Algebra (AS Level - Paper 2)

1. Absolute Value (Modulus)

What is Absolute Value?

The absolute value (also called modulus) of a number is its distance from zero on the number line, ignoring whether it's positive or negative. We write it using vertical bars: |x|.

Key idea: Absolute value is always positive or zero, never negative.

Examples:

• |5| = 5 (because 5 is 5 units from zero)
• |−5| = 5 (because −5 is also 5 units from zero)
• |0| = 0

Formal definition:

• If x ≥ 0, then |x| = x
• If x < 0, then |x| = −x

This means that for negative numbers, we flip the sign to make them positive.

Graphs of y = |ax + b|

When we sketch y = |ax + b|, we're drawing the graph of a straight line but making any negative parts positive (reflecting them above the x-axis).

Step-by-step method:

1. First, sketch y = ax + b (the line without the modulus)

   • Find where it crosses the x-axis by solving ax + b = 0
   • Find where it crosses the y-axis (when x = 0)
   • Draw the straight line

2. Apply the modulus

   • Keep the part of the line that's already above the x-axis (where y ≥ 0)
   • Reflect any part below the x-axis upward (flip it to make y positive)

Example: Sketch y = |2x − 4|

Step 1: Sketch y = 2x − 4

• When x = 0: y = −4 (y-intercept at (0, −4))
• When y = 0: 2x − 4 = 0, so x = 2 (x-intercept at (2, 0))
• This is a line with gradient 2

Step 2: Apply modulus

• For x ≥ 2: the line is already positive, so y = 2x − 4
• For x < 2: the line is negative, so we reflect it: y = −(2x − 4) = −2x + 4

The graph is V-shaped with the point (vertex) at (2, 0). The left branch has gradient −2, and the right branch has gradient 2.

Key features of y = |ax + b| graphs:

• They are always V-shaped
• The vertex (lowest point) is on the x-axis at the point where ax + b = 0
• The graph never goes below the x-axis
• The two branches have gradients that are equal in size but opposite in sign

Solving Equations Involving Absolute Value

Important relation: |a| = |b| ⟺ a² = b²

This means: if two numbers have the same absolute value, then their squares are equal (and vice versa).

Why? Because |a| = |b| means both numbers are the same distance from zero. This happens when:

• a = b (both numbers are identical), OR
• a = −b (both numbers are opposites)

Method 1: Using a² = b²

If |3x − 2| = |2x + 7|, then:

• Square both sides: (3x − 2)² = (2x + 7)²
• Expand and solve

Method 2: Consider two cases

Since |A| = |B| means A = B or A = −B:

Example: Solve |3x − 2| = |2x + 7|

Case 1: 3x − 2 = 2x + 7

• 3x − 2x = 7 + 2
• x = 9

Case 2: 3x − 2 = −(2x + 7)

• 3x − 2 = −2x − 7
• 3x + 2x = −7 + 2
• 5x = −5
• x = −1

Always check your answers by substituting back into the original equation.

Solving Inequalities Involving Absolute Value

Important relation: |x − a| < b ⟺ a − b < x < a + b (where b > 0)

This reads: "The distance from x to a is less than b" means "x is between a − b and a + b."

Similarly:

• |x − a| ≤ b ⟺ a − b ≤ x ≤ a + b
• |x − a| > b ⟺ x < a − b or x > a + b (outside the range)
• |x − a| ≥ b ⟺ x ≤ a − b or x ≥ a + b

Example 1: Solve |x + 1| < 5

This is the same as |x − (−1)| < 5, so a = −1 and b = 5.

Using the formula: −1 − 5 < x < −1 + 5 Answer: −6 < x < 4

Example 2: Solve 2x + 5 < |x + 1|

This type requires considering cases based on whether (x + 1) is positive or negative.

Case 1: x + 1 ≥ 0 (i.e., x ≥ −1) Then |x + 1| = x + 1, so:

• 2x + 5 < x + 1
• x < −4

But this contradicts x ≥ −1, so there's no solution in this region.

Case 2: x + 1 < 0 (i.e., x < −1) Then |x + 1| = −(x + 1) = −x − 1, so:

• 2x + 5 < −x − 1
• 3x < −6
• x < −2

This is consistent with x < −1, so the solution is x < −2.