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Mastering E2.5 Equations: A Complete Learning Guide

From basic expressions to complex quadratic equations - everything you need to know

📚 Learning Objectives

1 Construct expressions, equations and formulas
2 Solve linear equations in one unknown
3 Solve fractional equations
4 Solve simultaneous linear equations
5 Solve linear + non-linear simultaneous equations
6 Solve quadratic equations three ways
7 Change the subject of formulas

1. Construct Expressions, Equations, and Formulas

Expression

Mathematical phrase without equals sign

Example: 3x + 2, 5y² - 2y + 1
Syllabus Example: "Product of two consecutive even numbers"
Let first number be n, next is n + 2
Expression: n(n + 2)

Equation

Statement that two expressions are equal

Example: 3x + 2 = 11

Formula

Equation showing relationship between variables

Example: Area of circle A = πr²

🧠 Practice Questions

Q1: Write an expression for "three times a number plus five"

Answer: Let the number be x. The expression is 3x + 5

Q2: Write an expression for "the sum of two consecutive odd numbers"

Answer: Let the first odd number be n. The next consecutive odd number is n + 2. The expression is n + (n + 2) = 2n + 2

2. Solve Linear Equations in One Unknown

Equations where highest power of variable is 1. Key Rule: Whatever you do to one side, do to the other.

Example 1 (Simple):

1. 3x + 4 = 10
2. 3x = 6 ← Subtract 4 from both sides
3. x = 2 ← Divide both sides by 3

Example 2 (With brackets):

1. 5 - 2x = 3(x + 7)
2. 5 - 2x = 3x + 21 ← Expand bracket
3. 5 = 5x + 21 ← Add 2x to both sides
4. -16 = 5x ← Subtract 21 from both sides
5. x = -16/5 ← Divide by 5

🧠 Practice Questions

Q1: Solve 4x - 7 = 13

Answer: x = 5
Steps:
4x - 7 = 13
4x = 20 ← Add 7 to both sides
x = 5 ← Divide by 4

Q2: Solve 2(3x - 1) = 5x + 4

Answer: x = 6
Steps:
2(3x - 1) = 5x + 4
6x - 2 = 5x + 4 ← Expand brackets
6x - 5x = 4 + 2 ← Rearrange terms
x = 6

3. Solve Fractional Equations

Unknown appears in denominator. Key Strategy: Multiply by LCD to eliminate fractions.

Example 1:

1. x/(2x + 1) = 4
2. x = 4(2x + 1) ← Multiply by (2x + 1)
3. x = 8x + 4 ← Expand
4. -7x = 4 ← Solve
5. x = -4/7

Example 2:

1. 2/(x + 2) + 3/(2x - 1) = 1
2. 2(2x - 1) + 3(x + 2) = 1(x + 2)(2x - 1) ← Multiply by LCD
3. 4x - 2 + 3x + 6 = 2x² + 3x - 2 ← Expand
4. 7x + 4 = 2x² + 3x - 2 ← Simplify
5. 0 = 2x² - 4x - 6 ← Rearrange
6. 0 = x² - 2x - 3 ← Divide by 2
7. 0 = (x - 3)(x + 1) ← Factorise
8. x = 3 or x = -1
✓ Always check: Answers don't make original denominators zero!

🧠 Practice Questions

Q1: Solve 3/(x - 2) = 5

Answer: x = 13/5
Steps:
3/(x - 2) = 5
3 = 5(x - 2) ← Multiply both sides by (x - 2)
3 = 5x - 10 ← Expand
13 = 5x ← Add 10 to both sides
x = 13/5

Q2: Solve (x + 1)/(x - 3) = 2

Answer: x = 7
Steps:
(x + 1)/(x - 3) = 2
x + 1 = 2(x - 3) ← Multiply both sides by (x - 3)
x + 1 = 2x - 6 ← Expand
1 + 6 = 2x - x ← Rearrange
7 = x

4. Solve Simultaneous Linear Equations

Find values that satisfy both equations. Methods: Substitution or Elimination.

Example (Elimination Method):

Solve: 1) 2x + 3y = 7 and 2) 4x - y = 3

1. Multiply equation (2) by 3: 12x - 3y = 9
2. Add equations: (2x + 3y) + (12x - 3y) = 7 + 9
3. 14x = 16 → x = 16/14 = 8/7
4. Substitute into (2): 4(8/7) - y = 3
5. 32/7 - y = 3 → -y = -11/7
6. y = 11/7
Solution: x = 8/7, y = 11/7

🧠 Practice Questions

Q1: Solve: 1) 3x + 2y = 8 and 2) 2x - y = 3

Answer: x = 2, y = 1
Steps (Elimination):
Multiply equation (2) by 2: 4x - 2y = 6
Add to equation (1): 7x = 14 → x = 2
Substitute into (2): 4 - y = 3 → y = 1

Q2: Solve: 1) 5x - 3y = 1 and 2) 2x + y = 7

Answer: x = 2, y = 3
Steps (Substitution):
From (2): y = 7 - 2x
Substitute into (1): 5x - 3(7 - 2x) = 1
5x - 21 + 6x = 1 → 11x = 22 → x = 2
y = 7 - 4 = 3

5. Solve Simultaneous Equations (Linear + Non-linear)

One linear, one quadratic equation. Use substitution method.

Example:

Solve: 1) y = x + 1 (Linear) and 2) y = x² + 2x - 5 (Quadratic)

1. x + 1 = x² + 2x - 5 ← Set equal
2. 0 = x² + x - 6 ← Rearrange
3. 0 = (x + 3)(x - 2) ← Factorise
4. x = -3 or x = 2
5. Find y-values using linear equation:
When x = -3, y = -2
When x = 2, y = 3
Solutions: (-3, -2) and (2, 3)

🧠 Practice Questions

Q1: Solve: 1) y = 2x + 1 and 2) y = x² + 3

Answer: (2, 5) and (-1, -1)
Steps:
2x + 1 = x² + 3
0 = x² - 2x + 2
0 = (x - 2)(x + 1)
x = 2 or x = -1
When x = 2, y = 5
When x = -1, y = -1

6. Solve Quadratic Equations - All Methods

Equations of form ax² + bx + c = 0. Four main methods:

A) Solving by Factorisation

Find two numbers that multiply to a×c and add to b.

Example 1: x² - 3x - 10 = 0

1. Numbers that multiply to -10 and add to -3: -5 and +2
2. (x - 5)(x + 2) = 0
3. x - 5 = 0 or x + 2 = 0
Solutions: x = 5 or x = -2

Example 2: 2x² + 7x + 3 = 0

1. Multiply a×c: 2 × 3 = 6
2. Numbers that multiply to 6 and add to 7: 6 and 1
3. Split middle term: 2x² + 6x + x + 3 = 0
4. Factorise: 2x(x + 3) + 1(x + 3) = 0
5. (2x + 1)(x + 3) = 0
6. 2x + 1 = 0 or x + 3 = 0
Solutions: x = -1/2 or x = -3

B) Solving by Completing the Square

Write in the form (x + p)² = q and solve.

Example 1: x² + 6x + 5 = 0

1. Move constant: x² + 6x = -5
2. Complete square: (x + 3)² = x² + 6x + 9
3. Add 9 to both sides: x² + 6x + 9 = -5 + 9
4. (x + 3)² = 4
5. Square root: x + 3 = ±2
6. Solve: x = -3 + 2 = -1 or x = -3 - 2 = -5
Solutions: x = -1 or x = -5

C) Writing Quadratic in Completed Square Form: (x + a)² + b

Step-by-step process for writing x² + bx + c as (x + p)² + q:

1

Start with the quadratic

x² + bx + c (coefficient of x² must be 1)

2

Take half of b

Calculate: b/2

3

Square this value

Calculate: (b/2)²

4

Add and subtract the square

x² + bx + (b/2)² - (b/2)² + c

5

Group perfect square

(x + b/2)² - (b/2)² + c

6

Simplify constants

(x + b/2)² + [c - (b/2)²]

Example 1: Write x² + 8x + 10 in completed square form

1. x² + 8x + 10 ← Original
2. Half of 8 is 4
3. Square of 4 is 16
4. x² + 8x + 16 - 16 + 10 ← Add/subtract 16
5. (x + 4)² - 16 + 10 ← Group perfect square
6. (x + 4)² - 6 ← Simplify
Completed square form: (x + 4)² - 6

Example 2: Write x² - 6x + 2 in completed square form

1. x² - 6x + 2
2. Half of -6 is -3
3. Square of -3 is 9
4. x² - 6x + 9 - 9 + 2
5. (x - 3)² - 9 + 2
6. (x - 3)² - 7
Completed square form: (x - 3)² - 7

D) Solving Using Quadratic Formula

For ax² + bx + c = 0, use: x = [-b ± √(b² - 4ac)] / (2a)

Example 1: 2x² - 4x - 6 = 0

1. Identify coefficients: a = 2, b = -4, c = -6
2. Substitute into formula: x = [4 ± √(16 + 48)] / 4
3. x = [4 ± √64] / 4
4. x = [4 ± 8] / 4
5. x = 12/4 = 3 or x = -4/4 = -1
Solutions: x = 3 or x = -1

Example 2: x² - 6x + 9 = 0 (Perfect Square)

1. Identify coefficients: a = 1, b = -6, c = 9
2. Substitute: x = [6 ± √(36 - 36)] / 2
3. x = [6 ± √0] / 2
4. x = 6/2 = 3
Solution: x = 3 (repeated root)

🧠 Practice Questions - Quadratic Equations

Q1 (Factorisation): Solve x² + 5x + 6 = 0

Answer: x = -2 or x = -3
Steps:
x² + 5x + 6 = 0
(x + 2)(x + 3) = 0
x + 2 = 0 or x + 3 = 0
x = -2 or x = -3

Q2 (Completing Square): Write x² + 10x + 15 in form (x + a)² + b

Answer: (x + 5)² - 10
Steps:
x² + 10x + 15
Half of 10 is 5, square is 25
x² + 10x + 25 - 25 + 15
(x + 5)² - 10

Q3 (Quadratic Formula): Solve 2x² - 5x - 3 = 0

Answer: x = 3 or x = -0.5
Steps:
a = 2, b = -5, c = -3
x = [5 ± √(25 + 24)] / 4
x = [5 ± √49] / 4
x = [5 ± 7] / 4
x = 12/4 = 3 or x = -2/4 = -0.5

7. Change the Subject of Formulas

Rearrange to make a different variable the subject.

Example 1 (Simple):

Make r subject of A = πr²

1. A/π = r² ← Divide by π
2. r = √(A/π) ← Square root

Example 2 (Subject appears twice):

Make x subject of y = (x + 2)/(x - 1)

1. y(x - 1) = x + 2 ← Multiply by (x - 1)
2. yx - y = x + 2 ← Expand
3. yx - x = 2 + y ← Rearrange
4. x(y - 1) = y + 2 ← Factorise
5. x = (y + 2)/(y - 1) ← Divide

Example 3 (With root):

Make t subject of s = 3 + 5√t

1. s - 3 = 5√t ← Subtract 3
2. (s - 3)/5 = √t ← Divide by 5
3. t = [(s - 3)/5]² ← Square both sides

🧠 Practice Questions

Q1: Make h the subject of V = πr²h

Answer: h = V/(πr²)
Steps:
V = πr²h
h = V/(πr²) ← Divide both sides by πr²

Q2: Make a the subject of p = 2(a + b)

Answer: a = p/2 - b
Steps:
p = 2(a + b)
p/2 = a + b ← Divide both sides by 2
a = p/2 - b ← Subtract b from both sides

💡 Study Tips

  • Practice each method until you're comfortable with it
  • Always check your solutions by substituting back into the original equation
  • For fractional equations, verify denominators don't become zero
  • Use the quadratic formula when factorisation isn't obvious
  • Show all your working - you get marks for method even if the final answer is wrong!

Solving Quadratic Equations Graphically

Learn how to find solutions by plotting and analyzing quadratic graphs

Solving Quadratic Equations Graphically

Find where the parabola y = ax² + bx + c crosses the x-axis (y = 0).

Step-by-Step Graphical Method

1

Create a table of values

Choose x-values and calculate corresponding y-values using the equation

2

Plot the points

Draw the parabola through the plotted points on a coordinate grid

3

Identify x-intercepts

Find where the curve crosses the x-axis (where y = 0)

4

Read solutions

The x-coordinates of the intercepts are the solutions to the equation

Example 1: Solve x² - x - 6 = 0 graphically

Step 1: Create Table of Values
x -3 -2 -1 0 1 2 3 4
y = x² - x - 6 6 0 -4 -6 -6 -4 0 6
Step 2: Plot these points on a graph and draw a smooth curve through them
Step 3: Identify where the parabola crosses the x-axis
We can see the curve crosses at x = -2 and x = 3
Step 4: Verify by substitution:
(-2)² - (-2) - 6 = 4 + 2 - 6 = 0 ✓
(3)² - 3 - 6 = 9 - 3 - 6 = 0 ✓
Graphical Solutions: x = -2 and x = 3

Example 2: Solve x² - 4x + 4 = 0 graphically

Step 1: Create Table of Values
x 0 1 2 3 4
y = x² - 4x + 4 4 1 0 1 4
Step 2: Plot points - the parabola touches the x-axis at exactly one point
Step 3: Single x-intercept at x = 2
Step 4: This is a repeated root (perfect square)
Graphical Solution: x = 2 (repeated root)

Example 3: Solve x² + 2x + 5 = 0 graphically

Step 1: Create Table of Values
x -3 -2 -1 0 1
y = x² + 2x + 5 8 5 4 5 8
Step 2: Plot points - the parabola never crosses the x-axis
Step 3: No x-intercepts found
No real solutions (the equation has complex roots)
Graphical Method Tips:
  • Choose x-values that give a clear picture of the curve shape
  • Include points around where you expect the roots to be
  • The solutions are where y = 0 (x-axis crossings)
  • If parabola doesn't cross x-axis: no real solutions
  • If it touches x-axis at one point: repeated root
  • If it crosses at two points: two distinct real roots

🧠 Practice Questions

Q1: Solve x² - 2x - 3 = 0 graphically by creating a table of values for x = -2, -1, 0, 1, 2, 3, 4

Answer: x = -1 and x = 3
Table of Values:
x-2-101234
y50-3-4-305
Solutions: x-intercepts at x = -1 and x = 3

Q2: Solve x² + 4x + 4 = 0 graphically using x = -4, -3, -2, -1, 0

Answer: x = -2 (repeated root)
Table of Values:
x-4-3-2-10
y41014
Solution: Parabola touches x-axis at x = -2 only

Q3: Solve x² + 1 = 0 graphically using x = -2, -1, 0, 1, 2

Answer: No real solutions
Table of Values:
x-2-1012
y52125
Solution: Parabola never crosses x-axis - no real roots

💡 Graphical Method Advantages

  • Visual representation helps understand the nature of roots
  • Can estimate solutions even when exact values are difficult to find
  • Shows the relationship between the equation and its graph
  • Useful for checking solutions found by other methods
  • Helps identify when equations have no real solutions
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Introduction to Quadratic Expressions

A quadratic expression has the general form: ax² + bx + c

Where:

  • a = coefficient of x² (a ≠ 0)
  • b = coefficient of x
  • c = constant term

1. Factorisation when a = 1 (Product-Sum Method)

When a = 1, we have: x² + bx + c

Rule: Find two numbers m and n such that:

  1. m × n = c (product equals constant)
  2. m + n = b (sum equals coefficient of x)

Factorised form: x² + bx + c = (x + m)(x + n)

Example 1: Factorise x² + 7x + 12

We need m × n = 12 and m + n = 7

Possible pairs: (1,12), (2,6), (3,4)

3 + 4 = 7 ✓

Solution: x² + 7x + 12 = (x + 3)(x + 4)

Example 2: Factorise x² - 5x + 6

We need m × n = 6 and m + n = -5

Both must be negative since product is positive, sum is negative

(-2) × (-3) = 6 and (-2) + (-3) = -5 ✓

Solution: x² - 5x + 6 = (x - 2)(x - 3)

2. Factorisation when a ≠ 1 (PCM/AC Method)

For ax² + bx + c where a ≠ 1, follow these steps:

  1. Multiply a × c (product = P)
  2. Find two numbers m and n such that:
    • m × n = P
    • m + n = b
  3. Rewrite middle term: ax² + bx + c = ax² + mx + nx + c
  4. Factor by grouping
  5. Simplify

Example 3: Factorise 2x² + 7x + 3

Step 1: a = 2, c = 3 → P = 2 × 3 = 6

Step 2: Need m × n = 6 and m + n = 7 → m = 6, n = 1

Step 3: Rewrite: 2x² + 6x + 1x + 3

Step 4: Group: (2x² + 6x) + (1x + 3)

Step 5: Factor groups: 2x(x + 3) + 1(x + 3)

Step 6: Common factor: (x + 3)(2x + 1)

Check: (x + 3)(2x + 1) = 2x² + x + 6x + 3 = 2x² + 7x + 3 ✓

3. Completing the Square Method

When a = 1

Formula: x² + bx + c = (x + b/2)² - (b/2)² + c

Example 4: Complete the square for x² + 6x + 5

b = 6, so b/2 = 3, (b/2)² = 9

x² + 6x + 5 = (x + 3)² - 9 + 5

= (x + 3)² - 4

This can be factorised as difference of squares:

= (x + 3 - 2)(x + 3 + 2) = (x + 1)(x + 5)

When a ≠ 1

Steps:

  1. Factor out a: ax² + bx + c = a(x² + (b/a)x) + c
  2. Complete the square inside bracket
  3. Simplify

Example 5: Complete the square for 2x² + 8x + 6

Step 1: Factor 2: 2(x² + 4x) + 6

Step 2: Inside bracket: x² + 4x, b = 4, so b/2 = 2, (b/2)² = 4

Complete square: 2[(x + 2)² - 4] + 6

Step 3: Expand: 2(x + 2)² - 8 + 6 = 2(x + 2)² - 2

Factor further: = 2[(x + 2)² - 1] = 2(x + 2 - 1)(x + 2 + 1)

= 2(x + 1)(x + 3)

4. Practice Problems

Try factorising these expressions:

  1. x² + 9x + 20
  2. x² - 3x - 10
  3. 2x² + 11x + 5
  4. 3x² - 14x + 8
  5. 4x² - 12x + 9 (Perfect Square)
  6. x² - 16 (Difference of Squares)
Click to show answers
  1. (x + 4)(x + 5)
  2. (x - 5)(x + 2)
  3. (2x + 1)(x + 5)
  4. (3x - 2)(x - 4)
  5. (2x - 3)²
  6. (x - 4)(x + 4)

5. Tips and Common Errors

Common Errors to Avoid:

  • Forgetting to include both factors when a ≠ 1
  • Misidentifying signs of m and n
  • Not checking your answer by expanding
  • Incorrect grouping in PCM method

Check Your Factorisation:

Always expand (mx + p)(nx + q) to verify:

  • m × n = a
  • m × q + n × p = b
  • p × q = c

Product-Sum Method Practice Questions

Part A: When a = 1 (Simple Case)

Find two numbers m and n such that:

  • m × n = c
  • m + n = b

Then write: x² + bx + c = (x + m)(x + n)

Question 1: x² + 8x + 15

Find m and n where:

m × n = (15)

m + n = (8)

m = , n =

Question 2: x² + 11x + 24

Find m and n where:

m × n = (24)

m + n = (11)

m = , n =

Question 3: x² - 9x + 20

Find m and n where:

m × n = (20)

m + n = (-9)

m = , n =

Question 4: x² + 2x - 15

Find m and n where:

m × n = (-15)

m + n = (2)

m = , n =

Question 5: x² - 7x - 18

Find m and n where:

m × n = (-18)

m + n = (-7)

m = , n =

Part B: When a ≠ 1 (PCM Method)

First: Multiply a × c = P

Then find two numbers m and n such that:

  • m × n = P (product of a and c)
  • m + n = b (sum equals b)

Question 6: 2x² + 9x + 4

Step 1: a × c = (P)

Step 2: Find m and n where:

m × n = (P)

m + n = (9)

m = , n =

Question 7: 3x² + 14x + 8

Step 1: a × c = (P)

Step 2: Find m and n where:

m × n = (P)

m + n = (14)

m = , n =

Question 8: 4x² - 4x - 3

Step 1: a × c = (P)

Step 2: Find m and n where:

m × n = (P)

m + n = (-4)

m = , n =

Question 9: 5x² - 13x + 6

Step 1: a × c = (P)

Step 2: Find m and n where:

m × n = (P)

m + n = (-13)

m = , n =

Question 10: 6x² + 19x + 10

Step 1: a × c = (P)

Step 2: Find m and n where:

m × n = (P)

m + n = (19)

m = , n =

Answers Summary

Part A Answers (a = 1):

  1. x² + 8x + 15 = (x + 3)(x + 5) → m=3, n=5
  2. x² + 11x + 24 = (x + 3)(x + 8) → m=3, n=8
  3. x² - 9x + 20 = (x - 4)(x - 5) → m=-4, n=-5
  4. x² + 2x - 15 = (x + 5)(x - 3) → m=5, n=-3
  5. x² - 7x - 18 = (x - 9)(x + 2) → m=-9, n=2

Part B Answers (a ≠ 1):

  1. 2x² + 9x + 4 → P=8, m=8, n=1 → (2x+1)(x+4)
  2. 3x² + 14x + 8 → P=24, m=12, n=2 → (3x+2)(x+4)
  3. 4x² - 4x - 3 → P=-12, m=-6, n=2 → (2x-3)(2x+1)
  4. 5x² - 13x + 6 → P=30, m=-10, n=-3 → (5x-3)(x-2)
  5. 6x² + 19x + 10 → P=60, m=15, n=4 → (3x+2)(2x+5)
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