Linear and Quadratic Inequalities
This topic covers the methods for solving inequalities involving linear and quadratic expressions. Unlike equations, inequalities define a range of possible values, and the rules for manipulating them, especially with multiplication and division, are stricter.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The most important rule: when you multiply or divide both sides of an inequality by a negative number, you MUST flip the inequality sign (e.g., '>' becomes '<').
- To solve a quadratic inequality, first rearrange it so one side is zero, e.g., ax2 + bx + c > 0.
- Find the 'critical values' by solving the corresponding equation ax2 + bx + c = 0. These are the roots.
- Sketch the parabola. If the x2 term is positive, it's a U-shape; if negative, an n-shape. The critical values are where it crosses the x-axis.
- Use the sketch to identify the regions of x that satisfy the inequality (e.g., for '>' find where the graph is above the x-axis).
- Solutions for quadratics are often two separate regions (e.g., x < a or x > b) or a single region between the roots (e.g., a < x < b).
Diagram
Formulae
ax2 + bx + c > 0 (or <, ≤, ≥) This is the standard form for a quadratic inequality. The primary method is to find the roots of ax2 + bx + c = 0 and then analyse the sign of the quadratic in the intervals defined by these roots.
Definitions
- Critical Values
- The roots of the inequality's corresponding equation (i.e., where the expression equals zero). These values form the boundaries of the solution intervals on a number line.
Worked example
Find the set of values for x that satisfy the inequality 10 - 3x ≤ x2.
- 1
Rearrange the inequality to get a standard quadratic form with zero on one side:
0 ≤ x2 + 3x - 10.
- 2
Identify the critical values by solving the equation x2 + 3x - 10 = 0.
- 3
Factorise the quadratic:
(x + 5)(x - 2) = 0.
The critical values are x = -5 and x = 2.
- 4
Sketch the graph of y = x2 + 3x - 10.
Since the coefficient of x2 is positive (1), it is a U-shaped parabola crossing the x-axis at -5 and 2.
- 5
The inequality is x2 + 3x - 10 ≥ 0, so we need the regions where the graph is on or above the x-axis.
- 6
From the sketch, this occurs when x is less than or equal to -5, or when x is greater than or equal to 2.
- 7
State the final answer using 'or' to connect the two distinct regions.
Answer: x ≤ -5 or x ≥ 2
Common mistakes
- ×The most frequent error is forgetting to reverse the inequality sign when multiplying or dividing by a negative. For instance, converting -5x > 20 to x > -4 is incorrect; the correct answer is x < -4.
- ×When solving a quadratic inequality like (x-a)(x-b) > 0, students sometimes write the solution as a single interval (e.g. a < x < b). For a 'greater than' inequality with a U-shaped parabola, the solution is two separate outer regions (x < a or x > b).
- ×Incorrectly solving inequalities with variables in the denominator, like 1/x > 2. You cannot simply multiply by x because you don't know if x is positive or negative. The safe method is to multiply by x2, which is always non-negative.
No-calculator tips
- ✓A quick sketch is your best tool. You don't need an accurate plot, just the roots and the basic shape (U-shape or n-shape) to see whether the solution is 'between the roots' or 'outside the roots'.
- ✓Use test values. Once you have the critical values (e.g., -5 and 2), pick a simple number from each of the three regions (e.g., -6, 0, and 3) and substitute it into the inequality to see if it holds true. This confirms which regions are part of the solution.