Solve quadratic inequalities using signs and endpoints
Turn roots into intervals, test the sign in each interval and include equality only when requested.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Paper 1 is part of every route. Written practice is self-checked; it does not award official examination marks.
Before you begin
- Solve quadratic equations and place real roots in order.
- Understand <, ≤, > and ≥ and how a negative multiplier reverses them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For x = 0, what is the sign of (x + 2)(x − 4)?
What you will learn
- Solve strict and non-strict quadratic inequalities.
- Handle negative leading coefficients, repeated roots and model constraints.
Roots mark possible sign changes
Rearrange an inequality so one side is zero and find the roots of the quadratic. For (x + 2)(x − 4) < 0, the roots −2 and 4 divide the real line into three intervals. Testing x = −3, 0 and 5 gives positive, negative and positive values. Therefore the product is negative exactly for −2 < x < 4.
For two distinct roots of an upward-opening parabola, it is negative between the roots and positive outside. For a downward-opening parabola, the signs reverse. Testing a value in each interval is safer than applying a remembered pattern without checking the leading coefficient. Do not divide by a variable factor of unknown sign, because you can change the solution set.
Decide whether zero is included
A strict inequality < or > excludes the roots; a non-strict inequality ≤ or ≥ includes roots satisfying equality. An outside solution uses 'or', such as x ≤ 2 or x ≥ 3, rather than an impossible combined chain x ≤ 2 ≤ 3 ≤ x.
A repeated root does not force a sign change. Since (x − 3)² ≥ 0 always, (x − 3)² ≤ 0 has only x = 3, while (x − 3)² < 0 has no real solutions. If the quadratic has no real roots, its sign stays that of its non-zero leading coefficient. After finding the algebraic set, intersect it with any stated domain, such as non-negative integers.
Pause and explain
Solve x² − 5x + 6 ≥ 0.
Worked example
Solve (x + 2)(x − 4) < 0 over real x.
Show the worked solution
- The roots are x = −2 and x = 4.
- Test values −3, 0 and 5: the products are 7, −8 and 7, so only the middle interval is negative.
- The inequality is strict, so exclude both roots.
Answer −2 < x < 4.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve x² − 5x + 6 ≥ 0 by a sign test.
Give me a hint
Factor, mark the roots and test one point in each interval.
Compare my reasoning
- Factor to (x − 2)(x − 3), with roots 2 and 3.
- At x = 0, 2.5 and 4 the values are 6, −0.25 and 2, respectively.
- Choose the positive outside intervals and include both roots because equality is allowed.
x ≤ 2 or x ≥ 3.
Look for these in your work
- Identify signs on all three intervals.
- Include equality and join the two parts with 'or'.
Solve −x² + 2x + 8 > 0.
Give me a hint
You can multiply by −1 if you reverse the inequality sign.
Compare my reasoning
- Multiply by −1 to get x² − 2x − 8 < 0.
- Factor as (x − 4)(x + 2) < 0, with roots −2 and 4.
- The upward quadratic is negative between its roots; exclude them for a strict inequality.
−2 < x < 4.
Look for these in your work
- Reverse the sign when multiplying by −1.
- Give the interval rather than just its boundary roots.
A supplied profit model is P(q) = −q² + 12q − 20 units, with q a non-negative integer. Which values of q give profit at least 15 units?
Give me a hint
Solve P(q) ≥ 15, then apply the integer restriction.
Compare my reasoning
- Rearrange to −q² + 12q − 35 ≥ 0, or q² − 12q + 35 ≤ 0.
- Factor (q − 5)(q − 7) ≤ 0, giving the real interval 5 ≤ q ≤ 7.
- The non-negative integer inputs in that interval are 5, 6 and 7; their profits are 15, 16 and 15.
q = 5, 6 or 7.
Look for these in your work
- Use the requested threshold, not zero profit.
- Intersect the real interval with the stated integer domain.
Common mistakes
- Find intervals, not only the roots of the related equation.
- Use 'or' for separate outside intervals.
- Check the leading coefficient and do not assume the sign flips at a repeated root.
What can you explain now?
Solve (x − 3)² ≤ 0.
Compare with the explanation
x = 3 only.
A real square cannot be negative. Equality is possible only when its bracket is zero.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why the sign changes at a simple root but not necessarily at a repeated root. Sketch an outside solution with included endpoints.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.