Welcome to Chapter 4: Equations, Inequalities, and Graphs!

Hello future Additional Mathematics expert! This chapter is incredibly important because it brings together many skills you've learned—like solving quadratics—and applies them to new, complex areas, especially dealing with the Modulus Function and solving tricky equations using Substitution.

Don't worry if absolute values seem intimidating. We will break them down using simple rules and visual aids. Mastering this topic will significantly boost your problem-solving confidence!


Section 1: Solving Equations Involving Modulus (Absolute Value)

What is the Modulus Function?

The modulus (or absolute value) of a number, denoted by \(|x|\), is simply its non-negative value. Think of it as the distance of a number from zero on the number line. Since distance is always positive, the modulus output is never negative.

  • \(|5| = 5\)
  • \(|-5| = 5\)
  • Analogy: Modulus is like a security guard that strips away the negative sign.

1.1 Basic Modulus Equations: \(|ax + b| = c\)

If \(| \text{something} | = c\), it means the value inside the modulus could be \(c\) or \(-c\).

Step-by-Step Method:

  1. Remove the modulus bars.
  2. Set the expression equal to the positive value AND the negative value of the right side.
  3. Solve both resulting linear equations.

Example: Solve \(|2x - 1| = 5\).

  • Case 1: \(2x - 1 = 5 \implies 2x = 6 \implies x = 3\)
  • Case 2: \(2x - 1 = -5 \implies 2x = -4 \implies x = -2\)

The solution set is \(x = 3\) or \(x = -2\).

1.2 Complex Modulus Equations: Linear Functions

When solving equations where variables appear on both sides, or where both sides have modulus signs, you have two main algebraic methods:

Method A: Squaring Both Sides (Good for all forms)

If \(|A| = |B|\), then \(A^2 = B^2\). This is useful because squaring removes the modulus bars completely.

Example: Solve \(|3x - 2| = |x + 4|\).

  • \((3x - 2)^2 = (x + 4)^2\)
  • \(9x^2 - 12x + 4 = x^2 + 8x + 16\)
  • \(8x^2 - 20x - 12 = 0\)
  • Divide by 4: \(2x^2 - 5x - 3 = 0\)
  • \((2x + 1)(x - 3) = 0\)
  • Solutions: \(x = 3\) or \(x = -0.5\)

Crucial Note: If only one side has the modulus (e.g., \(|ax + b| = cx + d\)), squaring can introduce extraneous roots (solutions that don't actually work in the original equation). You must check your final answers by substituting them back into the original equation.

Method B: The Definition Method (Splitting Cases)

We split the equation based on when the expression inside the modulus is positive or negative.

Example: Solve \(|x| = x - 2\).

  1. Case 1: \(x \geq 0\) (Modulus has no effect)
    \(x = x - 2 \implies 0 = -2\). This is impossible. No solution in this case.
  2. Case 2: \(x < 0\) (Modulus changes sign)
    \(-x = x - 2 \implies 2 = 2x \implies x = 1\).
    BUT we assumed \(x < 0\). Since \(x = 1\) violates this condition, it is rejected.

Conclusion: There are no solutions for \(|x| = x - 2\).

1.3 Quadratic Equations Inside Modulus: \(|ax^2 + bx + c| = d\)

This follows the basic split method (Section 1.1):

Example: Solve \(|x^2 - 3x| = 2\).

  1. Set \(x^2 - 3x = 2\)
    \(x^2 - 3x - 2 = 0\). Solve using the quadratic formula.
  2. Set \(x^2 - 3x = -2\)
    \(x^2 - 3x + 2 = 0 \implies (x-1)(x-2) = 0\). Solutions: \(x=1\) and \(x=2\).
Quick Review: Modulus Equations

When solving equations involving modulus: Always split into two cases (or square both sides, remembering to check for extraneous roots if the equation is not \(|A| = |B|\)).


Section 2: Solving Modulus Inequalities

Solving inequalities involves the same core idea (the value inside is close to the positive or negative boundary), but the direction of the inequality sign matters greatly.

2.1 Linear Modulus Inequalities: The Rules

We use two very helpful mnemonics here. Let \(a\) be a positive constant.

Rule 1: Less ThAND (\(<\) or \(\leq\))

If \(|x| < a\), the solution lies between the negative and positive boundaries.
$$|x| < a \implies -a < x < a$$

Example: Solve \(|2x + 3| \leq 7\).

  • \(-7 \leq 2x + 3 \leq 7\)
  • Subtract 3 from all parts: \(-10 \leq 2x \leq 4\)
  • Divide by 2: \(-5 \leq x \leq 2\)
Rule 2: GreatOR Than (\(>\) or \(\geq\))

If \(|x| > a\), the solution is outside the boundaries, separated by "OR".
$$|x| > a \implies x < -a \quad \text{OR} \quad x > a$$

Example: Solve \(|4x - 1| > 5\).

  • Case 1 (Positive): \(4x - 1 > 5 \implies 4x > 6 \implies x > 1.5\)
  • Case 2 (Negative): \(4x - 1 < -5 \implies 4x < -4 \implies x < -1\)

Solution: \(x < -1\) OR \(x > 1.5\).

2.2 Inequalities with Variables on Both Sides

The safest algebraic method here is Squaring Both Sides, provided we ensure the expression is simplified to zero on one side before solving the resulting inequality.

Example: Solve \(|x - 5| < 2x\).

  1. Square both sides: \((x - 5)^2 < (2x)^2\)
  2. \(x^2 - 10x + 25 < 4x^2\)
  3. \(0 < 3x^2 + 10x - 25\)
  4. Factorise the quadratic: \(3x^2 + 10x - 25 = (3x - 5)(x + 5)\)

The quadratic inequality \((3x - 5)(x + 5) > 0\) is satisfied when \(x > 5/3\) or \(x < -5\).

Wait! Don't forget the restriction! Since the left side \(|x-5|\) must be \(\geq 0\), the right side \(2x\) must also be positive.

  • Restriction: \(2x > 0 \implies x > 0\).

We combine the quadratic solution with the restriction \(x > 0\):

  • \(x < -5\) (Rejected because it violates \(x > 0\)).
  • \(x > 5/3\) (Accepted).

Final Solution: \(x > 5/3\).

2.3 Solving Modulus Inequalities Graphically

You can solve any modulus inequality by sketching the graphs of the functions on either side and finding where one graph lies above or below the other.

Example: Solve \(|x| < x - 2\) (referring back to the previous example).

  • Sketch \(y = |x|\) (a V-shape at the origin).
  • Sketch \(y = x - 2\) (a line with y-intercept -2).

You will see that the graph \(y = |x|\) never lies below the line \(y = x - 2\). Therefore, there is no solution. (This confirms our algebraic result from Section 1.2!)

Key Takeaway: Modulus Inequalities

Use the "Less ThAND" (\( -a < x < a \)) and "GreatOR Than" (\( x < -a \) OR \( x > a \)) rules. If variables are on both sides, squaring is often fastest, but always remember to consider any non-negativity restrictions imposed by the modulus.


Section 3: Solving Related Equations using Substitution (4.3)

Sometimes, an equation looks complicated, but if you look closely, you can spot a pattern that resembles a standard quadratic equation \(ay^2 + by + c = 0\). This is where substitution saves the day!

The Substitution Strategy

The goal is to simplify a non-standard equation by letting a complex term equal a new variable, \(y\).

  1. Identify the core term: Look for an expression that appears twice, where one appearance is squared relative to the other (e.g., \(y\) and \(y^2\)).
  2. Substitute: Let \(y\) equal the core term.
  3. Solve the new quadratic: Solve for \(y\).
  4. Back-substitute: Replace \(y\) with the original expression and solve for \(x\).

3.1 Exponential Examples

Equations involving exponents are common substitution candidates.

Example: Solve \(3e^{2x} - 12 = 5e^{-x}\) (This is adapted from the syllabus example \(3e^{x} = 12 - 5e^{-x}\)).

  1. Multiply everything by \(e^x\) to remove the negative power:
    \(3e^{2x} \cdot e^x - 12e^x = 5e^{-x} \cdot e^x\)
    \(3e^{3x} - 12e^x = 5\)
    Wait! This isn't a simple quadratic. Let's use the syllabus example that forms a quadratic:
    Syllabus example: Solve \(3e^{2x} = 12 - 5e^{x}\)
    \(3e^{2x} + 5e^x - 12 = 0\)
  2. Let \(y = e^x\). Since \(e^{2x} = (e^x)^2\), this becomes:
    \(3y^2 + 5y - 12 = 0\)
  3. Solve for \(y\): \((3y - 4)(y + 3) = 0\).
    \(y = 4/3\) or \(y = -3\).
  4. Back-substitute:
    • Case 1: \(e^x = 4/3 \implies x = \ln(4/3)\)
    • Case 2: \(e^x = -3\). Since \(e^x\) must be positive, this case has no solution.

3.2 Logarithmic and Fractional Power Examples

Example 1 (Logarithmic): Solve \(2(\ln 5x)^2 + \ln 5x - 6 = 0\).

  • Let \(y = \ln 5x\).
  • The equation becomes \(2y^2 + y - 6 = 0\).
  • \((2y - 3)(y + 2) = 0\). So \(y = 3/2\) or \(y = -2\).
  • Back-substitute:
    \(\ln 5x = 3/2 \implies 5x = e^{3/2} \implies x = \frac{1}{5}e^{3/2}\)
    \(\ln 5x = -2 \implies 5x = e^{-2} \implies x = \frac{1}{5}e^{-2}\)

Example 2 (Fractional Power): Solve \(x^{2/3} + x^{1/3} - 12 = 0\).

  • Notice that \(x^{2/3} = (x^{1/3})^2\).
  • Let \(y = x^{1/3}\).
  • The equation becomes \(y^2 + y - 12 = 0\).
  • \((y + 4)(y - 3) = 0\). So \(y = -4\) or \(y = 3\).
  • Back-substitute:
    \(x^{1/3} = -4 \implies x = (-4)^3 = -64\)
    \(x^{1/3} = 3 \implies x = (3)^3 = 27\)
Did you know?

The substitution method is really powerful because it converts seemingly complex equations from areas like logarithms or indices into the familiar territory of quadratic algebra.

Key Takeaway: Substitution

Always look for a structure where one term is the square of another. Remember to check for restrictions in the final step (e.g., you cannot take the logarithm of a negative number, and \(e^x\) cannot be negative).


Section 4: Graphs of Cubic Polynomials and Inequalities (4.4 & 4.5)

In Additional Mathematics, you need to be able to sketch cubic graphs and understand the effect of the modulus on them, especially when the cubic is given in factorised form.

4.1 Sketching Cubic Polynomials in Factorised Form

A cubic polynomial \(f(x)\) defined by three linear factors looks like this:
$$f(x) = k(x-a)(x-b)(x-c)$$

  • The roots (x-intercepts) are simply \(x = a\), \(x = b\), and \(x = c\). These are the points where \(f(x)=0\).
  • The y-intercept is found by calculating \(f(0)\).
  • The overall shape depends on the sign of the leading coefficient (the product of all coefficients, including \(k\)).
    • If the leading coefficient is Positive, the graph starts low (from Quadrant 3) and ends high (in Quadrant 1). (It looks like a stretched 'N').
    • If the leading coefficient is Negative, the graph starts high (from Quadrant 2) and ends low (in Quadrant 4). (It looks like a stretched 'S').

Example Sketch: \(f(x) = (x+1)(x-2)(x-4)\)

  • Roots: -1, 2, 4.
  • Y-intercept: \(f(0) = (1)(-2)(-4) = 8\).
  • Shape: Positive leading coefficient, so it starts low, passes through -1, peaks, passes through 2, troughs, passes through 4, and goes up forever.

4.2 Sketching Modulus of a Cubic: \(y = |f(x)|\)

The graph of \(y = |f(x)|\) is created by taking the sketch of \(y = f(x)\) and applying a transformation:

Reflection Rule: Any part of the graph of \(y = f(x)\) that is below the x-axis must be reflected upwards (mirror-imaged) above the x-axis. The parts already above the x-axis remain unchanged.

The intercepts (roots) remain the same, but the resulting graph will have sharp points, called cusps, at the x-intercepts.

4.3 Solving Cubic Inequalities Graphically (4.5)

The syllabus requires you to solve cubic inequalities of the form \(f(x) \geq d\) or \(|f(x)| < d\), where \(f(x)\) is the product of three linear factors, by using a sketch.

Step-by-Step for \(f(x) > d\):

  1. Sketch the graph of \(y = f(x)\).
  2. Draw the horizontal line \(y = d\).
  3. Find the x-coordinates of all intersection points (by solving \(f(x) = d\)).
  4. Identify the intervals where the sketch of \(f(x)\) is above the line \(y = d\).

Example: Use your sketch of \(f(x) = (x+1)(x-2)(x-4)\) to solve \(f(x) \leq 8\).

  1. We found \(f(0) = 8\). So the graph already touches \(y=8\) at \(x=0\).
  2. We solve \((x+1)(x-2)(x-4) = 8\). (Expand this carefully and set equal to zero).
    \(x^3 - 5x^2 + 2x + 8 = 8\)
    \(x^3 - 5x^2 + 2x = 0\)
    \(x(x^2 - 5x + 2) = 0\)
  3. The intersection points are \(x=0\) and the roots of \(x^2 - 5x + 2 = 0\). (Let's call these roots \(x_1\) and \(x_2\)).
  4. Looking at the sketch, we need where the curve is below or equal to the line \(y=8\). The curve is below \(y=8\) for \(x \leq x_1\), and between \(x_2\) and 0.
  5. Final solution will be expressed as ranges of \(x\).

Important Tip for Inequalities: When solving graphically, if the question asks for an accurate solution, you must ensure you algebraically calculate the exact points of intersection. The sketch guides your understanding of which intervals to choose.

Key Takeaway: Graphs

For cubic sketches, identify the roots and the end behaviour (positive coefficient starts low, negative starts high). The modulus transformation \(y = |f(x)|\) simply means "mirror" any part below the x-axis upwards. Use horizontal lines (\(y=d\)) to solve inequalities.