Further Mathematics | Inequalities

Welcome to Inequalities in Further Mathematics 2 (FP2)!

Hello future Mathematician! You've already mastered basic inequalities involving quadratics and simultaneous equations. In FP2, we level up! This chapter focuses on tackling two major types of advanced inequalities: Algebraic Fractions (Rational Inequalities) and Modulus Inequalities, especially when they involve variables on both sides.

Don't worry if this seems tricky at first. We will break down these complex problems using systematic, step-by-step methods that guarantee success. Mastering this chapter is crucial for later topics involving complex numbers and geometry, so let's get started!

Section 1: Solving Inequalities with Algebraic Fractions (Rational Inequalities)

A rational inequality is one that involves an algebraic fraction (a polynomial divided by another polynomial), such as \(\frac{x+1}{x-2} > 3\).

The Crucial Mistake to Avoid

In basic algebra, if you have \(\frac{x}{5} > 2\), you multiply by 5. In FP2, NEVER multiply by a denominator that contains the variable \(x\) (like \(x-2\)).

Why is this a deadly mistake?
If the expression you multiply by is positive, the inequality sign stays the same.
If the expression you multiply by is negative, the inequality sign must FLIP.
Since the sign of \(x-2\) depends on the value of \(x\), you don't know whether to flip the sign or not! You would have to split the problem into two separate cases, which is inefficient and highly prone to error.

The Golden Rule for Rational Inequalities: Always rearrange the inequality so that zero is on one side.

Step-by-Step Method: Critical Values and Sign Analysis

Let's solve the general rational inequality.

Step 1: Compare the Inequality to Zero

Move all terms to one side. For example, to solve \(\frac{x+1}{x-2} > 3\): $$ \frac{x+1}{x-2} - 3 > 0 $$

Step 2: Combine the Fractions into a Single Fraction

Find a common denominator (in this case, \(x-2\)): $$ \frac{x+1}{x-2} - \frac{3(x-2)}{x-2} > 0 $$ $$ \frac{(x+1) - (3x - 6)}{x-2} > 0 $$ $$ \frac{-2x + 7}{x-2} > 0 $$

Step 3: Find the Critical Values (CVs)

The critical values are the \(x\)-values where the expression might change sign. These occur when the Numerator is zero OR when the Denominator is zero.

1. Numerator CV: \(-2x + 7 = 0 \implies x = \frac{7}{2}\) (or \(x = 3.5\))
2. Denominator CV: \(x - 2 = 0 \implies x = 2\)

Important Memory Aid: The critical value derived from the denominator (the asymptote) must always be excluded from the solution set, regardless of whether the original inequality was \(> \), \(<\), \(\ge\), or \(\le\).

Step 4: Use a Number Line or Sign Table (The Test Region Method)

The CVs divide the number line into regions. We test a value in each region to see if the overall fraction is positive or negative (i.e., satisfies the inequality \(\frac{-2x + 7}{x-2} > 0\)).

Regions to test:

1. \(x < 2\) (e.g., test \(x=0\)): $$ \frac{-2(0) + 7}{0 - 2} = \frac{7}{-2} = -3.5 \quad (< 0, \text{ Fails}) $$ 2. \(2 < x < 3.5\) (e.g., test \(x=3\)): $$ \frac{-2(3) + 7}{3 - 2} = \frac{1}{1} = 1 \quad (> 0, \text{ Succeeds}) $$ 3. \(x > 3.5\) (e.g., test \(x=4\)): $$ \frac{-2(4) + 7}{4 - 2} = \frac{-1}{2} = -0.5 \quad (< 0, \text{ Fails}) $$

Step 5: Write the Final Solution

Since we required the expression to be \(> 0\), the solution is the middle region where the test succeeded: $$ 2 < x < 3.5 $$

Quick Review: Rational Inequalities
1. Rearrange to compare to 0. 2. Combine into a single fraction \(\frac{P(x)}{Q(x)}\). 3. Find Critical Values (where \(P(x)=0\) and \(Q(x)=0\)). 4. Use test points to determine the sign in each region. 5. Denominator critical values are always excluded.

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Section 2: Solving Modulus Inequalities in FP2

In FP2, we often encounter modulus inequalities where both sides contain a variable and are placed inside a modulus sign, such as \(|x+1| \ge |2x-3|\).

Review: What the Modulus Does

The modulus (or absolute value) function, \(|A|\), simply measures the distance of \(A\) from zero, making the result non-negative.

Did you know? The term 'modulus' comes from the Latin word meaning 'a small measure'.

Method: Squaring Both Sides

When you have the form \(|A| > |B|\) or \(|A| < |B|\), the safest and most efficient method is squaring both sides.

Since both \(|A|\) and \(|B|\) are guaranteed to be non-negative, squaring them does not change the direction of the inequality sign.

Key Principle: \(|A| > |B| \iff A^2 > B^2\)

Example: Solving \(|x+1| \ge |2x-3|\)

Step 1: Square Both Sides

$$ (x+1)^2 \ge (2x-3)^2 $$

Step 2: Rearrange to Compare to Zero

We want to avoid expanding and dealing with a complicated quadratic. Instead, move everything to the left side: $$ (x+1)^2 - (2x-3)^2 \ge 0 $$

Step 3: Use the Difference of Two Squares Identity (\(a^2 - b^2 = (a-b)(a+b)\))

Let \(A = (x+1)\) and \(B = (2x-3)\). $$ \left[ (x+1) - (2x-3) \right] \left[ (x+1) + (2x-3) \right] \ge 0 $$

Watch out for the signs in the first bracket! $$ [ x+1 - 2x + 3 ] [ x+1 + 2x - 3 ] \ge 0 $$ $$ [ -x + 4 ] [ 3x - 2 ] \ge 0 $$

Step 4: Find the Critical Values (CVs)

Set each factor to zero:
1. \(-x + 4 = 0 \implies x = 4\)
2. \(3x - 2 = 0 \implies x = \frac{2}{3}\)

Step 5: Use a Sign Analysis (Number Line)

We are testing the sign of the product \([-x + 4] [ 3x - 2 ]\). We need this product to be \(\ge 0\) (positive or zero).

1. \(x < \frac{2}{3}\) (e.g., test \(x=0\)):
\([-0 + 4] [ 3(0) - 2 ] = (4)(-2) = -8 \quad (< 0, \text{ Fails})\)

2. \(\frac{2}{3} < x < 4\) (e.g., test \(x=1\)):
\([-1 + 4] [ 3(1) - 2 ] = (3)(1) = 3 \quad (> 0, \text{ Succeeds})\)

3. \(x > 4\) (e.g., test \(x=5\)):
\([-5 + 4] [ 3(5) - 2 ] = (-1)(13) = -13 \quad (< 0, \text{ Fails})\)

Step 6: Write the Final Solution

Since the inequality was \(\ge\), we include the critical points. $$ \frac{2}{3} \le x \le 4 $$

Alternative Method: Critical Points (Region Analysis)

Sometimes, especially if the inequality is mixed (e.g., \(|x+1| > x+5\)), the squaring method might be overly complex or inappropriate (if the right side isn't guaranteed non-negative).

The critical points method involves finding the values of \(x\) where the expressions inside the modulus signs change from negative to positive.

Example Revisited: \(|x+1| \ge |2x-3|\)

1. Find CVs where the contents of the modulus equal zero:
\(x+1=0 \implies x = -1\)
\(2x-3=0 \implies x = 1.5\)

2. These CVs divide the number line into three regions: \(x < -1\), \(-1 \le x < 1.5\), and \(x \ge 1.5\).

3. Solve the inequality for each region by rewriting the expression without the modulus sign:

Region A: \(x < -1\)

Both expressions inside the modulus are negative. We use negative signs to remove the moduli: $$ -(x+1) \ge -(2x-3) $$ $$ -x - 1 \ge -2x + 3 $$ $$ x \ge 4 $$ Contradiction: We assumed \(x < -1\), but found \(x \ge 4\). No solutions in this region.

Region B: \(-1 \le x < 1.5\)

\(x+1\) is positive, \(2x-3\) is negative. $$ (x+1) \ge -(2x-3) $$ $$ x + 1 \ge -2x + 3 $$ $$ 3x \ge 2 \implies x \ge \frac{2}{3} $$ Intersection: We need to satisfy both \(-1 \le x < 1.5\) AND \(x \ge \frac{2}{3}\). The solution is: $$ \frac{2}{3} \le x < 1.5 $$

Region C: \(x \ge 1.5\)

Both expressions are positive. $$ (x+1) \ge (2x-3) $$ $$ 4 \ge x $$ Intersection: We need to satisfy both \(x \ge 1.5\) AND \(x \le 4\). The solution is: $$ 1.5 \le x \le 4 $$

4. Combine the solutions: We join the results from Region B and Region C: $$ \left( \frac{2}{3} \le x < 1.5 \right) \cup \left( 1.5 \le x \le 4 \right) $$
Since 1.5 is included in the second set and approached in the first, we can write the combined solution smoothly: $$ \frac{2}{3} \le x \le 4 $$

Key Takeaway: The squaring method is usually much faster and less prone to case-based errors when dealing with \(|A| > |B|\). The critical points method (Region Analysis) is vital for mixed inequalities like \(|A| > B\).

Common Pitfalls and Summary

Mistakes to Avoid

1. Multiplying by the Variable Denominator: (Rational Inequalities) If you do this, you lose solutions or get the signs wrong. ALWAYS compare to zero.

2. Excluding Numerator CVs: (Rational Inequalities) If the original inequality includes equality (\(\ge\) or \(\le\)), the numerator critical value must be included in the final solution. Only the denominator CV is always excluded.

3. Sign Errors During Squaring: (Modulus Inequalities) When using \(a^2 - b^2 = (a-b)(a+b)\), be incredibly careful with the negative sign when calculating \((a-b)\). Example: \((x) - (2x-3)\) becomes \(x - 2x + 3\).

Quick Review Checklist

Rational Inequalities (\(\frac{P}{Q} \ge 0\))

• Rearrange to zero.
• Combine fractions.
• Identify CVs from P and Q.
• Use sign analysis (number line).
• Check exclusion rules (Q=0 is never allowed).

Modulus Inequalities (\(|A| \ge |B|\))

• Square both sides: \(A^2 \ge B^2\).
• Rearrange to use Difference of Two Squares: \((A-B)(A+B) \ge 0\).
• Find CVs from factors.
• Use sign analysis (number line) to find the final region.

You now have the fundamental tools needed to solve complex FP2 inequalities. Remember, practice makes perfect! Focus on maintaining a consistent process, and these problems will become second nature. Good luck!