + 1 = 6

(c) Replace every x with (x+1):
f(x+1) = 3(x+1)² − 2(x+1) + 1
= 3(x²+2x+1) − 2x − 2 + 1
= 3x² + 6x + 3 − 2x − 1 = 3x² + 4x + 2

Composite Functions

Composite Function fg(x): Apply g first, then apply f to the result.
fg(x) = f(g(x)) — read right to left: g first, then f.
Note: fg(x) ≠ gf(x) in general — order matters.

📐 Worked Example 7 — Composite Functions

f(x) = 2x + 1 and g(x) = x². Find: (a) fg(x) (b) gf(x) (c) fg(3)

(a) fg(x) = f(g(x)) = f(x²) — apply f to x²:
fg(x) = 2(x²) + 1 = 2x² + 1

(b) gf(x) = g(f(x)) = g(2x+1) — apply g to (2x+1):
gf(x) = (2x+1)² = 4x² + 4x + 1

(c) fg(3) — apply g first: g(3) = 9. Then apply f: f(9) = 2(9)+1 = 19
Check: fg(x) = 2x²+1 → fg(3) = 2(9)+1 = 19 ✓

Inverse Functions

Inverse Function f⁻¹(x): Reverses the effect of f. If f maps x to y, then f⁻¹ maps y back to x.
f(f⁻¹(x)) = x and f⁻¹(f(x)) = x always.
Method: Write y = f(x), swap x and y, solve for y — that gives f⁻¹(x).

📐 Worked Example 8 — Inverse Functions

Find f⁻¹(x) for: (a) f(x) = 3x − 5 (b) f(x) = (2x+1)/(x−3)

(a) Write y = 3x − 5. Swap x and y: x = 3y − 5.
Solve for y: 3y = x + 5 → y = (x+5)/3
f⁻¹(x) = (x + 5)/3

(b) Write y = (2x+1)/(x−3). Swap: x = (2y+1)/(y−3).
Multiply both sides by (y−3): x(y−3) = 2y+1
Expand: xy − 3x = 2y + 1
Collect y terms: xy − 2y = 3x + 1
Factorise: y(x − 2) = 3x + 1
f⁻¹(x) = (3x + 1)/(x − 2)

Domain Restrictions: Some functions have restricted domains. For example:
• f(x) = √x requires x ≥ 0 (cannot square root a negative number)
• f(x) = 1/(x−2) requires x ≠ 2 (denominator cannot be zero)
Always state domain restrictions if asked.

4. Graphs of Functions

Understanding the shape and key features of graphs is essential for Cambridge examinations. You must recognise and sketch graphs for standard function types.

Function Type	Equation Form	Key Features
Linear	y = mx + c	Straight line. Gradient m, y-intercept c. Positive m → rising left to right.
Quadratic	y = ax² + bx + c	Parabola. a > 0 → U-shape (minimum). a < 0 → ∩-shape (maximum). Axis of symmetry at x = −b/2a.
Cubic	y = ax³ + ...	S-shaped curve. a > 0 → rises left to right overall. May have up to 2 turning points.
Reciprocal	y = k/x	Hyperbola with asymptotes along both axes. Two separate branches. xy = k (constant product).
Exponential	y = aˣ (a > 0)	Always positive. Passes through (0,1). a > 1 → growth. 0 < a < 1 → decay. Never crosses x-axis.
Square Root	y = √x	Starts at origin. Defined only for x ≥ 0. Increases but at a decreasing rate.

Finding Key Features of a Quadratic Graph

📐 Worked Example 9 — Sketching a Quadratic

For y = x² − 4x − 5, find: (a) y-intercept (b) x-intercepts (c) turning point (d) axis of symmetry. Then sketch.

(a) y-intercept: Set x = 0: y = 0 − 0 − 5 = −5. Point: (0, −5)

(b) x-intercepts: Set y = 0 and factorise:
x² − 4x − 5 = 0 → (x−5)(x+1) = 0
x = 5 or x = −1. Points: (5, 0) and (−1, 0)

(c) Axis of symmetry: x = −b/2a = 4/2 = 2. Substitute:
y = 4 − 8 − 5 = −9. Turning point (minimum): (2, −9)

Since a = 1 > 0, the parabola is U-shaped (minimum at base). Sketch using all four key points above.

Gradient of a Curve — Tangent Lines

Gradient of a curve at a point is found by drawing the tangent to the curve at that point and calculating the gradient of the tangent line (rise ÷ run). This is an estimation method used in O Level — exact calculus is in Additional Mathematics.

Solving Equations Graphically

Graphical Solutions: To solve an equation of the form f(x) = g(x) graphically, draw both curves y = f(x) and y = g(x). The x-coordinates of the intersection points are the solutions.

To solve f(x) = k graphically: draw y = f(x) and the horizontal line y = k. Solutions are the x-values where they intersect.

5. Sequences — Advanced Topics

Quadratic Sequences

A quadratic sequence has a constant second difference. The nth term contains n².

📐 Worked Example 10 — nth Term of a Quadratic Sequence

Find the nth term of: 3, 8, 15, 24, 35, ...

Find first differences: 5, 7, 9, 11 (not constant → not linear)
Find second differences: 2, 2, 2 (constant → quadratic)

Since second difference = 2, the nth term starts with (2/2)n² = n².
Subtract n² from each term: 3−1=2, 8−4=4, 15−9=6, 24−16=8 → remainders: 2, 4, 6, 8 → linear: 2n

nth term = n² + 2n
Check: n=1: 1+2=3 ✓ n=3: 9+6=15 ✓ n=5: 25+10=35 ✓

Geometric Sequences

Geometric Sequence Formulae

nth term: Tₙ = ar^(n−1)

Sum of n terms: Sₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1

where a = first term, r = common ratio

📐 Worked Example 11 — Geometric Sequence

A geometric sequence has first term 5 and common ratio 3. Find (a) the 6th term (b) the sum of the first 6 terms.

(a) T₆ = 5 × 3^(6−1) = 5 × 243 = 1215

(b) S₆ = 5(3⁶ − 1)/(3−1) = 5(729−1)/2 = 5 × 364 = 1820

6. Distance-Time and Speed-Time Graphs

These are practical applications of graphs that appear frequently in Cambridge examinations.

Distance-Time Graph

Gradient = speed at that point
Steep line → fast speed
Horizontal line → stationary (speed = 0)
Negative gradient → moving back toward start
Curved line → changing speed (acceleration)

Speed-Time Graph

Gradient = acceleration
Positive gradient → accelerating
Negative gradient → decelerating
Horizontal line → constant speed
Area under graph = distance travelled

📐 Worked Example 12 — Speed-Time Graph

A car accelerates from rest to 30 m/s in 10 seconds, travels at 30 m/s for 20 seconds, then decelerates to rest in 15 seconds. Find: (a) the acceleration in the first stage (b) the total distance travelled.

(a) Acceleration = gradient = change in speed ÷ time
a = (30 − 0)/10 = 3 m/s²

(b) Total distance = area under graph = area of trapezium:
The shape is a trapezium with parallel sides 20 and (10+20+15)=45 seconds, height 30 m/s.
Area = ½ × (20 + 45) × 30 = ½ × 65 × 30 = 975 m

📝 Exam Practice Questions

Q1 [3 marks] — Solve by factorisation: 6x² + x − 12 = 0

Working:
a×c = 6×(−12) = −72. Find two numbers that multiply to −72, add to 1: 9 and −8.
6x² + 9x − 8x − 12 = 0
3x(2x + 3) − 4(2x + 3) = 0
(3x − 4)(2x + 3) = 0
x = 4/3 or x = −3/2

Q2 [3 marks] — Solve 2x² + 3x − 4 = 0. Give your answers correct to 3 significant figures.

Working:
a=2, b=3, c=−4. Δ = 9 + 32 = 41
x = (−3 ± √41)/4
x = (−3 + 6.403)/4 = 3.403/4 = 0.851
x = (−3 − 6.403)/4 = −9.403/4 = −2.35

Exam Tip: Show the discriminant calculation separately. If Δ < 0 you must state "no real solutions." Never leave it blank.

Q3 [4 marks] — Write x² − 10x + 18 in the form (x + p)² + q. Hence find the minimum value of x² − 10x + 18 and the value of x at which it occurs.

Working:
x² − 10x + 18 = (x − 5)² − 25 + 18 = (x − 5)² − 7
Minimum value = −7 when x = 5

Exam Tip: (x−5)² is always ≥ 0. So the minimum of (x−5)²−7 occurs when (x−5)²=0, i.e. x=5, giving minimum value = −7.

Q4 [3 marks] — y is directly proportional to the square of x. When x = 4, y = 48. Find y when x = 5. Find x when y = 75.

Working:
y = kx². Substituting: 48 = k(16) → k = 3. So y = 3x²
When x=5: y = 3(25) = 75
When y=75: 75 = 3x² → x² = 25 → x = 5

Q5 [4 marks] — f(x) = 2x − 3 and g(x) = x² + 1. Find (a) f⁻¹(x) (b) gf(x) (c) x if fg(x) = 7

(a) y = 2x−3 → x = 2y−3 → y = (x+3)/2 → f⁻¹(x) = (x+3)/2

(b) gf(x) = g(2x−3) = (2x−3)²+1 = 4x²−12x+9+1 = 4x²−12x+10

(c) fg(x) = f(x²+1) = 2(x²+1)−3 = 2x²−1
2x²−1 = 7 → 2x² = 8 → x² = 4 → x = ±2

Q6 [3 marks] — The nth term of a sequence is n² + 3n − 1. Find the first three terms. Find the value of n for which the nth term equals 49.

First three terms:
n=1: 1+3−1 = 3
n=2: 4+6−1 = 9
n=3: 9+9−1 = 3, 9, 17

nth term = 49:
n²+3n−1 = 49 → n²+3n−50 = 0
(n+...): use formula: n = (−3 ± √(9+200))/2 = (−3 ± √209)/2
n = (−3 + 14.46)/2 ≈ 5.73 — not integer. Let us check n=5: 25+15−1=39, n=6: 36+18−1=53. No integer solution exists, so 49 is not a term in this sequence.

Q7 [4 marks] — A particle travels in a straight line. Its speed-time graph shows it starts at rest, accelerates uniformly to 20 m/s over 8 seconds, maintains 20 m/s for 12 seconds, then decelerates uniformly to rest in 5 seconds. Find (a) the deceleration in the final stage (b) the total distance travelled.

(a) Deceleration = change in speed ÷ time = (20−0)/5 = 4 m/s²

(b) Total distance = area under graph (trapezium):
Parallel sides: 12 (constant speed section) and 12+8+5=25 (total time base)
Area = ½ × (12 + 25) × 20 = ½ × 37 × 20 = 370 m

Exam Tip: Always draw a sketch of the speed-time graph first if one is not provided. Label all key points with time and speed values. Area = distance is the most important rule for speed-time graphs.

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