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Inequalities are statements about order — that one quantity is strictly less than, strictly greater than, or no greater than another. At GCSE you solved short linear inequalities and represented their solutions on a number line. The Edexcel 9MA0 A-Level specification dramatically extends the range and depth: you must solve linear, quadratic, and rational inequalities; manipulate compound (double) inequalities; reason carefully about the inversion that occurs when multiplying or dividing by a negative; and present solutions fluently in inequality, set-builder, and interval notations. The reasoning skills built here recur throughout the course — in domain analysis, in finding ranges of validity for binomial expansions, in optimisation problems with constraints, and in convergence arguments at university level.
This lesson develops a unified framework. We treat an inequality as a problem about signs of an underlying expression. We always factorise to a product (or quotient) of linear factors, locate the critical values where each factor changes sign, and then read off the solution from a sign diagram. This single method handles every Edexcel-style inequality you will meet at A-Level — quadratic, cubic, or rational.
Edexcel 9MA0 specification, Pure Mathematics Paper 1 — Section 2.7 covers linear and quadratic inequalities in a single variable and interpret such inequalities graphically, including inequalities with brackets and fractions. Express solutions through correct use of 'and' and 'or', or through set notation. Represent linear and quadratic inequalities such as y>x+1 and y>ax2+bx+c graphically (refer to the official specification document for exact wording).
This sub-strand connects directly to section 2.4 (quadratic functions and the discriminant — covered in lesson 2), section 2.6 (simultaneous equations — lesson 3, since combined inequalities behave as systems), section 2.8 (polynomials — lesson 5, where higher-degree inequalities are first met), and section 2.10 (modulus functions — lesson 9, where inequalities involving ∣f(x)∣ require case-splitting). Differentiation problems in section 7 frequently end with an inequality such as f′(x)>0 characterising intervals of increase, so the techniques of this lesson power large parts of the calculus syllabus too.
The Edexcel formula booklet does not list any inequality-specific formulae — you are expected to manipulate by inspection. The discriminant b2−4ac, however, is the principal analytic tool when a quadratic inequality has no real roots and you must reason about whether the parabola lies entirely above or below the x-axis.
A linear inequality has the form ax+b>0 (or ≥, <, ≤). Solve it as you would a linear equation, with one critical exception: multiplying or dividing both sides by a negative number reverses the inequality sign.
Example 1. Solve 3x−5≤7.
Add 5 to both sides: 3x≤12. Divide by 3 (positive — sign unchanged): x≤4.
Example 2. Solve 2−5x>17.
Subtract 2: −5x>15. Divide by −5 — reverse the inequality — to obtain x<−3.
The reversal is not optional. Geometrically, multiplying by a negative number reflects the real line through the origin, so the order of any two points is swapped. Algebraically: if a>b and c<0, then ac<bc. This is one of the order axioms of R, and it underpins all subsequent inequality manipulation.
A safer alternative — strongly recommended in exam conditions — is to avoid negative coefficients on the variable in the first place. Add 5x to both sides of 2−5x>17 to obtain 2>17+5x, then subtract 17: −15>5x, divide by 5 (positive!): −3>x, i.e. x<−3. No reversal needed; no risk of slip.
A double inequality such as −3<2x+1≤7 asserts both −3<2x+1 and 2x+1≤7 simultaneously. Solve by performing identical operations on all three parts.
Example 3. Solve −3<2x+1≤7.
Subtract 1 from each part: −4<2x≤6. Divide each part by 2: −2<x≤3.
The solution is the intersection of the two single inequalities. If at any stage you must multiply or divide by a negative, all three parts reverse simultaneously.
Example 4. Solve 1≤5−2x<9.
Subtract 5: −4≤−2x<4. Divide by −2 and reverse both inequalities: 2≥x>−2, which is more naturally written −2<x≤2.
A quadratic inequality is one of the form ax2+bx+c>0 (or <, ≥, ≤). The strategy is always the same:
Example 5. Solve x2−x−6>0.
Factorise: (x−3)(x+2)>0. The roots are x=−2 and x=3. The leading coefficient is positive, so the parabola opens upwards; the curve is above the x-axis when x<−2 or x>3, and below between the roots.
The solution is therefore x<−2 or x>3. In set notation: {x∈R:x<−2}∪{x∈R:x>3}. In interval notation: (−∞,−2)∪(3,∞).
The sketch is the single most reliable tool here — it eliminates almost all sign errors. Always draw it, even when the algebra looks obvious.
Example 6. Solve 2x2+5x−3≤0.
Factorise: (2x−1)(x+3)≤0. Roots: x=−3 and x=21. Leading coefficient positive, so the parabola opens upwards; the curve is below or on the x-axis between the roots inclusive. Solution: −3≤x≤21.
When the coefficient of x2 is negative, the parabola opens downwards. You can either reason from a downward parabola directly, or — preferred in exams — multiply through by −1 first (reversing the inequality) and reduce to Type 1.
Example 7. Solve −x2+4x+5≥0.
Multiply by −1 and reverse: x2−4x−5≤0. Factorise: (x−5)(x+1)≤0. Roots −1 and 5, upward parabola; below or on the axis between the roots.
Solution: −1≤x≤5, i.e. x∈[−1,5].
The temptation, when seeing −x2+4x+5≥0, is to forget that the parabola opens downwards and write the solution as the exterior of the roots. Multiplying by −1 at the start removes this risk entirely.
If the discriminant Δ=b2−4ac<0, the quadratic ax2+bx+c does not cross the x-axis. The expression therefore has constant sign — entirely positive (if a>0) or entirely negative (if a<0).
Example 8. Solve x2+x+1>0.
Discriminant: 1−4=−3<0. No real roots. Leading coefficient positive, so x2+x+1>0 for all real x. Solution: x∈R.
A common A vs A* slip is to write "no solution" because the quadratic does not factorise — the right reading is that it has every real number as a solution.
A critical value is a value of x where the underlying expression changes sign — for a polynomial, the roots; for a rational expression, the roots of numerator and denominator both. Between consecutive critical values the sign is constant, so testing one point in each interval determines the sign across the whole interval. This is the sign-diagram method.
Edexcel accepts solutions in three notations, and you must be fluent in all three.
| Inequality form | Set-builder notation | Interval notation |
|---|---|---|
| x>3 | {x∈R:x>3} | (3,∞) |
| x≤−1 | {x∈R:x≤−1} | (−∞,−1] |
| −2<x≤5 | {x∈R:−2<x≤5} | (−2,5] |
| x<0 or x>4 | {x:x<0}∪{x:x>4} | (−∞,0)∪(4,∞) |
| All real x | R | (−∞,∞) |
| Empty | ∅ | ∅ |
A round bracket ( or ) excludes the endpoint (corresponds to strict < or >); a square bracket [ or ] includes it (corresponds to ≤ or ≥). The infinite endpoints −∞ and ∞ always take round brackets — infinity is not a real number and so cannot be "included".
Edexcel asks candidates to represent inequalities on a number line. The conventions are:
For x<−2 or x>3, you draw two open circles and shade outwards from each: an arrow leftwards from −2 and an arrow rightwards from 3, with the segment [−2,3] left blank.
For −1≤x≤5 you draw two closed circles and shade the segment between them.
Two inequalities can be combined either by intersection (both must hold — written "and") or by union (at least one must hold — written "or").
"and" — intersection. x≥2 and x≤7 gives 2≤x≤7. The solution is the overlap of the two solution sets.
"or" — union. x<0 or x>4 gives the disjoint union (−∞,0)∪(4,∞) — there is no overlap and the inequalities cannot be merged into a single double inequality.
A common confusion: the solution to a quadratic with positive leading coefficient and f(x)>0 is always an "or" (exterior of roots), while f(x)<0 is always an "and" (interior of roots). Conflating these — writing the exterior with "and" — produces the empty set, which is almost never the intended answer.
Inequalities involving an algebraic fraction such as x+2x−1>0 are an A-Level extension. The naive move — multiplying both sides by x+2 — is dangerous, because the sign of x+2 depends on x. If you multiply by a negative quantity without reversing the inequality, the answer is wrong. There are two safe methods.
Method A: Critical values and sign diagram. Mark the critical values where the numerator is zero (x=1, the fraction equals zero) and where the denominator is zero (x=−2, the fraction is undefined and must be excluded). These split the real line into three intervals: x<−2, −2<x<1, x>1. Test one point per interval.
Method B: Multiply by the square of the denominator. Since (x+2)2≥0 for all x=−2, multiplying through preserves the inequality direction (provided we exclude x=−2). Then x+2x−1>0 becomes (x−1)(x+2)>0 subject to x=−2, which is now a standard quadratic inequality.
Example 9 — rational inequality. Solve x+2x−1>0.
By Method B: (x−1)(x+2)>0 with x=−2. Roots 1 and −2, upward parabola; the product is positive outside the roots. Solution: x<−2 or x>1.
The exclusion x=−2 matters because at x=−2 the original fraction is undefined. The strict inequality >0 at x=1 also forces x=1 (the fraction is zero there, which is not greater than zero).
| Interval | Sign of x−1 | Sign of x+2 | Sign of fraction |
|---|---|---|---|
| x<−2 | − | − | + |
| −2<x<1 | − | + | − |
| x>1 | + | + | + |
The fraction is positive precisely in the first and third rows: x<−2 or x>1, in agreement with Method B.
Example 10 — combined system. Find the set of values of x for which both 2x−3≥1 and x2−5x+6<0 hold.
Linear part: 2x≥4, i.e. x≥2.
Quadratic part: factorise x2−5x+6=(x−2)(x−3). Roots 2 and 3, upward parabola, expression negative between the roots: 2<x<3.
Intersection (both must hold): x≥2 and 2<x<3 gives 2<x<3, since the strict >2 from the quadratic dominates the weak ≥2 from the linear.
Solution: x∈(2,3).
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