Quadratic Inequalities (AA SL)
A quadratic inequality doesn't ask for exact roots - it asks for a whole range of \(x\)-values, everywhere the parabola sits above or below the \(x\)-axis. This page covers the sign-diagram method for solving them, with worked examples and the mistakes that cost marks under exam conditions. It's part of the broader Quadratics & Equations topic.
11 questions on this sub-topic.
Finding and using the critical values
Covered under IB syllabus reference SL2.7: solution of quadratic equations and inequalities using factorisation, completing the square, and the quadratic formula, plus the discriminant and the nature of the roots. You don't need to memorise the quadratic formula - it's in the booklet - but you do need to turn its roots into a sign diagram.
Find the critical values
\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
Factorise where possible, or fall back on the quadratic formula, to find where the expression equals zero.
Read the sign
An upward-opening parabola (\(a>0\)) is negative between its roots and positive outside them; a downward-opening parabola is the reverse.
Sketch the shape first, then read the inequality straight off the sketch rather than guessing.
Need the full syllabus wording, formula-booklet table, or calculator methods? See Quadratics & Equations (calculator steps under Using your GDC).
Worked examples
Solve \((x-1)(x-5)<0\).
Worked solution
The roots are \(x=1\) and \(x=5.\) A1
An upward parabola is negative between its roots. R1
Solution: \(1<x<5.\) A1
Solve \(x^2 - x - 6 > 0\).
Worked solution
Factorising, \((x-3)(x+2)>0\), so the critical values are \(x=-2\) and \(x=3\). M1 A1
An upward parabola is positive outside its roots. R1
Solution set: \(x<-2\) or \(x>3\). A1
A ball is thrown so that its height above the ground after \(t\) seconds is \(h(t)=-5t^2+20t+3\) (metres).
Find the values of \(t\) for which the ball is more than 18 m above the ground.
Worked solution
\(-5t^2+20t+3>18.\) M1
\(-5t^2+20t-15>0\Rightarrow t^2-4t+3<0\) (dividing by \(-5\) and flipping the inequality). M1
\((t-1)(t-3)<0\), roots \(t=1,3.\) A1
An upward parabola is negative between its roots. R1
\(1<t<3.\) A1
Common mistakes
- Forgetting the \(\pm\) when the roots come from completing the square. A square root always gives two possible values, so both critical values must go onto the sign diagram, not just one.
- Mixing up "between" and "outside" the roots. For an upward parabola, \(<0\) or \(\le0\) holds between the roots and \(>0\) or \(\ge0\) holds outside them - sketch the graph rather than guessing which way round it goes.
- Getting the boundary wrong for a strict inequality. \(<\) and \(>\) exclude the critical values themselves; \(\le\) and \(\ge\) include them. The final answer should use the same type of inequality symbol as the question.
Ready to practise properly?
11 quadratic-inequality questions, marked instantly like the real exam.
Quick answers
How do you solve a quadratic inequality?
Rearrange so one side is zero, factorise (or use the quadratic formula) to find the critical values, sketch the parabola, then read the sign you need straight off the sketch.
Are the roots included in the solution of a quadratic inequality?
Only if the inequality is not strict. \(\le\) and \(\ge\) include the roots; \(<\) and \(>\) exclude them.