Modulus Functions - Solving Equations (AQA A-Level Mathematics): Revision Notes

Example 1: Simple Modulus Equation

Solve the equation $\ |x| = 4$.

Solution:

The equation $\ |x| = 4$ means that $\ x$ can be either $4$ or -$4$ because the modulus function removes any negative sign:

$x = 4 \quad \text{or} \quad x = -4$

Thus, the solutions are $\ ( x = 4 )\ and \ ( x = -4 )$.

Example 2: Modulus Equation with a Linear Function

Solve $\ |2x - 3| = 5$.

Solution:

Here, the equation states that the modulus of $\ 2x - 3$ is equal to $5$. This can happen in two cases:

Case 1: $\ 2x - 3 \geq 0$

$2x - 3 = 5$ Add $3$ to both sides: $2x = 8$ Divide by $2$: $\ x = 4$

Case 2: $\ 2x - 3 < 0$

$-(2x - 3) = 5$ Distribute the negative sign: $-2x + 3 = 5$ Subtract 3 from both sides: $-2x = 2$ Divide by -$2$: $x = -1$

Thus, the solutions are $\ x = 4\ and \ x = -1$.

Example 3: Solving Modulus Equations with Two Modulus Terms

Solve $\ |x - 1| = |2x + 3|$.

Solution:

For this equation, we consider the following cases:

1. Case 1: Both expressions are non-negative $( x - 1 \geq 0 )\ and \ ( 2x + 3 \geq 0 )$ This implies $\ x \geq 1 .$

Solve: $x - 1 = 2x + 3$ Subtract $x$ from both sides: $-1 = x + 3$ Subtract $3$ from both sides: $x = -4$ But this contradicts $\ x \geq 1$, so no solution in this case.

2. Case 2: Both expressions are negative $( x - 1 < 0 ) and \ (2x + 3 < 0)$ This implies $\ x < 1 and \ x < -\frac{3}{2} \ (because \ 2x + 3 < 0 ).$

Solve: $-(x - 1) = -(2x + 3)$ Simplify: $x - 1 = 2x + 3$ Subtract $x$ from both sides: $-1 = x + 3$ Subtract $3$ from both sides: $x = -4$

$\ x = -4$ is less than $1$ and$\ -\frac{3}{2}$, so this is a valid solution.

3. Case 3: $\ ( x - 1 \geq 0 )\ and \ ( 2x + 3 < 0 )$ Here, $\ x \geq 1$ and $\ x < -\frac{3}{2}$ which is a contradiction, so no solution in this case.

4. Case 4: $\ x - 1 < 0$ and $\ 2x + 3 \geq 0$ Similarly, this case results in a contradiction, so no solution here either.

Thus, the only solution is $\ x = -4 .$

Exam Tip:

When dealing with modulus functions in an exam:

• Always split the equation into different cases based on where the expressions inside the modulus change sign.
• Consider the validity of each case based on the initial conditions you assume.
• Ensure you check each solution in the original equation to confirm its validity.