Chapter 3: Equations, Inequalities and Graphs
3.1 Modulus Equations
Recall the definition of modulus:
\[
|x| = \begin{cases}
x & \text{if } x \ge 0 \\
-x & \text{if } x < 0
\end{cases}
\]
Key property: For \(c \ge 0\),
\[
|x| = c \quad \iff \quad x = c \ \text{ or } \ x = -c
\]
Proof: Why \(|x| = c\) gives two solutions (for \(c > 0\))
If \(x \ge 0\): \(|x| = x = c \Rightarrow x = c\)
If \(x < 0\): \(|x| = -x = c \Rightarrow x = -c\)
Both solutions are valid provided \(c \ge 0\). If \(c = 0\), both cases give \(x = 0\) (one solution).
For equations of the form \(|ax + b| = |cx + d|\):
We can square both sides (since both sides are non-negative):
\[
(ax + b)^2 = (cx + d)^2
\]
This gives:
\[
[(a - c)x + (b - d)] \cdot [(a + c)x + (b + d)] = 0
\]
Proof: Squaring both sides is valid
For any real numbers \(p\) and \(q\): \(|p| = |q| \iff p^2 = q^2\).
Because \(|p| = |q|\) implies \(|p|^2 = |q|^2\), and \(|p|^2 = p^2\).
Conversely, if \(p^2 = q^2\), then \(|p| = |q|\) (since square roots are non-negative).
Worked Example 3.1 (Simple modulus equation – non-calculator)
Solve \(|2x - 5| = 9\).
Solution:
Case 1: \(2x - 5 = 9 \Rightarrow 2x = 14 \Rightarrow x = 7\)
Case 2: \(2x - 5 = -9 \Rightarrow 2x = -4 \Rightarrow x = -2\)
Answer: \(x = 7\) or \(x = -2\)
Worked Example 3.2 (Modulus equals modulus – non-calculator)
Solve \(|3x - 2| = |x + 4|\).
Solution (using squaring):
Square both sides:
\[
(3x - 2)^2 = (x + 4)^2
\]
\[
9x^2 - 12x + 4 = x^2 + 8x + 16
\]
\[
8x^2 - 20x - 12 = 0
\]
\[
2x^2 - 5x - 3 = 0
\]
\[
(2x + 1)(x - 3) = 0
\]
\[
x = -\frac{1}{2} \quad \text{or} \quad x = 3
\]
Answer: \(x = -\frac{1}{2}\) or \(x = 3\)
Mauritian context – distance problem
A bus travels from Port Louis to Mahébourg. The distance \(d\) (in km) from Port Louis satisfies \(|d - 25| = 8\). Find the two possible positions of the bus.
Solution:
\(d - 25 = 8 \Rightarrow d = 33\) km from Port Louis
\(d - 25 = -8 \Rightarrow d = 17\) km from Port Louis
Answer: 17 km or 33 km from Port Louis.
3.2 Modulus Inequalities
For \(|X| > c\) (with \(c \ge 0\)):
\[
|X| > c \quad \iff \quad X < -c \ \text{ or } \ X > c
\]
For \(|X| \le c\) (with \(c > 0\)):
\[
|X| \le c \quad \iff \quad -c \le X \le c
\]
Proof: Why \(|X| \le c\) is equivalent to \(-c \le X \le c\)
\(|X| \le c\) means the distance from \(X\) to 0 is at most \(c\).
On the number line, this is the interval from \(-c\) to \(c\) inclusive.
Algebraically:
If \(X \ge 0\): \(|X| = X \le c \Rightarrow 0 \le X \le c\)
If \(X < 0\): \(|X| = -X \le c \Rightarrow -c \le X < 0\)
Combining: \(-c \le X \le c\).
Worked Example 3.4 (Simple modulus inequality – non-calculator)
Solve \(|3x - 5| \le 7\).
Solution:
\[
-7 \le 3x - 5 \le 7
\]
Add 5 to all parts:
\[
-2 \le 3x \le 12
\]
Divide by 3:
\[
-\frac{2}{3} \le x \le 4
\]
Answer: \(\{x : -\frac{2}{3} \le x \le 4\}\)
Worked Example 3.5 (Modulus inequality with > – non-calculator)
Solve \(|2x + 1| > 9\).
Solution:
\[
2x + 1 < -9 \quad \text{or} \quad 2x + 1 > 9
\]
First inequality: \(2x < -10 \Rightarrow x < -5\)
Second inequality: \(2x > 8 \Rightarrow x > 4\)
Answer: \(\{x : x < -5\} \cup \{x : x > 4\}\)
3.3 Cubic Graphs in Factorised Form
A cubic polynomial in factorised form is given as:
\[
y = k(x - a)(x - b)(x - c)
\]
where \(k \neq 0\) and \(a, b, c\) are the roots (x-intercepts).
Key features of the graph:
| Feature |
How to find |
| Roots (x-intercepts) |
Set \(y = 0\) \(\rightarrow\) \(x = a, b, c\) |
| y-intercept |
Substitute \(x = 0\): \(y = k(-a)(-b)(-c) = -k \cdot abc\) |
| End behaviour (as \(x \to \pm\infty\)) |
Sign of \(k\) \(\times\) sign of \(x^3\) term |
| Shape between roots |
Test a point in each interval |
End behaviour rule:
- If \(k > 0\): as \(x \to \infty\), \(y \to \infty\); as \(x \to -\infty\), \(y \to -\infty\)
- If \(k < 0\): as \(x \to \infty\), \(y \to -\infty\); as \(x \to -\infty\), \(y \to \infty\)
Proof: Sign of \(k\) determines end behaviour
For large \(|x|\), \(y \approx k x^3\) (since \((x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + \cdots\)).
If \(k > 0\) and \(x \to \infty\): \(k x^3 \to \infty\)
If \(k > 0\) and \(x \to -\infty\): \(k x^3 \to -\infty\) (because \((-\infty)^3 = -\infty\))
If \(k < 0\), the signs reverse.
Worked Example 3.7 (Sketching a cubic – non-calculator)
Sketch \(y = (x - 1)(x - 3)(x - 5)\) for \(0 \le x \le 6\). Label the roots.
Solution:
Roots at \(x = 1, 3, 5\).
\(k = 1 > 0\) \(\rightarrow\) graph comes from bottom left, goes to top right.
Sign analysis (for \(k = 1\)):
| Interval |
\(x < 1\) |
\(1 < x < 3\) |
\(3 < x < 5\) |
\(x > 5\) |
| Sign of \(y\) |
\(-\) |
\(+\) |
\(-\) |
\(+\) |
y-intercept: \(x = 0\) \(\rightarrow\) \(y = (-1)(-3)(-5) = -15\)
3.4 Cubic Inequalities Solved Graphically
To solve a cubic inequality like \(k(x - a)(x - b)(x - c) \le d\):
- Sketch the graph of \(y = k(x - a)(x - b)(x - c)\)
- Draw the horizontal line \(y = d\)
- Read off the x-values where the cubic is below (or above) the line
For \(\ge 0\) or \(\le 0\) (i.e., \(d = 0\)):
Use the sign diagram.
Worked Example 3.9 (Simple cubic inequality – non-calculator)
Solve \((x - 1)(x - 3)(x - 5) \ge 0\).
Solution (using sign diagram from Example 3.7):
Signs: \(x < 1\): negative; \(1 < x < 3\): positive; \(3 < x < 5\): negative; \(x > 5\): positive.
We want \(\ge 0\) (positive or zero). Include the roots where \(y = 0\).
Answer: \(\{x : 1 \le x \le 3\} \cup \{x : x \ge 5\}\)
Chapter 3 Summary
| Concept |
Key points |
| \(|X| = c\) |
\(X = c\) or \(X = -c\) (for \(c \ge 0\)) |
| \(|X| = |Y|\) |
Square both sides \(\rightarrow\) \((X-Y)(X+Y) = 0\) |
| \(|X| > c\) |
\(X < -c\) or \(X > c\) (for \(c \ge 0\)) |
| \(|X| \le c\) |
\(-c \le X \le c\) (for \(c > 0\)) |
| Cubic graph \(y = k(x-a)(x-b)(x-c)\) |
Roots at \(a, b, c\); end behaviour from sign of \(k\) |
| Cubic inequality |
Use sign diagram (for \(\ge 0\) / \(\le 0\)) or graphical method |
Exercises – Chapter 3
Easy (Drill – Non-Calculator)
- Solve:
(a) \(|x - 5| = 3\)
(b) \(|2x + 1| = 9\)
(c) \(|4 - 3x| = 8\)
- Solve the inequalities:
(a) \(|x + 4| > 6\)
(b) \(|3x - 2| \le 5\)
- Sketch \(y = (x - 2)(x - 4)(x - 6)\) for \(0 \le x \le 8\). Label the roots.
Medium (Examination Style)
- Solve \(|3x - 2| = |x + 6|\).
- Mauritian context: A delivery truck's distance \(d\) (in km) from a warehouse in Rose Hill satisfies \(|d - 12| \le 4\). Find the range of possible distances.
- Solve \((x - 1)(x - 3)(x - 5) < 0\).
- Solve \(|2x - 5| \le |x + 3|\).
- Sketch \(y = -2(x + 1)(x - 2)(x - 4)\). Label all intercepts.
Hard (Challenge for A*)
- Solve \(|x^2 - 5x + 6| = 2\).
Hint: treat as two separate quadratic equations after removing modulus.
- The cubic \(y = k(x - 1)(x - 3)(x - 5)\) passes through the point \((0, -30)\).
(a) Find the value of \(k\).
(b) Sketch the graph.
(c) Solve \(k(x - 1)(x - 3)(x - 5) \le 0\).
- Mauritian context: A factory's daily production cost \(C\) (in thousands of rupees) is modelled by:
\[
C(x) = (x - 2)(x - 5)(x - 8) + 10
\]
where \(x\) is the number of hours of operation (in hours).
(a) Sketch the graph of \(y = (x - 2)(x - 5)(x - 8)\) for \(0 \le x \le 10\).
(b) By drawing a suitable horizontal line on your sketch, estimate the range of \(x\) for which the cost is at most 10 thousand rupees.
(c) For what range of \(x\) is \(C(x) \le 10\) exactly? (Give your answer in terms of the roots.)