Week 4 · The sign of a constant, linear and quadratic function, and quadratic inequalities in one unknown · 3 lessons
3
Solving a quadratic inequality by studying the sign of its function and writing the solution as intervals
Mathematics Sec 1 Term 1, Unit 1, Lessons 4–5
GoalSolve quadratic inequalities with one unknown and write the solution set as intervals.
New words
quadratic inequality · an inequality such as x² − 3x − 10 > 0 with a quadratic expressionsolution set · all numbers that make the inequality trueempty set · ∅: the set with no numbers (the inequality is never true)
Explanation
To solve a quadratic inequality, study the sign of the quadratic function. Step 1: move everything to one side so that the other side is zero. Step 2: find the roots and decide the sign of the function on the number line (the sign of a outside the roots, the opposite between them, or the sign of a everywhere if there are no roots). Step 3: choose the intervals where the sign matches the inequality. With a strict sign (> or <) the roots are not solutions and the brackets are open; with ≤ or ≥ the roots are solutions and the brackets are closed. Never divide both sides by x or by any expression that can be negative: the inequality might flip. Sometimes the answer is the empty set ∅ (never true) or the whole set R.
Step
What to do
1
Move everything to one side; the other side is 0
2
Find the roots and the sign of the function (the sign of a, the roots)
3
Keep the intervals that match the inequality sign; ≤ and ≥ include the roots
Worked examples
Solve x² − 3x − 10 > 0. Roots: (x − 5)(x + 2) = 0 gives −2 and 5. a > 0, so f > 0 outside the roots. The solution is ]−∞, −2[ ∪ ]5, ∞[. With ≤ 0 instead, the solution is the closed interval [−2, 5].
Solve 2x² < 7x + 4. Move: 2x² − 7x − 4 < 0, which factorises as (2x + 1)(x − 4) < 0. Roots −1/2 and 4, a > 0, so f < 0 between them. The solution is ]−1/2, 4[.
Solve x² > 10x − 25. Move: x² − 10x + 25 > 0, that is (x − 5)² > 0. A square is positive except where it is zero, so every x except 5: R − {5}. Also x² + 2x + 5 < 0 has no solution: its discriminant 4 − 20 < 0 and a > 0 make it always positive, so the solution set is ∅.
Solve −x² + 5x − 4 > 0. Roots 1 and 4. Now a = −1 < 0, so the function is positive between the roots: ]1, 4[. Test x = 2: −4 + 10 − 4 = 2 > 0 ✓.
Class activity
Boundary check. For x² − 3x − 10 ≥ 0 test x = −2, x = 0 and x = 5 in the inequality. Which are solutions? Which bracket types does this suggest for ≥ and for a strict “greater than”?
