Edexcel IAL - Pure Maths 1- 1.9 Solutions of Linear and Quadratic Inequalities- Study notes - New syllabus
Edexcel IAL – Pure Maths 1- 1.9 Solutions of Linear and Quadratic Inequalities -Study notes- New syllabus
Edexcel IAL – Pure Maths 1- 1.9 Solutions of Linear and Quadratic Inequalities -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- Solutions of Linear and Quadratic Inequalities
Solutions of Linear and Quadratic Inequalities
Solving inequalities involves finding the range of \( x \)-values for which an expression is greater or smaller than another. Linear inequalities produce intervals on the number line; quadratic inequalities produce one or two intervals depending on the discriminant and the sign of the leading coefficient.
Key Types of Inequalities
| Type | General Form | Method |
| Linear inequality | \( ax + b > cx + d \) | Rearrange to isolate \( x \) |
| Quadratic inequality compared with 0 | \( px^2 + qx + r \gtrless 0 \) | Solve quadratic equation; test intervals |
| Quadratic compared with linear | \( px^2 + qx + r < ax + b \) | Bring to one side → quadratic inequality |
General Method for Quadratic Inequalities
- Rearrange inequality so the right-hand side is zero.
- Solve the related quadratic equation to find critical points.
- Sketch the parabola (opening up or down).
- Use the graph to determine intervals where the inequality holds.
Example
Solve the linear inequality \( 3x + 5 > x + 9 \).
▶️ Answer / Explanation
Rearrange:
\( 3x + 5 > x + 9 \Rightarrow 2x > 4 \Rightarrow x > 2 \)
Solution: \( x > 2 \)
Example
Solve the quadratic inequality \( x^2 – 3x – 10 \ge 0 \).
▶️ Answer / Explanation
Step 1: Factorise:
\( x^2 – 3x – 10 = (x – 5)(x + 2) \)
Step 2: Roots are \( x = 5 \) and \( x = -2 \).
Step 3: Parabola opens upward → expression is ≥ 0 outside the roots.
Solution:
\( x \le -2 \quad \text{or} \quad x \ge 5 \)
Example
Solve the inequality \( 2x^2 + 3x – 7 < x + 5 \).
▶️ Answer / Explanation
Step 1: Bring all terms to one side:
\( 2x^2 + 3x – 7 – x – 5 < 0 \)
\( 2x^2 + 2x – 12 < 0 \)
Step 2: Factor out 2 for simplicity:
\( 2(x^2 + x – 6) < 0 \)
\( x^2 + x – 6 < 0 \)
Step 3: Solve quadratic:
\( x^2 + x – 6 = (x + 3)(x – 2) \)
Roots: \( x = -3 \), \( x = 2 \)
Step 4: Parabola opens upward → expression is < 0 between the roots.
Solution:
\( -3 < x < 2 \)