Quadratics
Quadratic inequalities and how to write the answer
On this page
What you need to be able to do
- 1.1Solve quadratic inequalities in one unknown: find the critical values, sketch, choose the region.
- 1.1Write the answer in a form the mark scheme accepts, with strict or non-strict signs as the question needs.
- 1.1Use this to finish discriminant questions (set of values of \(k\)) and calculus questions (where a function increases or decreases).
Before you start: The discriminant. Next: Where a line meets a curve.
What an inequality asks, why its answer is always between or outside the roots, and how to write it, with our own examples. Past-paper questions come in Part 2.
What you should already know
Linear inequalities
Solve them like equations, with one extra rule: multiplying or dividing by a negative number reverses the sign. \(-2x>6\Rightarrow x<-3\).
Quick check: solve \(5-3x\le11\)
\(-3x\le6\Rightarrow x\ge-2\)
Number lines
\(x>2\) is everything to the right of 2, with an open dot (2 itself is not included). \(x\ge2\) uses a filled dot. \(-1<x<4\) is one stretch between two numbers.
Quick check: is \(x=4\) in \(-1<x<4\)? in \(-1<x\le4\)?
no; yes
Factorising and the formula
You need the roots of the quadratic, so all of Solving quadratic equations is used here.
Quick check: roots of \(2x^2-x-3\)?
\((2x-3)(x+1)\): \(\tfrac32\) and \(-1\)
Sign of a product
Positive × positive and negative × negative are positive; mixed signs give a negative product.
Quick check: sign of \((x-3)(x+1)\) at \(x=0\)?
\((-3)(1)=-3\): negative
What a quadratic inequality asks
An equation like \(x^2-x-6=0\) has a few answers (here \(x=3\) and \(x=-2\)). An inequality like \(x^2-x-6<0\) asks for every \(x\) that makes the left side negative, and the answer is a whole range of numbers. Test a few values:
| \(x\) | \(-3\) | \(-2\) | \(0\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|
| \(x^2-x-6=(x-3)(x+2)\) | \(6\) | \(0\) | \(-6\) | \(-4\) | \(0\) | \(6\) |
| sign | + | 0 | − | − | 0 | + |
The sign changes only at \(-2\) and \(3\), the roots. Everything between them is negative, so \(x^2-x-6<0\) for \(-2<x<3\). Testing values shows the pattern; the graph below explains it, and is how you will actually solve these.
The idea: a quadratic only changes sign at its roots
\(y=(x+4)(3x-2)\) is a ∪-shaped curve crossing the \(x\)-axis at \(x=-4\) and \(x=\tfrac23\). Those two numbers are the critical values. Between them the curve is below the axis (negative); outside them it is above (positive). So an inequality only ever has two kinds of answer: between the roots, or outside them.
The method
- Rearrange so one side is 0, and make the \(x^2\) coefficient positive (multiply through by −1 if needed, and reverse the inequality sign when you do).
- Find the critical values: solve the quadratic = 0 (factorise, formula or completing the square).
- Sketch a ∪ through the critical values. Ten seconds; it decides the region, so never skip it.
- Choose the region: \(<0\) (or \(\le0\)) is between the roots; \(>0\) (or \(\ge0\)) is outside.
- Write the set with the right signs (table below).
- Solve \(x^2-5x+6\le0\).
Answer
roots \(2,3\); between, ends included: \(2\le x\le3\) - Solve \(x^2>16\).
Answer
\((x-4)(x+4)>0\): \(x<-4,\ x>4\) - Solve \(3-2x-x^2\ge0\).
Answer
\(\times(-1)\): \(x^2+2x-3\le0\Rightarrow(x+3)(x-1)\le0\): \(-3\le x\le1\)
Writing the answer correctly
| Region | Write | Also correct | Wrong |
|---|---|---|---|
| between the roots (one interval) | \(-3<x<\tfrac12\) | \(x>-3,\ x<\tfrac12\); \(x>-3\) and \(x<\tfrac12\); \(\left(-3,\tfrac12\right)\) | \(x>-3\) or \(x<\tfrac12\) (that is every real number); \(\tfrac12>x>-3\) written the wrong way round is risky |
| outside the roots (two pieces) | \(x<-3,\ x>\tfrac12\) | \(x<-3\) or \(x>\tfrac12\); \((-\infty,-3)\cup\left(\tfrac12,\infty\right)\) | \(\tfrac12<x<-3\) (impossible); \(-3>x>\tfrac12\); just the critical values |
| strict (< or >) | keep the ends out | \(\le\) or \(\ge\): the end-points don't satisfy a strict inequality | |
| includes = (≤, ≥) | \(-3\le x\le\tfrac12\) | \(\left[-3,\tfrac12\right]\) | strict signs (the end-points are solutions) |
Why "or" is wrong for one interval: "\(x>-3\) or \(x<\tfrac12\)" is true for every number (\(x=100\) is \(>-3\)). Between the roots means both at once, so write one chain, \(-3<x<\tfrac12\). For two pieces, "or" (or a comma) is right. And answer in the letter of the question: an inequality in \(k\) has its answer in \(k\).
Case by case
1 Between the roots
Find the set of values of \(x\) for which \(2x^2+5x-3<0\).
Askedset of xKnowa ∪ quadratic, < 0Routecritical values, then between
- \(2x^2+5x-3=0\Rightarrow(2x-1)(x+3)=0\): \(x=\tfrac12,\ -3\)
- \(<0\) is between the roots
- \(\mathbf{-3<x<\tfrac12}\)
2 Outside the roots, after a negative x²
Find the set of values of \(x\) for which \(4+5x-6x^2<0\).
Askedset of xKnowthe \(x^2\) coefficient is negativeRoutemultiply by −1 (reverse the sign), then outside
- \(\times(-1)\): \(6x^2-5x-4>0\) (sign reversed)
- \((3x-4)(2x+1)=0\Rightarrow x=\tfrac43,\ -\tfrac12\)
- \(>0\) is outside: \(\mathbf{x<-\tfrac12,\ x>\tfrac43}\)
3 Not factorisable
Solve \(x^2-4x-1\ge0\).
Askedset of xKnow≥ 0, roots are surdsRouteformula or completing the square, then outside, ends included
- \((x-2)^2-5=0\Rightarrow x=2\pm\sqrt5\)
- \(\ge0\) is outside, ends included
- \(\mathbf{x\le2-\sqrt5,\ x\ge2+\sqrt5}\)
4 An inequality in k from a discriminant
The discriminant work ends with \((k+2)(k-6)>0\). Solve it for \(k\).
Askedset of kKnowcritical values −2 and 6Route∪ in k, > 0, outside
- Critical values \(k=-2,\ 6\)
- \(\mathbf{k<-2,\ k>6}\) (strict, as the original inequality was)
5 Both sides need moving
Solve \(x(x+2)<15\).
Askedset of xKnownot = 0 yetRouteexpand, move 15 across, factorise
- \(x^2+2x-15<0\Rightarrow(x+5)(x-3)<0\)
- \(\mathbf{-5<x<3}\)
"x < 15 or x + 2 < 15" is the trap: a product is only compared with 0, never with another number.
- Solve \(x^2<2x+8\).
Answer
\(x^2-2x-8<0\Rightarrow(x-4)(x+2)<0\): \(-2<x<4\) - Find the set of values of \(k\) for which \(k^2-3k-10\ge0\).
Answer
\((k-5)(k+2)\ge0\): \(k\le-2,\ k\ge5\)
Key facts at a glance
- A quadratic only changes sign at its roots (the critical values).
- Make the \(x^2\) coefficient positive first; multiplying by \(-1\) reverses the sign.
- ∪ shape: \(<0\) between the roots, \(>0\) outside.
- Between: one chain \(a<x<b\), never "or". Outside: \(x<a,\ x>b\).
- Strict stays strict; \(\le,\ \ge\) include the ends. Never square-root both sides.
How inequalities are examined, what the mark scheme accepts, then real Paper 1 questions worked through slowly.
How this is examined
A quadratic inequality is rarely a question on its own: it finishes a discriminant question (the set of \(k\)) or a calculus one (where a function decreases). The marks: M1 A1 for the critical values with a method shown, A1 for the final set in an accepted form. The March 2022 report: "where a sketch was used the correct regions were nearly always selected". Schemes give B0 for \(\le\) or \(\ge\) when the question is strict.
The "or" rule has tightened: the 2022 scheme condoned "\(-8<k\) or \(k<8\)" 9709/12/O/N/22 Q3, but the 2025 schemes say "Do not allow 'or'" for a single interval 9709/12/F/M/25 Q1 and "not \(x>-4\) or \(x<\tfrac23\)" 9709/13/M/J/25 Q7. Write a single interval as one chain, \(a<x<b\), and you can never be marked down. Use the right letter too: if the question is about \(k\), the answer is in \(k\) ("Must be k", 9709/11/M/J/25 Q6).
Mark-scheme focus
Find the set of values of x for which f(x) is decreasing
![A function f is such that f′(x) = 6(2x − 3)² − 6x for x ∈ R. (a) Determine the set of values of x for which f(x) is decreasing. [4]](/notes-content/9709/quadratics/img/9709_s24_12_q09_a.png)
- \(6(2x-3)^2-6x<0\) (less than 0; \(\le0\) condoned here) B1*
- expanded to a 3-term quadratic: \(4x^2-13x+9<0\) M1
- critical values \(1\) and \(\tfrac94\) B1
- \(1<x<\tfrac94\), or "\(x>1\) and \(x<\tfrac94\)" DB1FT
- "\(x>1\) or \(x<\tfrac94\)", or "\(x>1,\ x<\tfrac94\)" (both refused on this scheme)
- \(f'(x)>0\): that is where f increases
- the critical values with no inequality
"Decreasing" means f′(x) < 0, so the answer is where the quadratic is negative: between the roots, written as one chain.
consistent \(\le\) and \(\ge\) (\(1\le x\le\tfrac94\)) was condoned on this scheme; strict signs are always safe.
From: 9709/12/M/J/24 Q9(a).
Find the set of values of k for which the line and curve do not meet
![A curve has equation y = 5 + 3x − 2x² and a straight line has equation y = kx + 13, where k is a constant. Find the set of values of k for which the curve and the line do not meet. [4]](/notes-content/9709/quadratics/img/9709_m25_12_q01.png)
- \(2x^2+(k-3)x+8=0\) B1; \(b^2-4ac<0\) M1
- critical values \(-5\) and \(11\) A1
- \(-5<k<11\) A1
- "\(k>-5\) or \(k<11\)" ("Do not allow 'or'")
- \(-5\le k\le11\) ("A0 if \(\le\) sign or signs used")
- only the critical values (the March 2025 report: some omitted the interval entirely)
From: 9709/12/F/M/25 Q1 and the March 2025 report.
- The final line of every discriminant question is a quadratic inequality (2020-25: about 14 questions).
- Increasing / decreasing: dy/dx > 0 or < 0, a quadratic inequality, 3-4 marks (Differentiation).
- The last A1 is for the notation: strict signs, no "or" for one interval, the right letter.
Worked examples
Example 1 · Where a curve is decreasing (Paper 1, 3 marks) ★ Easy
![A curve is such that dy/dx = 3x² + 10x − 8. (a) Find the set of values of x for which y decreases as x increases. [3]](/notes-content/9709/quadratics/img/9709_s25_13_q07_a.png)
We are told the gradient. The curve goes down wherever the gradient is negative. So we need the \(x\)-values that make \(3x^2+10x-8\) negative.
- decreases as x increases\(\dfrac{dy}{dx}<0\)
- set of valuesan inequality in x
- (a)set of \(x\)
- \(\dfrac{dy}{dx}=3x^2+10x-8\)
- Findset of x
- Use\(3x^2+10x-8<0\)
- Roots\((3x-2)(x+4)=0\)
- Region< 0: between
- Use\(3x^2+10x-8<0\)
Which two numbers multiply to −24 and add to 10?
12 and −2: \(3x^2+12x-2x-8=3x(x+4)-2(x+4)\).
Between or outside?
Between: a ∪ is below the axis between its roots.
5 Now solve it have a go on paper first, then open the worked solution
- \(3x^2+10x-8<0\) *B1
- \((3x-2)(x+4)=0\Rightarrow x=\tfrac23,\ -4\) B1
- \(\mathbf{-4<x<\tfrac23}\) DB1
Test \(x=0\) (inside): \(-8<0\) ✓ decreasing. Test \(x=1\): \(5>0\), increasing, correctly excluded.
Example 2 · A negative x² and outside regions (Paper 1, 4 marks) ★★ Medium
![The equation of a curve is y = 4 + 5x + 6x² − 3x³. (a) Find the set of values of x for which y decreases as x increases. [4]](/notes-content/9709/quadratics/img/9709_w24_11_q09_a.png)
Differentiate, then ask where the gradient is negative. The gradient is a quadratic with a negative \(x^2\) term, so flip it first.
- decreases\(\dfrac{dy}{dx}<0\)
- \(-9x^2\)multiply by −1 and reverse the sign
- (a)set of \(x\)
- \(y=4+5x+6x^2-3x^3\)
- Findset of x
- First\(\tfrac{dy}{dx}=5+12x-9x^2\)
- Use\(5+12x-9x^2<0\Leftrightarrow9x^2-12x-5>0\)
- Region> 0: outside
- Trapforgetting to reverse the sign
Factorise 9x² − 12x − 5.
\((3x-5)(3x+1)\): check \(-15x+3x=-12x\) ✓.
Why is the answer now two pieces?
After flipping we need \(>0\): outside the roots.
5 Now solve it have a go on paper first, then open the worked solution
- \(\dfrac{dy}{dx}=5+12x-9x^2\) B1
- Critical values: \(9x^2-12x-5=0\Rightarrow(3x+1)(3x-5)=0\) M1
- \(x=-\tfrac13,\ \tfrac53\) A1
- \(\mathbf{x<-\tfrac13,\ x>\tfrac53}\) A1FT
\(x=2\): \(5+24-36=-7<0\) ✓ decreasing. \(x=0\): \(5>0\), excluded ✓.
Example 3 · Two distinct intersections: an inequality in m (Paper 1, 5 marks) ★★ Medium
![A curve has equation y = 3x² − 4x + 4 and a straight line has equation y = mx + m − 1, where m is a constant. Find the set of values of m for which the curve and the line have two distinct points of intersection. [5]](/notes-content/9709/quadratics/img/9709_w20_13_q04.png)
Where the line meets the curve, their \(y\)-values are equal. That gives a quadratic in \(x\) with \(m\) in it. Two distinct meeting points means two distinct roots: discriminant \(>0\). The last step is an inequality in \(m\).
- two distinct points of intersection\(b^2-4ac>0\)
- y = mx + m − 1m appears in b and in c
- set of \(m\)
- \(y=3x^2-4x+4\)curve
- \(y=mx+m-1\)line
- Findset of m
- First\(3x^2-(4+m)x+(5-m)=0\)
- Use\((4+m)^2-12(5-m)>0\)
- Then\(m^2+20m-44>0\)
- Regionoutside
Collect terms: what is the constant term?
\(4-(m-1)=5-m\).
Factorise m² + 20m − 44.
\((m+22)(m-2)\).
5 Now solve it have a go on paper first, then open the worked solution
- \(3x^2-4x+4=mx+m-1\Rightarrow3x^2-(4+m)x+(5-m)=0\) M1
- \(b^2-4ac=(4+m)^2-4(3)(5-m)\) M1 \(=m^2+20m-44\) A1
- \((m+22)(m-2)>0\): critical values \(-22,\ 2\) A1
- \(\mathbf{m<-22,\ m>2}\) A1
The November 2020 report: common errors were "errors in collecting terms at the beginning, omitting a term or making sign errors" and leaving the answer as "−22, 2 rather than applying correct inequalities".
\(m=3\): \(3x^2-7x+2=0\), \(D=49-24=25>0\) ✓ two points.
Summary
- Make the \(x^2\) coefficient positive (reverse the sign if you multiply by a negative), find the critical values with a method shown, sketch.
- \(<0\): between, \(a<x<b\). \(>0\): outside, \(x<a,\ x>b\).
- Strict unless the question includes \(=\); never "or" for a single interval; answer in the question's letter.
| If you see… | solve |
|---|---|
| "decreasing" / "y decreases as x increases" | \(\tfrac{dy}{dx}<0\) |
| "increasing" | \(\tfrac{dy}{dx}>0\) |
| "do not meet" / "no real roots" | \(b^2-4ac<0\): usually between |
| "two distinct points" | \(b^2-4ac>0\): usually outside |
| \(x^2>9\), \(x^2\le16\) | move everything to one side and factorise; never "\(\pm\)" an inequality |
Can I do this?
- I can find critical values and decide between/outside with a sketch.
- I reverse the inequality when I multiply by a negative number.
- I write the answer in an accepted form (strict signs, no "or" for one interval, the right letter).
- I can solve inequalities from discriminants and from dy/dx.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
The multiple-choice and guided questions need JavaScript. The concept checks and exam practice below work without it.
Concept checks
Answer each one in your head first, then open it.
1Why can't you divide both sides of \(x^2>3x\) by \(x\)?Show answer
\(x\) might be negative (which would reverse the sign) or zero. Write \(x^2-3x>0\Rightarrow x(x-3)>0\Rightarrow x<0,\ x>3\).
2Is \(x^2+4>0\) true for all \(x\)?Show answer
Yes: \(x^2\ge0\), so \(x^2+4\ge4\). Its solution set is every real number.
3An answer reads \(5<x<-1\). What went wrong?Show answer
No number is both more than 5 and less than −1. The intended answer was probably \(x<-1,\ x>5\) (outside).
Exam-style practice
New questions, not used above. Do them on paper, then mark your work against the official mark scheme.
![Find the set of values of k for which the curve y = kx² − 3x and the line y = x − k do not meet. [3]](/notes-content/9709/quadratics/img/9709_w16_13_q01.png)
Show the mark scheme
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(kx^2-3x=x-k\Rightarrow kx^2-4x+k=0\) | M1 | eliminate y, 3-term quadratic | |
| \(b^2-4ac=16-4k^2\) (or \((4-2k)(4+2k)\)) | M1 | discriminant seen or implied | |
| \(k<-2,\ k>2\) | A1 | do not meet: \(16-4k^2<0\), outside regions; \((-\infty,-2)\cup(2,\infty)\) accepted | |
| Total: 3 marks | |||
Examiner tip Watch the direction: the inequality is \(16-4k^2<0\), i.e. \(k^2>4\). That is OUTSIDE \(\pm2\), not between.
![A line has equation y = 3kx − 2k and a curve has equation y = x² − kx + 2, where k is a constant. (i) Find the set of values of k for which the line and curve meet at two distinct points. [4]](/notes-content/9709/quadratics/img/9709_w19_13_q06_i.png)
Show the mark scheme
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| (i) | \(x^2-4kx+2k+2=0\) | B1 | the \(kx\) terms combined correctly |
| \(b^2-4ac=16k^2-4(2k+2)\) | M1 | a quadratic in k formed | |
| critical values \(1\) and \(-\tfrac12\) | A1 | from \(2k^2-k-1=(2k+1)(k-1)\) | |
| \(k>1,\ k<-\tfrac12\) | A1 | outside regions, strict signs | |
| Total: 4 marks | |||
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