Algebra (A2 Level)

1. Modulus Functions

What is Modulus?

The modulus of a number (written as |x|) means its distance from zero, ignoring whether it's positive or negative. Distance is always positive or zero, so modulus values are never negative.

Examples:

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

Key definition:

This means: if the number inside is positive or zero, leave it alone. If it's negative, make it positive by multiplying by -1.

Sketching y = |ax + b|

To sketch the graph of y = |ax + b|:

Step 1: Sketch the original line y = ax + b (a straight line)

Step 2: Find where the line crosses the x-axis (where y = 0). This happens when ax + b = 0, so x = -b/a

Step 3: Any part of the line that is below the x-axis (negative y-values) gets "reflected" upwards to make it positive

The result is a V-shaped graph where the point of the V is at x = -b/a.

Example: Sketch y = |2x - 4|

• Original line: y = 2x - 4
• Crosses x-axis when 2x - 4 = 0, so x = 2
• For x < 2, the original line is negative, so we reflect it upwards
• The graph is a V-shape with the point at (2, 0)

Useful Relations

Relation 1: |a| = |b| ⟺ a² = b²

This means: two numbers have the same modulus if and only if their squares are equal.

Why? If |a| = |b|, both numbers are the same distance from zero. When you square them, you get the same result.

Example: |3| = |-3| because 3² = (-3)² = 9

Relation 2: |x - a| < b ⟺ a - b < x < a + b (where b > 0)

This means: if the distance from x to a is less than b, then x must be within b units of a on both sides.

Example: |x - 3| < 5 means -2 < x < 8 (x is within 5 units of 3, so between 3-5=−2 and 3+5=8)

Solving Modulus Equations

Method for |expression 1| = |expression 2|:

Use the fact that |a| = |b| means a² = b².

Example: Solve |3x - 2| = |2x + 7|

Square both sides: (3x - 2)² = (2x + 7)²

9x² - 12x + 4 = 4x² + 28x + 49

5x² - 40x - 45 = 0

x² - 8x - 9 = 0

(x - 9)(x + 1) = 0

So x = 9 or x = -1

Always check your answers in the original equation to make sure they work.

Solving Modulus Inequalities

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

We need to consider two cases based on whether (x + 1) is positive or negative.

Case 1: x + 1 ≥ 0 (so x ≥ -1)

Here |x + 1| = x + 1

So: 2x + 5 < x + 1

x < -4

But this contradicts x ≥ -1, so there are no solutions in this case.

Case 2: x + 1 < 0 (so x < -1)

Here |x + 1| = -(x + 1) = -x - 1

So: 2x + 5 < -x - 1

3x < -6

x < -2

Combined with x < -1, the solution is x < -2.