Strand 4 — Algebra

Inequalities

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to solve linear inequalities and represent their solutions on a number line and using interval notation.
  • By the end of this lesson students will be able to solve quadratic inequalities using graphical methods or sign analysis.
  • By the end of this lesson students will be able to solve modulus inequalities (Higher Level) using algebraic methods or squaring both sides.
  • By the end of this lesson students will be able to solve rational inequalities (Higher Level) by finding critical values and using sign analysis.

Key concepts

Introduction to Inequalities

An inequality is a mathematical statement that compares two expressions using an inequality symbol: < (less than) > (greater than) ≤ (less than or equal to) ≥ (greater than or equal to) ≠ (not equal to) Solving an inequality means finding the range of values for the variable that make the statement true. The solution is often represented on a number line or using interval notation.

Linear Inequalities

Linear inequalities involve a variable raised to the power of one. They are solved using similar algebraic operations to linear equations, with one crucial difference: if you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality sign.

Quadratic Inequalities

Quadratic inequalities involve a quadratic expression (e.g., ax² + bx + c) and an inequality symbol. The most common methods for solving them involve: 1. **Finding the roots:** Solve the corresponding quadratic equation (ax² + bx + c = 0) to find the critical values. 2. **Sketching the graph:** Sketch the parabola y = ax² + bx + c. If a > 0, the parabola opens upwards; if a < 0, it opens downwards. Identify the regions where the graph is above or below the x-axis, corresponding to the inequality. 3. **Sign analysis (test points):** Use the critical values to divide the number line into intervals. Pick a test point from each interval and substitute it into the original inequality to determine if the inequality holds true for that interval.

Modulus Inequalities (Higher Level)

Modulus inequalities involve the absolute value (modulus) of an expression. The modulus of a number is its distance from zero, always non-negative. For a positive constant 'a': 1. **|x| < a** means -a < x < a (x is between -a and a). 2. **|x| > a** means x < -a or x > a (x is outside the range [-a, a]). These principles can be applied to expressions within the modulus. Alternatively, squaring both sides can be used, but care must be taken to ensure all terms are on one side before squaring, and to correctly factorise the resulting quadratic.

|x| < a \iff -a < x < a \text{ (for } a > 0) \newline |x| > a \iff x < -a \text{ or } x > a \text{ (for } a > 0)
Rational Inequalities (Higher Level)

Rational inequalities involve a rational expression (a fraction where the numerator and/or denominator contain variables) and an inequality symbol. The key steps are: 1. **Move all terms to one side:** Ensure one side of the inequality is zero. 2. **Combine into a single fraction:** Find a common denominator if necessary. 3. **Find critical values:** Determine the values of x that make the numerator zero and the values that make the denominator zero. These are the critical values. 4. **Sign analysis:** Use the critical values to divide the number line into intervals. Test a value from each interval in the simplified rational expression to determine its sign. Pay attention to whether the inequality includes equality (≤ or ≥) and if the denominator can be zero (which it cannot).

Key facts to remember

  • 1When multiplying or dividing an inequality by a negative number, always reverse the inequality sign.
  • 2Solutions to inequalities can be represented on a number line (open circle for < or >, closed circle for ≤ or ≥) or using interval notation.
  • 3For quadratic inequalities, find the roots (critical values) and use a graph sketch or sign table to determine the solution intervals.
  • 4For modulus inequalities |x| < a, the solution is -a < x < a. For |x| > a, the solution is x < -a or x > a (for a > 0).
  • 5For rational inequalities, bring all terms to one side, combine into a single fraction, find critical values (roots of numerator and denominator), and use sign analysis.
  • 6Critical values from the denominator of a rational inequality are never included in the solution set, as they make the expression undefined.
  • 7The union symbol 'U' is used to combine disjoint intervals in a solution set.
  • 8Always check the original inequality with test points from your solution intervals to verify correctness.

Worked examples

Example 1

Solve the inequality 3(x - 2) + 5 ≤ 2x + 7 and represent the solution on a number line.

IExpand the bracket: 3x - 6 + 5 ≤ 2x + 7
IISimplify the left side: 3x - 1 ≤ 2x + 7
IIISubtract 2x from both sides: 3x - 2x - 1 ≤ 7
IVSimplify: x - 1 ≤ 7
VAdd 1 to both sides: x ≤ 7 + 1
VIFinal solution: x ≤ 8
VIITo represent on a number line: Draw a number line. Place a closed circle at 8 (because x can be equal to 8) and draw an arrow extending to the left, indicating all values less than 8.

Answer

x ≤ 8

Remember to keep the inequality sign consistent throughout the steps unless multiplying or dividing by a negative number.

Example 2

Solve the quadratic inequality x² - x - 12 > 0.

IFind the roots of the corresponding quadratic equation x² - x - 12 = 0.
IIFactorise the quadratic: (x - 4)(x + 3) = 0
IIIThe roots (critical values) are x = 4 and x = -3.
IVSketch the graph of y = x² - x - 12. Since the coefficient of x² is positive (a = 1), the parabola opens upwards. It intersects the x-axis at x = -3 and x = 4.
VWe are looking for where x² - x - 12 > 0, which means where the graph is above the x-axis.
VIFrom the sketch, the graph is above the x-axis when x < -3 or when x > 4.
VIIThe solution in interval notation is (-∞, -3) U (4, ∞).

Answer

x < -3 or x > 4

A sign table can also be used: Test points in intervals (-∞, -3), (-3, 4), (4, ∞). For example, x = -4 gives (-8)(-1) = 8 > 0 (True). x = 0 gives (-4)(3) = -12 > 0 (False). x = 5 gives (1)(8) = 8 > 0 (True).

Example 3

Solve the modulus inequality |2x - 3| ≥ 5 (Higher Level).

IFor an inequality of the form |expression| ≥ a, we have two separate inequalities: expression ≤ -a or expression ≥ a.
IISo, we have: 2x - 3 ≤ -5 OR 2x - 3 ≥ 5.
IIISolve the first inequality: 2x - 3 ≤ -5
IVAdd 3 to both sides: 2x ≤ -5 + 3
V2x ≤ -2
VIDivide by 2: x ≤ -1.
VIISolve the second inequality: 2x - 3 ≥ 5
VIIIAdd 3 to both sides: 2x ≥ 5 + 3
92x ≥ 8
10Divide by 2: x ≥ 4.
11Combine the solutions: x ≤ -1 or x ≥ 4.
12In interval notation: (-∞, -1] U [4, ∞).

Answer

x ≤ -1 or x ≥ 4

Remember that for 'greater than or equal to' modulus inequalities, the solution is typically two separate intervals, not a single 'between' interval.

Example 4

Solve the rational inequality (x + 1) / (x - 2) < 3 (Higher Level).

IBring all terms to one side to make the right side zero: (x + 1) / (x - 2) - 3 < 0.
IIFind a common denominator: (x + 1) / (x - 2) - 3(x - 2) / (x - 2) < 0.
IIICombine into a single fraction: (x + 1 - 3(x - 2)) / (x - 2) < 0.
IVSimplify the numerator: (x + 1 - 3x + 6) / (x - 2) < 0.
V(-2x + 7) / (x - 2) < 0.
VIFind the critical values by setting the numerator and denominator to zero.
VIINumerator: -2x + 7 = 0 => 2x = 7 => x = 7/2 or x = 3.5.
VIIIDenominator: x - 2 = 0 => x = 2.
9These critical values (x = 2 and x = 3.5) divide the number line into three intervals: (-∞, 2), (2, 3.5), (3.5, ∞).
10Perform sign analysis for each interval for the expression (-2x + 7) / (x - 2):
11Interval 1: x < 2 (e.g., x = 0). Numerator: 7 (positive). Denominator: -2 (negative). Fraction: positive / negative = negative. So, (-2x + 7) / (x - 2) < 0 is TRUE.
12Interval 2: 2 < x < 3.5 (e.g., x = 3). Numerator: -6 + 7 = 1 (positive). Denominator: 3 - 2 = 1 (positive). Fraction: positive / positive = positive. So, (-2x + 7) / (x - 2) < 0 is FALSE.
13Interval 3: x > 3.5 (e.g., x = 4). Numerator: -8 + 7 = -1 (negative). Denominator: 4 - 2 = 2 (positive). Fraction: negative / positive = negative. So, (-2x + 7) / (x - 2) < 0 is TRUE.
14The intervals where the inequality is true are x < 2 or x > 3.5.
15In interval notation: (-∞, 2) U (3.5, ∞). Note that x cannot be 2 because it makes the denominator zero.

Answer

x < 2 or x > 3.5

Never multiply by a variable expression (like x-2) in an inequality unless you are certain of its sign, as this can lead to incorrect solutions. Always use sign analysis after bringing all terms to one side.

Common mistakes

  • Forgetting to reverse the inequality sign when multiplying or dividing by a negative number.
  • Incorrectly handling the endpoints of intervals (e.g., using a closed circle instead of an open circle for strict inequalities).
  • Multiplying across by a variable expression (like x-2) in rational inequalities without considering its sign, which can lead to losing solutions or introducing extraneous ones.
  • Incorrectly applying the rules for modulus inequalities, especially confusing the 'and' and 'or' conditions.
  • Algebraic errors when simplifying expressions, particularly with negative signs or distributing terms.

Exam tips

  • Always show all steps of your working clearly, especially for sign analysis or graphical methods, as partial credit is awarded.
  • Use a number line to visualise the solution intervals, even if not explicitly asked for, as it helps prevent errors.
  • For Higher Level questions, clearly state the critical values and show your sign analysis or reasoning for choosing intervals.
  • Double-check your final answer by substituting a value from your solution set (and one outside it) into the original inequality to ensure it holds true.

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