What is 2.5: Linear and quadratic inequalities?

An inequality solution is a set of values; endpoints are included for ≤ or ≥ and excluded for . For a quadratic or rational inequality, locate every zero and undefined value, then use a sign diagram or graph on the resulting intervals.

How many marks are 2.5 questions worth?

2.5 practice drills on this page are worth 2–5 marks each; marks in a real exam vary by question.

Which past exam papers has 2.5 appeared in?

We have not yet indexed a verified past-paper appearance for 2.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.5

Solve linear and quadratic inequalities in one variable and interpret them graphically, including with brackets and fractions; express solutions using 'and'/'or' or set notation; represent linear and quadratic inequalities graphically.

Draft — not yet indexed

Linear and quadratic inequalities

Worked answers and methods for 2.5 on Edexcel A-level Maths 9MA0.

Explanation

An inequality solution is a set of values; endpoints are included for $\leq$ or $\geq$ and excluded for $<$ or $>$ . For a quadratic or rational inequality, locate every zero and undefined value, then use a sign diagram or graph on the resulting intervals.
For example, $(x-2)(x+3)<0$ between its roots, so $-3<x<2$ .
Multiplying by an expression of unknown sign can reverse the inequality unpredictably; bring a fractional inequality to one side and analyse signs instead.
Mark open or closed endpoints carefully and exclude any value that makes an original denominator zero, even when it appears as a boundary on the sign diagram.
For a two-variable graphical inequality, draw a solid boundary for $\leq$ or $\geq$ , a dashed boundary for $<$ or $>$ , test a point and shade the correct region.

Boundary curves and a test point determine the shaded region of a two-variable inequality.

Worked example

Solve $(x-4)(x+1)\leq0$ .

1.The boundary values are $x=-1$ and $x=4$ .
2.The upward-opening quadratic is non-positive between these roots, and equality includes both endpoints.
3.Therefore $-1\leq x\leq4$ .

Answer: $-1\leq x\leq4$

Common mistakes

Don't divide an inequality by a negative quantity without reversing the inequality sign.
Don't cross-multiply a rational inequality by an expression of unknown sign, so the direction of the inequality is unjustified.

Exam tip

For a quadratic or rational inequality, show the critical values and a sign diagram before writing the final intervals.

Worked practice

Tier 1 · Easy

Solve

5-2x<11

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$x>-3$	2
Notes
Subtract $5$ to obtain $-2x<6$ . Dividing by the negative number $-2$ reverses the inequality, giving $x>-3$ .

(2 marks)

Tier 2 · Standard

Solve the compound inequality

3(2x-1)\leq5-x<4x+10

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$-1<x\leq\dfrac87$	4
Notes
From $3(2x-1)\leq5-x$ , $6x-3\leq5-x$ , so $7x\leq8$ and $x\leq8/7$ . From $5-x<4x+10$ , $-5<5x$ , so $x>-1$ . Both conditions must hold, giving $-1<x\leq8/7$ .

(4 marks)

Tier 3 · Hard

Solve

\dfrac{x+2}{x-3}>2

and give the result in set notation.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\{x\in\mathbb R:3<x<8\}$	4
Notes
Move everything to one side: $\dfrac{x+2}{x-3}-2=\dfrac{8-x}{x-3}>0$ . The critical values are $3$ , where the expression is undefined, and $8$ , where it is zero. A sign check shows the fraction is positive only for $3<x<8$ . Both endpoints are excluded, so the set is $\{x\in\mathbb R:3<x<8\}$ .

(4 marks)

Tier 1 · Easy

Solve

4(x-3)\leq2x+6

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x\leq9$	2
Notes
Expanding gives $4x-12\leq2x+6$ . Hence $2x\leq18$ , so $x\leq9$ .

(2 marks)

Tier 2 · Standard

Solve

x^2-5x-14<0

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$-2<x<7$	3
Notes
Factorise to obtain $(x-7)(x+2)<0$ . The upward-opening quadratic is negative between its roots, so the solution is $-2<x<7$ .

(3 marks)

Tier 3 · Hard

Solve

\dfrac{(x-4)(x+1)}{x-2}\leq0

, giving your answer as inequalities.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x\leq-1$ or $2<x\leq4$	5
Notes
The critical values are $-1$ and $4$ , where the numerator is zero, and $2$ , where the expression is undefined. A sign check on the four intervals gives negative, positive, negative, positive respectively. Include the numerator zeros because equality is allowed, but exclude $x=2$ . Hence $x\leq-1$ or $2<x\leq4$ .

(5 marks)

Tier 2 · Standard

Sketch and shade the region satisfying

x\geq1

y\geq0

and

x+2y\leq7

. State the coordinates of all vertices of the region.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	The closed triangular region bounded by $x=1$ , $y=0$ and $x+2y=7$ , with vertices $(1,0)$ , $(7,0)$ and $(1,3)$ .	4
Notes
All three boundaries are solid because equality is included. The lines $x=1$ and $y=0$ meet at $(1,0)$ . Setting $y=0$ in $x+2y=7$ gives $(7,0)$ , and setting $x=1$ gives $y=3$ , producing $(1,3)$ . Testing a point inside these boundaries identifies the closed triangular region containing, for example, $(2,1)$ .

(4 marks)

Tier 3 · Hard

Find the real values of

x

that satisfy both

x^2-5x+4\leq0

and

2x^2+x-6>0

. Give your answer as inequalities.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$\dfrac32<x\leq4$	5
Notes
The first inequality is $(x-1)(x-4)\leq0$ , so $1\leq x\leq4$ . The second is $(2x-3)(x+2)>0$ , so $x<-2$ or $x>3/2$ . Intersecting these two solution sets gives $3/2<x\leq4$ .

(5 marks)

Tier 3 · Hard

Solve

\sqrt{2x+3}>x

, giving your answer as an interval.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\left[-\dfrac32,3\right)$	5
Notes
The square root requires $x\geq-3/2$ . If $-3/2\leq x<0$ , the left-hand side is non-negative and the right-hand side is negative, so every such $x$ works. If $x\geq0$ , both sides are non-negative and squaring preserves the inequality: $2x+3>x^2$ . Thus $(x-3)(x+1)<0$ , giving $-1<x<3$ ; intersecting with $x\geq0$ gives $0\leq x<3$ . Combining the cases yields $-3/2\leq x<3$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Solve

x^4-10x^2+9\leq0

, giving your answer as inequalities.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$-3\leq x\leq-1$ or $1\leq x\leq3$	4
Notes
Let $u=x^2$ , where $u\geq0$ . The inequality becomes $u^2-10u+9\leq0$ , or $(u-1)(u-9)\leq0$ , so $1\leq u\leq9$ . Hence $1\leq x^2\leq9$ , which gives $-3\leq x\leq-1$ or $1\leq x\leq3$ .

(4 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 2.5. Browse the Edexcel A-level Maths 9MA0 past papers directly.

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