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Solving linear and quadratic inequalities, including ones with brackets and fractions, and writing the solution with 'and', 'or' or set notation. You also show inequalities as regions on a graph. Quadratic inequalities often appear as the final step of a discriminant question.
Key facts
Multiply or divide by a negative: reverse the inequality sign
\(\gt 0\): curve above the \(x\)-axis; \(\lt 0\): curve below
\(\cup\) shape, roots \(\alpha \lt \beta\): \(\gt 0 \Rightarrow x \lt \alpha\) or \(x \gt \beta\)
'or' / \(\cup\) for two separate regions; 'and' / \(\cap\) for an overlap
\(\{x : x \gt 3\}\) means 'the set of values of \(x\) such that \(x \gt 3\)'
Boundary included for \(\le, \ge\) (solid); excluded for \(\lt, \gt\) (dashed)
Fraction inequality: multiply by the denominator squared
Notes
Linear inequalities
Solve like an equation. If you multiply or divide both sides by a negative number, reverse the sign: \(-3x \lt 12\) gives \(x \gt -4\)
You can avoid this by collecting the \(x\) terms on the side where their coefficient is positive.
With fractions, multiply every term by a positive common denominator: \(\dfrac{x + 1}{4} \le \dfrac{2x - 3}{3}\) gives \(3x + 3 \le 8x - 12\), so \(x \ge 3\)
Quadratic inequalities
diagram
Rearrange so that one side is 0, then solve the equation to find the critical values.
Sketch the curve. '\(\gt 0\)' means where the curve is above the \(x\)-axis; '\(\lt 0\)' means below.
x² − x − 12 ≤ 0 where the curve is on or below the x-axis, so −3 ≤ x ≤ 4.
For a \(\cup\)-shaped curve with roots \(\alpha \lt \beta\): \(\gt 0\) gives \(x \lt \alpha\) or \(x \gt \beta\); \(\lt 0\) gives \(\alpha \lt x \lt \beta\)
Example: \(x^2 - x - 12 \le 0\) gives \((x - 4)(x + 3) \le 0\), so \(-3 \le x \le 4\)
Never divide both sides by an expression in \(x\) that could be negative or zero.
Writing the answer
One interval: \(-3 \le x \le 4\), or \(\{x : -3 \le x \le 4\}\)
Two intervals: \(x \lt -2\) or \(x \gt 5\), or \(\{x : x \lt -2\} \cup \{x : x \gt 5\}\). Never join them into one inequality.
Values that satisfy both of two inequalities are the overlap, written with 'and' or \(\cap\)
Graphs and regions
\(y \gt x + 1\) is the region above the line \(y = x + 1\); \(y \lt x^2 - 4\) is the region below the curve.
A boundary is included for \(\le\) or \(\ge\) (usually drawn solid) and not included for \(\lt\) or \(\gt\) (usually drawn dashed).
For \(\dfrac{a}{x - b} \gt c\), multiply both sides by \((x - b)^2\), which is positive. Collect terms on one side, factorise and use a sketch. Remember \(x \ne b\) Grade A or above only
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find the set of values of x for a quadratic inequality
4 marksGrade D
Rearrange to a quadratic compared with 0.
Solve the quadratic equation to find the critical values.
Sketch the parabola and choose the correct region.
Write the answer with 'or' or set notation.
Example. Find the set of values of \(x\) for which \(2x^2 + 5x - 3 \gt 0\)
Show the model answerHide the model answer
\((2x - 1)(x + 3) = 0\)
(M1)
Critical values \(x = \dfrac{1}{2}\) and \(x = -3\)
(A1)
Choosing the outside region
(M1)
\(x \lt -3\) or \(x \gt \dfrac{1}{2}\)
(A1)
Find values that satisfy two inequalities
5 marksGrade C
Solve each inequality separately.
Draw both solutions on a number line.
Write down the overlap.
Example. Find the set of values of \(x\) for which both \(3(x - 2) \lt x + 4\) and \(x^2 - 2x - 8 \gt 0\)
Find where the boundaries meet by solving their equations together.
Decide whether the region is above or below each boundary.
Use \(\le\), \(\ge\) if the boundaries are included.
Example. The curve \(C\) has equation \(y = x^2 - 4x\) and the line \(l\) has equation \(y = x\). The finite region \(R\) is bounded by \(C\) and \(l\), and includes its boundaries. (a) Find the \(x\)-coordinates of the points where \(l\) meets \(C\). (b) Write down inequalities that define \(R\).
Show the model answerHide the model answer
(a) \(x^2 - 4x = x\), so \(x(x - 5) = 0\)
(M1)
\(x = 0\) and \(x = 5\)
(A1)
(b) \(y \ge x^2 - 4x\)
(B1)
\(y \le x\)
(B1)
Solve an inequality with x in the denominator
5 marksGrade A
Multiply both sides by the denominator squared.
Collect all terms on one side and factorise.
Find the critical values, including the value that makes the denominator 0.
Use a sketch to choose the region, excluding that value.
Example. Find the set of values of \(x\) for which \(\dfrac{6}{x - 1} \lt 2\), \(x \ne 1\)
Writing two regions as one inequality, e.g. \(\dfrac{1}{2} \lt x \lt -3\), or joining them with 'and'.
Square rooting both sides: \(x^2 \gt 9\) gives \(x \lt -3\) or \(x \gt 3\), not just \(x \gt 3\)
Multiplying by \(x - 1\), which might be negative, instead of \((x - 1)^2\)
Stopping at the critical values without choosing the region.
Forgetting to reverse the sign when dividing by a negative number.
More tips
Memory tricks
Always sketch: a five-second parabola stops you choosing the wrong region.
Test a value: pick a number in your answer and check it satisfies the original inequality.
'Find the set of values' is exam language for 'solve and give the answer as inequalities or a set'.
For two inequalities, draw the solution lines one above the other on the same number line.
Exam technique
Answer in the form asked for. For set notation use braces and \(\cup\) or \(\cap\) correctly.
Show your sketch or sign diagram: it earns the method mark for choosing the region.
Carry strict (\(\lt\)) or inclusive (\(\le\)) signs through to the final answer exactly as the question needs.
In discriminant questions, the final inequality in \(k\) is solved exactly like this.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Solve a linear inequalitySolve it like an equation, but reverse the sign if you multiply or divide by a negative number.
Grade D
Solve a quadratic inequality using a sketchFind the critical values, sketch the parabola and choose the part above or below the \(x\)-axis.
Grade C
Write solutions using set notatione.g. \(\{x : x \lt -2\} \cup \{x : x \gt 5\}\) for two regions, \(\{x : -2 \lt x \lt 5\}\) for one.
Grade C
Find values satisfying two inequalities togetherShow both solution sets on a number line and find where they overlap.
Grade C
Define a region using inequalitiesDecide which side of each line or curve the region is on, and whether the boundary is included.
Grade B
Solve inequalities using graphs\(f(x) \gt g(x)\) where the graph of \(f\) is above the graph of \(g\); find the intersections first.
Grade A
Solve inequalities with x in the denominatorMultiply both sides by the square of the denominator, which is positive, then solve.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Find the set of values of \(x\) for which \(5(x + 2) \lt x + 30\)[2]
(b) Find the set of values of \(x\) for which \(x^{2} - 4x - 12 \lt 0\)[3]
(c) Hence find the set of values of \(x\) for which both inequalities are satisfied.[1]
Show the answer and mark scheme
(a)Answer: \(x \lt 5\)
M1 for expanding the bracket and collecting the \(x\) terms: \(4x \lt 20\)
A1 for \(x \lt 5\)
Worked solution: \(5x + 10 \lt x + 30\), so \(4x \lt 20\), so \(x \lt 5\).
(b)Answer: \(-2 \lt x \lt 6\)
M1 for finding the critical values, e.g. \((x + 2)(x - 6) = 0\)
A1 for \(x = -2\) and \(x = 6\)
A1ft for choosing the inside region: \(-2 \lt x \lt 6\)
Worked solution: \(x^{2} - 4x - 12 = (x + 2)(x - 6)\), so the critical values are \(x = -2\) and \(x = 6\). The graph of \(y = x^{2} - 4x - 12\) is \(\cup\)-shaped, so it is below the \(x\)-axis between the roots: \(-2 \lt x \lt 6\).
(c)Answer: \(-2 \lt x \lt 5\)
B1ft for \(-2 \lt x \lt 5\)
Worked solution: Combining \(x \lt 5\) with \(-2 \lt x \lt 6\) (a number line helps) gives \(-2 \lt x \lt 5\). In set notation: \(\{x : -2 \lt x \lt 5\}\).
Question 2Medium6 marks
A student is asked to find the set of values of \(x\) for which \[\frac{x - 1}{x + 2} \le 2\]The student writes: \(x - 1 \le 2(x + 2)\) \(x - 1 \le 2x + 4\) \(-5 \le x\), so the answer is \(x \ge -5\).
(a) Show that \(x = -3\) satisfies the student’s answer but does not satisfy the original inequality.[1]
(b) Explain the error in the student’s method.[1]
(c) Find the correct set of values of \(x\) for which \(\dfrac{x - 1}{x + 2} \le 2\).[4]
Show the answer and mark scheme
(a)Answer: \(-3 \ge -5\), but \(\frac{-3 - 1}{-3 + 2} = 4\), which is not \(\le 2\).
B1 for evaluating \(\frac{-4}{-1} = 4 \gt 2\) and noting \(-3 \ge -5\)
Worked solution: \(-3 \ge -5\), so \(x = -3\) is in the student’s answer. But \(\dfrac{-3 - 1}{-3 + 2} = \dfrac{-4}{-1} = 4\), and \(4 \gt 2\).
(b)Answer: The student multiplied by \(x + 2\), which is negative when \(x \lt -2\); multiplying by a negative number reverses the inequality.
B1 for the student multiplied both sides by \(x + 2\), which can be negative (reversing the inequality), without considering its sign
Worked solution: Multiplying an inequality by a negative number reverses it. The student multiplied by \(x + 2\), which is negative for \(x \lt -2\), so the method fails for those values.
(c)Answer: \(x \le -5\) or \(x \gt -2\)
M1 for multiplying both sides by \((x + 2)^2\) (or forming a single fraction \(\frac{-x - 5}{x + 2} \le 0\))
A1 for critical values \(-5\) and \(-2\), e.g. from \((x + 2)(x + 5) \ge 0\)
M1 for choosing the outside region
A1 for \(x \le -5\) or \(x \gt -2\) (strict at \(-2\))
Worked solution: Multiply both sides by \((x + 2)^2 \gt 0\) (for \(x \ne -2\)): \((x - 1)(x + 2) \le 2(x + 2)^2\). So \(0 \le 2(x + 2)^2 - (x - 1)(x + 2) = (x + 2)(2x + 4 - x + 1) = (x + 2)(x + 5)\). \((x + 2)(x + 5) \ge 0\) for \(x \le -5\) or \(x \ge -2\), and \(x = -2\) is excluded, so \(x \le -5\) or \(x \gt -2\).
Question 3Hard5 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The diagram shows a sketch of the curve \(C\) with equation \(y = -\frac{4}{x}\), \(x \ne 0\), and the line \(l\) with equation \(y = -2x + 2\).
[object Object]
Using algebra and the diagram, find the set of values of \(x\) for which \(-\frac{4}{x} \lt -2x + 2\)[5]
Show the answer and mark scheme
Answer: \(x \lt -1\) or \(0 \lt x \lt 2\)
M1 for multiplying through by \(x\) to obtain a quadratic equation: \(-2x^{2} + 2x + 4 = 0\)
A1 for the critical values \(x = -1\) and \(x = 2\)
B1 for identifying \(x = 0\) as a further critical value (the vertical asymptote)
M1 for using the diagram to choose regions where \(C\) is below \(l\)
A1 for \(x \lt -1\) or \(0 \lt x \lt 2\)
Worked solution: Multiplying \(-\frac{4}{x} = -2x + 2\) by \(x\): \(-2x^{2} + 2x + 4 = 0\), so \(-2(x + 1)(x - 2) = 0\) and \(x = -1\) or \(x = 2\). The curve is not defined at \(x = 0\), so \(x = 0\) is also a critical value. From the diagram, \(C\) is below \(l\) when \(x \lt -1\) or \(0 \lt x \lt 2\).