Quadratic inequalities, interpreted and represented graphically

~45 min · WMA11 · 1.7

WMA11 · 1.7 · 45 min

A quadratic inequality only ever asks one question: is the curve above the axis, or below it, at this x? Pearson's own examiner report on the first time this content was ever examined calls it plainly 'a new topic within the qualification' and records that 'it was disappointing that only a minority of candidates gained full marks' (Jan 2019, Q4) — not because the algebra is unfamiliar (the roots are found exactly as before), but because the final line has to be read correctly off a picture, in the right variable. A second series shows the same picture doing work a purely algebraic approach cannot: skipping the sketch is where a sign case goes missing.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What a quadratic inequality is actually asking — and why the graph answers it directly

Spec 1.7 names the skill precisely: 'interpret linear and quadratic inequalities graphically.' An inequality like $f(x) > 0$ is a question about height — at this value of $x$ , is the curve above the x-axis or below it? $f(x) < 0$ asks the opposite question. This connects directly to work already done on the : the roots of $f$ are exactly the x-values where the curve's height is zero, the only places it can cross from one side of the axis to the other. Nothing about finding those roots is new here — factorisation, the formula and all still apply, in exactly the roles they had before.

What IS new is what counts as 'the answer.' In the discriminant lesson, the roots themselves — or a range of values for a parameter — were the final thing being asked for. Here the roots are only the BOUNDARY of the answer; the actual answer is a whole interval, or a pair of separate intervals, of x. Stating that interval correctly, in the right variable and with the right strictness, is where this topic's marks are actually lost — not in finding the boundary itself.

Worth taking at face value: Pearson's own examiner report on the first sitting to include this content states plainly that this is 'a new topic within the qualification' and records that 'it was disappointing that only a minority of candidates gained full marks' (Jan 2019, Q4). That is a specific, documented statement about candidate performance on exactly this material, not a generic warning — a real signal that the difficulty here is genuinely in the reading-off step, not in unfamiliar algebra.

Two steps, every time: find the boundary, then read the region off the shape

Step one is unchanged from the discriminant/completing-the-square lesson: rearrange so one side is zero, then solve $f(x) = 0$ to find the critical values, by whichever of the three equally-creditable routes suits the numbers — factorisation, the formula, or completing the square.

Step two is the genuinely new part (spec 1.9): use the SHAPE of the curve, decided entirely by the sign of $a$ , to choose which side of the boundary the inequality wants. Two facts to hold in mind for now — the mechanism block just below derives both from one argument, so neither needs to be memorised as a separate rule: an upward-opening parabola ( $a>0$ ) is positive OUTSIDE its roots and negative BETWEEN them; a downward-opening one ( $a<0$ ) is the exact reverse.

A worked mini-example. Solve $x^2 - x - 6 > 0$ . Factorise: $(x-3)(x+2) = 0$ gives critical values $x = 3$ and $x = -2$ . Here $a = 1 > 0$ , so $f(x) > 0$ is the 'outside the roots' case: $x < -2$ or $x > 3$ . Check with $x=0$ , which sits between the roots: $f(0) = -6$ , correctly negative, confirming that 'between' is the negative region and 'outside' is where the strict inequality above is satisfied. This is exactly the curve plotted as C1 in the diagram just below, with that same check point at $x=0$ marked on it — the dip to $f(0)=-6$ between the roots and the rise back through zero at $x=-2$ and $x=3$ are the actual shape being sketched there, not just a description in words.

If the inequality compares two DIFFERENT expressions rather than one expression to zero — a curve and a line, say, or two curves — rearrange everything onto one side first, so the question becomes '(something) compared to 0.' That single step reduces every version of this question type to the same two-step method above; the worked-chain and marked-solution blocks below both use it.

Mechanism

Why 'outside' and 'between' aren't two rules to memorise — it's one sign pattern

Write the quadratic in factorised form, $f(x) = a(x-p)(x-q)$ , with $p < q$ the two DISTINCT real roots (which exist only when $b^2-4ac > 0$ , strictly — the same discriminant condition from the previous lesson, tightened here to its strict form because $p<q$ itself already rules out the repeated-root case $b^2-4ac=0$ , where the curve only touches the axis and there is no 'between the roots' region to speak of). Consider the product $(x-p)(x-q)$ on its own, ignoring $a$ for a moment. For $x < p$ : both $(x-p)$ and $(x-q)$ are negative, so their product is positive. For $p < x < q$ : $(x-p)$ is positive but $(x-q)$ is still negative, so the product is negative. For $x > q$ : both factors are positive, so the product is positive again. That sign pattern — positive, negative, positive — is a plain consequence of two linear expressions each changing sign only at their own zero and nowhere else; it has nothing specifically to do with quadratics, and it holds for ANY $p < q$ , not particular ones. Now bring $a$ back in: multiplying by a positive $a$ leaves every sign exactly as it was, giving the 'upward: outside positive, between negative' rule stated above. Multiplying by a negative $a$ flips every sign, giving 'downward: outside negative, between positive' — exactly its reverse. A single test point substituted into ANY interval settles that interval's sign directly, and is often faster in the exam than trying to recall which shape goes with which case.

Diagram — Two shapes, one sign pattern — the four regions a quadratic inequality can ask about

x-axis: x · y-axis: y

C1 · y = x² − x − 6 (a = 1 > 0), roots p = −2, q = 3: Crosses the x-axis at p=−2 and q=3 — the same curve as the worked mini-example above. Between them the curve dips below the axis — f(x)<0, down to a minimum of −6.25 at x=0.5 — because the two linear factors (x+2) and (x−3) carry opposite signs there. Outside them, both factors share a sign, so the product — and f(x) — is positive: f(−3)=6, f(4)=6.
C2 · y = −x² + 5x − 4 (a = −1 < 0), roots r = 1, s = 4: Crosses the x-axis at r=1 and s=4. Every sign from C1 is reversed here, because a=−1 flips the product's sign at every x without moving where the curve actually crosses the axis: f(x)>0 between the roots, up to a maximum of 2.25 at x=2.5, and f(x)<0 outside them: f(−3)=−28, f(5)=−4.

p = −2 — C1's left root: One of the two x-values where f(x)=0 — the boundary the whole inequality question hinges on, and the same value the worked mini-example above finds by factorising (x+2)(x−3).
q = 3 — C1's right root: The other x-value where f(x)=0. Nothing about finding p and q changes for an inequality question — it is the same discriminant/factorisation/formula calculation as before; what's new is reading the region off the shape between them.
r = 1 — C2's left root: The downward parabola's roots, found the same three ways as any other quadratic. C2's shape is the mirror image of C1's around the x-axis: same method for the roots, opposite sign pattern either side of them.
s = 4 — C2's right root: C2's other root. With r and s fixed, the sign either side is decided purely by a being negative — the same factorised-form argument as C1, run with the opposite sign for a.
One sign test per interval: A single value picked from an interval is enough to fix the sign of the whole interval, because a continuous function can only change sign at a root. f(0)=−6, strictly between p=−2 and q=3, confirms C1 is negative there — a check that works even if the 'outside/between' rule is misremembered.
The same one derivation, run on C2: g(2)=2, strictly between r=1 and s=4, is positive — the mirror image of C1's negative dip, from exactly the same factorised-form argument with a flipped in sign. All four sign patterns on this diagram come from that one derivation; none of them needs memorising as an independent case.

Common error: Reading the boundary correctly off the sketch — say, x = 16 — and then writing the final line of the answer in the wrong letter, or against the label of a shaded region on the diagram rather than the axis it actually sits on.

Correct: The boundary lives on the x-axis, however the question is dressed up (a shaded region, a named 'R', a picture with y on the vertical axis) — the final answer is stated in x (or whichever letter the axis and the question itself use), never in y and never in a region's label.

examiner-report · Jan 2019 · Q4

Representing the answer: a number line, or a shaded region with a solid or dashed boundary

Spec 1.8 is explicit that representing an inequality graphically needs two conventions together: shading (which side or region satisfies it) and a dotted-versus-solid boundary line — verbatim, 'Represent linear and quadratic inequalities graphically (shading, dotted/solid line convention required).' The rule is the same one used anywhere strict and non-strict inequalities are drawn: a SOLID boundary means the boundary itself is included ( $\leq$ or $\geq$ ); a DOTTED or DASHED boundary means it is excluded ( $<$ or $>$ ). On a number line this is usually a filled circle (included) versus an open circle (excluded) at each critical value, doing exactly the same job.

The practical version for this exam: before drawing or writing anything, look back at the ORIGINAL inequality sign. If it was strict, both the picture (dashed line, open circle) and the final written inequality ( $<$ or $>$ ) must stay strict all the way through; if it was not, both stay non-strict ( $\leq$ / $\geq$ , solid line, filled circle). Losing track of which one the question started with, partway through several lines of working, costs an easy mark far more often than any error in the algebra itself does.

Honest note on the evidence: this research pass's examiner-report material did not surface a specific documented WMA11 script error tied to the dotted/solid convention on its own — unlike the wrong-variable and skipped-sketch traps below, which are directly quoted. It is taught here as real, required spec content (1.8's own wording, quoted above) rather than left out for want of a matching quote.

Spec 1.7 and 1.8 also run in the OTHER direction from every example so far: instead of solving an inequality and then sketching the answer, a question can show the picture FIRST and ask you to WRITE DOWN the inequalities that define a shaded or unshaded region. This is a confirmed, recurring real WMA11 question type — seen independently in two separate series testing exactly this skill — and it usually needs inequalities in BOTH x and y together: one read straight off each boundary line or curve (no calculation needed — an independent mark, since 'above this curve' or 'below this line' is a fact about the picture, not something to solve for), plus one found by actually solving where a boundary curve crosses an axis (a genuine method mark). The worked example just below shows the full mark-by-mark structure real mark schemes use for this question type.

Marked, line by line

A region $R$ is bounded by the line $m$ with equation $2y = x$ , the curve $D$ with equation $y = 3x - \frac{1}{6}x^2$ , and a line $n$ , parallel to the y-axis, passing through the point where $D$ crosses the positive x-axis. Given that $R$ lies below $m$ , above $D$ , and to the left of $n$ , identify the inequalities that define $R$ . (3 marks) — VERIDIAN-original, structurally modelled on a real WMA11 question type confirmed independently in TWO series: Jan 2019 Q4 (a line, a quadratic curve and a vertical boundary from the curve's own positive x-intercept — exactly this shape) and Oct 2021 Q3(ii) (a line, a quadratic curve and a boundary from where the two meet — the same skill in a different specific configuration). Both real mark schemes credit this skill with the same three-mark-type structure used below: a B1 for reading an inequality straight off the picture, an M1 for finding a numeric boundary by a genuine method, and an A1 for the complete, correctly-strict answer. See the closing flag for exactly what is real and what is original here.

3 marks available

01
Either $2y \leq x$ or $y \geq 3x - \frac{1}{6}x^2$
Independent mark for correctly reading EITHER ONE of the two boundary inequalities straight off the picture — no method is needed here, since 'below m' and 'above D' are facts about the sketch, not calculations. This is the mark type spec 1.7's own wording ('interpret ... graphically') is actually testing: a B mark, not an M mark, because there is nothing to solve at this stage.
B1
02
$3x - \frac{1}{6}x^2 = 0 \Rightarrow x = 18 \Rightarrow x < ... \text{ or } x \leq ...$
Method mark for finding the third boundary: set the curve's own equation to zero to find where it meets the x-axis, then use that value to state a bound on x. This is the only one of the three inequalities that needs a genuine calculation — the other two are read directly off the sketch, which is exactly why this line, and only this line, earns an M rather than a B.
M1
03
$x < 18$ , $2y \leq x$ and $y \geq 3x - \frac{1}{6}x^2$
Accuracy mark for the complete, correctly-strict SET of all three inequalities stated together, in x and y — the variables the region actually lives in — never collapsed into, or replaced by, a single label for the region itself.
A1

Worked, in full

Solve $x^2 - 2x - 15 \leq 0$ and $3x - 1 > 5$ together — the interval a number line catches and pure algebra can lose

01
Solve the quadratic inequality on its own first. $x^2-2x-15=(x-5)(x+3)$ , so the critical values are $x=-3$ and $x=5$ . Since $a=1>0$ , $\leq 0$ is the 'between the roots' case established above — and it is non-strict, so both endpoints are included: $-3 \leq x \leq 5$ .
Earns: M1 A1 — a correct method for the critical values (any of the three named routes) and the correct inclusive region.
02
Solve the linear inequality on its own, completely separately. $3x-1>5 \Rightarrow 3x>6 \Rightarrow x>2$ — an ordinary linear inequality, solved exactly like a linear equation, since the coefficient of x stays positive throughout (spec 1.9 covers this case too, not only the quadratic one).
Earns: M1 — a correct method for the linear inequality.
03
Put both solution sets on the same number line rather than trying to combine the two inequalities symbolically. Mark $-3$ and $5$ as filled (included) points, since the quadratic's inequality was non-strict; mark $2$ as an open (excluded) point, since the linear one was strict. The quadratic's shaded region is the closed block from $-3$ to $5$ ; the linear's is everything to the right of $2$ .
Earns: (M1) — attempts to combine the two regions using a sketch or number line, which is the graphical step this topic is actually testing, not just the two separate algebra steps above it.
04
Read off where the two shaded regions overlap — that overlap is the actual answer, since 'and' means both conditions hold at once. It starts just after $2$ (excluded, from the strict linear inequality) and runs up to and including $5$ (included, from the non-strict quadratic one): $2 < x \leq 5$ .
Earns: A1 — the correct combined interval, with the correct strict/non-strict boundary carried through from whichever original inequality supplied it.
05
Check with one value from inside the answer and one just outside it. $x=3$ : $3^2-2(3)-15=-12\leq 0$ ✓ and $3(3)-1=8>5$ ✓ — correctly inside. $x=2$ : $3(2)-1=5$ , and $5>5$ is false, so $x=2$ is correctly excluded, confirming the open boundary found in stage 4.
Earns: Nothing on the mark scheme — a check, not a scored line, and the fastest way to catch a boundary written the wrong way round.

Beyond spec

A lesson that only ever asked for a single quadratic inequality on its own would never rehearse the specific skill — combining two regions correctly — that this topic's own spec items (1.7-1.9, plural 'inequalities') require, and that no single available examiner finding for this topic happens to be about directly.

This exact system is VERIDIAN-original — built to rehearse a general finding from a real WMA11 examiner report on a different question: 'it was rare to see a sketch graph, which would have helped students to see that x > 0 was required' (Oct 2021, Q3(i)). The real question there was a single reciprocal inequality, 3/x > 4, not two inequalities combined — its trap was losing the necessary x > 0 case when rearranging without a sketch, not losing a branch of a two-condition system. The equation is not this one and the trap is not identical, but the diagnosis still generalises honestly: skipping the sketch is exactly where a purely symbolic approach can quietly lose a condition — a case a single inequality needs, or a branch a combined one does — and a number line is the check that catches it before the final line is written.

Complete it yourself

Complete the chain — solve $x^2 - 4x - 5 \geq 0$ and $2x + 3 < 11$ together, including the branch that disappears entirely

01
Solve the quadratic inequality alone. $x^2-4x-5=(x-5)(x+1)$ , critical values $x=-1$ and $x=5$ . Since $a=1>0$ , $\geq 0$ is the 'outside the roots' case, non-strict: $x\leq -1$ or $x\geq 5$ .
02
Solve the linear inequality alone. $2x+3<11 \Rightarrow 2x<8 \Rightarrow x<4$ .

Marked, line by line

The curve $C$ has equation $y = x^2 - x - 6$ and the line $l$ has equation $y = x + 2$ . (a) Find the x-coordinates of the two points where $C$ and $l$ intersect. (3) (b) Hence write down the set of values of $x$ for which $C$ lies below $l$ . (2) — VERIDIAN-original question, built to rehearse a real, verified WMA11 finding: an examiner report on the first sitting of this content records candidates finding the correct boundary values but then stating the final answer in the wrong variable or against an unrequested label from the diagram (Jan 2019, Q4) — see the commonWrongPath below, and the closing flag for exactly which parts of this example are real and which are original.

5 marks available

(a) — 3 marks

01
$x^2 - x - 6 = x + 2$
Method mark for setting the curve's expression equal to the line's — at any intersection point the two heights agree, so this equation's solutions are exactly the x-coordinates being asked for.
M1
02
$x^2 - 2x - 8 = 0$
Accuracy mark for correctly collecting everything on one side. This is the step every version of this question type shares, whatever the two original expressions were: rearrange to (something) = 0 before applying any quadratic technique.
A1
03
$(x-4)(x+2) = 0 \Rightarrow x = 4 \text{ or } x = -2$
Accuracy mark for both roots, by any of the three routes the general principles credit equally (factorisation, shown here, the formula, or completing the square).
A1

(b) — 2 marks

101
$C - l = (x^2-x-6) - (x+2) = x^2 - 2x - 8 = (x-4)(x+2)$ — an upward parabola ( $a=1>0$ ) with roots at $x=-2$ and $x=4$ , negative BETWEEN them by the sign pattern established above.
Method mark for reasoning about the sign of the DIFFERENCE between the two expressions, and correctly identifying which region — between or outside the roots — corresponds to 'below.' C lies below l exactly where C's height minus l's height is negative.
M1
102
$-2 < x < 4$
Accuracy mark, correct answer only, stated in x — the variable the question actually asked for the answer in — and strict at both ends, matching a strict 'below' (not 'below or equal to') throughout.
A1

Named traps

final-answer-stated-in-the-wrong-variable: Confirmed directly, and by Pearson's own account a genuinely widespread error on the first sitting of this content: "stating y < 16 was a frequent error" (Jan 2019, Q4) — the number 16 there is the real, verified boundary value from that question; candidates had correctly found it and then written the final line of the answer using y, the OUTPUT variable, instead of x, the one actually being solved for. The two letters sit on the same picture, which is exactly why the swap is easy to make and easy to miss when checking your own work.
region-label-used-as-the-answer-variable: Confirmed on the same question: "others stated inequalities involving R, which gained no credit" (Jan 2019, Q4) — a diagram will sometimes name a shaded region with a letter, most often R, purely as a caption for the picture. R is not a coordinate and does not belong in an inequality about x; treating a label as if it were algebra scores nothing, however correct the reasoning behind it was.
sketch-skipped-a-sign-case-goes-unnoticed: Confirmed, though on a different underlying question type from this lesson's own worked examples: "it was rare to see a sketch graph, which would have helped students to see that x > 0 was required" (Oct 2021, Q3(i) — solving 3/x > 4 alone, a single reciprocal inequality, not a quadratic one and not itself two inequalities combined). The specific equation is not this topic's, and the real trap there was a dropped CASE (the necessary x > 0, missed by candidates who rearranged straight to x < 3/4), not a dropped branch — but the report's own explanation generalises: skipping the picture is what let a sign condition slip past a purely algebraic approach. This lesson's worked-chain and chain-drill blocks rehearse the same discipline — checking a combined region on a number line rather than symbolically — on genuinely quadratic material, adapted from that diagnosis rather than reproducing its structure.
between-and-outside-the-roots-reversed: Not a specific quoted WMA11 finding for this topic in the research pass this lesson draws on — flagged honestly as a natural, mathematically-grounded slip rather than a documented script error. Swapping which region belongs to an upward parabola and which to a downward one is exactly the failure the factorised-form derivation above exists to make unnecessary: $(x-p)(x-q)$ is negative between its roots and positive outside them regardless of any picture, and multiplying by $a$ only flips which of those counts as 'positive' — a single test point rebuilds the correct region in one line, faster than trying to recall which shape goes with which case.
boundary-strict-vs-non-strict-mismatched: Not a specific quoted WMA11 finding for this exact topic either — the same discipline IS documented on a closely related question type in this course's discriminant lesson ('some gave their final answer as 0 ≤ k ≤ 5/9, losing the final mark', Jan 2023 Q4), and it transplants directly here: a strict original inequality ( $<$ or $>$ ) demands a strict final answer, and a non-strict one ( $\leq$ or $\geq$ ) demands the boundary be included. This lesson's own marked-solution and worked-chain examples are deliberately built with different strictness at each end for exactly this reason — copying one convention onto both ends of an answer is a fast way to lose a mark that has nothing to do with the algebra.

In your own words

In one sentence: why does an upward parabola's inequality $f(x) > 0$ land OUTSIDE the roots, while a downward parabola's $f(x) > 0$ lands BETWEEN them — using the factorised form, not the picture?

Beyond the spec

Spec 1.7-1.9 explicitly restrict this technique to LINEAR and QUADRATIC inequalities — no P1 spec item authorises solving a cubic or higher-degree one, and it is not part of what WMA11 examines. It is included here because watching the same sign-pattern argument survive completely unchanged when a third root is added is the clearest possible proof that the 'outside/between' rule above was never really about quadratics specifically — it is about how many times a continuous function can change sign, and where.

Take $(x+2)(x-1)(x-3) > 0$ , a cubic with three real roots at $-2$ , $1$ and $3$ . Each factor is linear and changes sign only at its own root, nowhere else — so the whole product can only change sign AT $x=-2$ , $x=1$ or $x=3$ , and stays one constant sign on every interval between consecutive roots. Test one point per interval to find which sign. At $x=-3$ (below all three roots): every factor is negative, and a product of three negatives is negative. Crossing $x=-2$ flips exactly one factor's sign, giving positive on $-2<x<1$ . Crossing $x=1$ flips a second factor, giving negative on $1<x<3$ . Crossing $x=3$ flips the third, giving positive for $x>3$ . The pattern — negative, positive, negative, positive — alternates at every simple root, for exactly the same reason a quadratic's two-root pattern does: nothing in the argument above ever depended on there being only two factors.

Retrieval — with feedback on every choice

Question 1

3 marks

Solve $x^2 + 2x - 15 < 0$ .

Question 2

3 marks

Solve $-x^2 + 5x - 4 \geq 0$ .

Question 3

3 marks

Which is the correct solution set for $x^2 - 9 \leq 0$ and $x > 1$ , taken together?

Question 4

4 marks

Which is the correct solution set for $x^2 - x - 6 \geq 0$ and $x < 0$ , taken together?

Question 5

2 marks

Based on what real WMA11 examiner reports record about this exact topic, which of the following is the most reliable habit for protecting the marks on a quadratic-inequality question?

Reference — not a study method, a lookup

f(x) > 0: curve above the x-axis. f(x) < 0: below. The roots are the boundary — find them the usual three ways.
Upward parabola (a>0): positive OUTSIDE the roots, negative BETWEEN them. Downward (a<0): reversed.
Two inequalities combined: always sketch a number line — a whole branch of one can vanish against the other, invisibly in pure algebra.
State the answer in the variable the question named (usually x) — never y, and never a region's label (e.g. R).
Strict (< / >): boundary excluded. Non-strict (≤ / ≥): boundary included. Carry the original's strictness into the final line.

CORRECTION (mark-scheme-bullet coverage audit, 2026-09-14): a fresh, direct re-extraction of the real Jan 2019 Q4 and Oct 2021 Q3 mark schemes (not a re-read of this file's own prior paraphrase) found that Oct 2021 Q3(ii) — previously described just below as 'a different sub-question... does NOT carry this quote' — is in fact the SAME real question type as this lesson's own primary anchor, Jan 2019 Q4: both hand a candidate a picture of a region bounded by a line and a quadratic curve and ask for the SET of inequalities (several in y, one in x from a solved root) that define it, and both real mark schemes credit it with an identical B1 (sight of an inequality, no method) / M1 (find the numeric boundary) / A1 (complete, correctly-strict answer) structure. That specific skill — reading several simultaneous inequalities in x AND y straight off one sketch, closed off by one calculated bound — was entirely absent from this lesson before this pass, despite being the most literal reading of spec 1.7 ('interpret ... graphically') and 1.8 ('represent ... graphically') and despite this lesson's own primary anchor testing exactly it; every prequestion, worked example and MCQ here solved only a single inequality in one variable, and no mark line in this file had ever used a B-code before this pass. Fixed with a new marked-solution block (immediately after the 'Representing the answer' teach block), built on VERIDIAN-original numbers — a line $2y=x$ and curve $y=3x-\frac{1}{6}x^2$ , neither the real Jan 2019 pairing ( $2y=x$ , $y=2x-\frac{1}{8}x^2$ ) nor the real Oct 2021 pairing ( $y=2x^2-50$ , a gradient-3 line) — but genuinely modelled on both real mark schemes' own B1/M1/A1 split, and reusing the SAME two real, verbatim-quoted traps below (wrong variable, region label) in their true originating context — a multi-inequality region — rather than only the simplified single-range-of-x proxy the rest of this lesson uses them in. Full accounting in WMA11-verified-facts.md's Lesson-audit log. Not affiliated with or endorsed by Pearson Edexcel. This topic sits outside the research bank's own §6 shortlist of 'richest' WMA11 topics — it is drawn from §4.1's fuller per-topic examiner-report material, and its evidence base is genuinely thinner than the six lessons built from that shortlist: only two series were found addressing this spec area (Jan 2019 Q4 and Oct 2021 Q3), and unlike, for example, the discriminant-with-a-parameter or hidden-quadratic-substitution lessons — where two series independently confirm the SAME named trap — these two series document two DIFFERENT failure modes on two DIFFERENT question types. Jan 2019 Q4 is the source for every 'wrong variable' and 'region label' quote in this lesson, and its one verified numeric fact (the boundary value 16) is used honestly in the prequestion, built around a VERIDIAN-original quadratic chosen only so that 16 is a genuine root of it — the rest of that real question's wording and structure is not available in the research bank and is not reproduced or guessed at here. Oct 2021 Q3 is quoted directly for its general finding about skipping a sketch, but the finding is from part (i) alone — solving 3/x > 4, a single reciprocal (a/x-type) inequality, not two inequalities combined — where the trap was a dropped case (the necessary x > 0), not a dropped branch of a combined system; a separate sub-question, part (ii), does involve a curve and a line but carries no quote used anywhere in this lesson. So every combined-inequality worked example in this lesson (the worked-chain, the chain-drill, and MCQ3/MCQ4) is VERIDIAN-original, built to rehearse the general discipline the report's finding points to — checking with a sketch or number line rather than trusting pure algebra — on genuinely quadratic, genuinely combined material the real question does not itself contain, not to reproduce the real question. The curve-and-line marked-solution question, its mark allocations, and every other numeric example in this lesson are likewise VERIDIAN-original — modelled on the verified mark-scheme conventions established in the research bank's §3 (M/A/B definitions, the cao rule, the three equally-creditable routes to a three-term-quadratic method mark) rather than transcribed from a real mark scheme, which for an original question does not exist. Every root, critical value and boundary below was checked by direct substitution back into the original expression before being written in.

Question 13 marks

Solve $x^2 + 2x - 15 < 0$ .

$-5 < x < 3$
Correct. $(x+5)(x-3)$ has critical values $-5$ and $3$ ; since $a=1>0$ this is the 'between the roots' case for the strict ' $<0$ ' condition, non-inclusive at both ends.
B $x < -5$ or $x > 3$
This is the region where $(x+5)(x-3)>0$ — outside the roots — the opposite of the condition asked for. For an upward parabola, 'less than zero' is always the region BETWEEN the roots, not outside them.
C $-5 < y < 3$
The boundary values are right and the letter is wrong. The question is about $x^2+2x-15$ , a function of x; y never appears in the question at all, and stating the answer in a letter the question never used is exactly the error a real WMA11 examiner report records on this topic.
D $-5 \leq x \leq 3$
The original inequality is strict ( $<0$ , not $\leq0$ ), so both boundary values must be excluded — at $x=-5$ or $x=3$ the expression equals exactly zero, which does not satisfy 'less than zero.'

Traps tested: Between and outside the roots reversed · Final answer stated in the wrong variable · Boundary strict vs non strict mismatched

Question 23 marks

Solve $-x^2 + 5x - 4 \geq 0$ .

$1 \leq x \leq 4$
Correct. $-x^2+5x-4 = -(x-1)(x-4)$ : a downward parabola with roots $1$ and $4$ . For a downward parabola, $\geq 0$ — above or touching the axis — is the region BETWEEN the roots, inclusive since the inequality is non-strict.
B $x \leq 1$ or $x \geq 4$
This is the 'outside the roots' region, which is where a DOWNWARD parabola is negative, not positive. The reversal that happens for $a<0$ is exactly what makes this case easy to get backwards without the sign-pattern derivation, or a sketch, to check against.
C $1 < x < 4$
The region is right and the boundary is wrong: the original inequality is $\geq 0$ , non-strict, so both $x=1$ and $x=4$ — where the expression equals exactly zero — belong in the answer, not just the values strictly between them.
D $1 \leq y \leq 4$
The region and both boundaries are correct; only the letter is wrong. The inequality was given entirely in terms of x, and the final answer has to be too.

Traps tested: Between and outside the roots reversed · Boundary strict vs non strict mismatched · Final answer stated in the wrong variable

Question 33 marks

Which is the correct solution set for $x^2 - 9 \leq 0$ and $x > 1$ , taken together?

$1 < x \leq 3$
Correct. $x^2-9\leq0$ gives $-3\leq x\leq3$ ; combined with $x>1$ , only the part of that interval above 1 survives — the boundary at $x=1$ itself is excluded (from the strict ' $>$ '), while $x=3$ is kept (from the non-strict ' $\leq$ ').
B $-3 \leq x \leq 3$
This is the quadratic inequality's solution set on its own, with the linear condition $x>1$ never applied. 'And' means both conditions hold at once — $x=-2$ satisfies the quadratic inequality but fails $x>1$ , so it cannot belong to the combined answer.
C $x > 1$
This is the linear inequality's solution set on its own, with the quadratic's upper bound never applied. $x=10$ satisfies $x>1$ but fails $x^2-9\leq0$ badly, so it cannot belong to the combined answer either.
D $1 \leq x \leq 3$
The interval is otherwise right, but the lower boundary is wrong: $x=1$ comes from the STRICT inequality $x>1$ , so it must be excluded, not included. Only the upper boundary, $x=3$ , is entitled to be non-strict, since that one came from $x^2-9\leq0$ .

Traps tested: Ignores one condition when combining · Boundary strict vs non strict mismatched

Question 44 marks

Which is the correct solution set for $x^2 - x - 6 \geq 0$ and $x < 0$ , taken together?

$x \leq -2$
Correct. The quadratic gives two separate pieces, $x\leq-2$ or $x\geq3$ . Only the first can ever satisfy $x<0$ as well — there is no number that is both at least 3 and less than 0, so the entire $x\geq3$ branch is eliminated, not narrowed, and the surviving piece is unchanged by the extra condition since every value with $x\leq-2$ already satisfies $x<0$ .
B $x \leq -2$ or $x \geq 3$
This is the quadratic inequality's solution set with the condition $x<0$ never applied. The $x\geq3$ branch cannot survive being combined with $x<0$ at all, whatever a purely algebraic shortcut might suggest by carrying the 'or' straight through unchecked.
C $-2 \leq x < 0$
This describes the region BETWEEN the quadratic's roots, which is where $x^2-x-6$ is NEGATIVE — the opposite of the $\geq0$ condition actually asked for. The correct region for this inequality is outside the roots, in two separate pieces, not a single interval between them.
D $x < 0$
This is the linear condition on its own, with the quadratic inequality's own constraint dropped entirely. $x=-1$ satisfies $x<0$ but fails $x^2-x-6\geq0$ (since $(-1)^2-(-1)-6=-4$ , negative), so it cannot belong to the combined answer.

Traps tested: Ignores one condition when combining · Between and outside the roots reversed

Question 52 marks

Based on what real WMA11 examiner reports record about this exact topic, which of the following is the most reliable habit for protecting the marks on a quadratic-inequality question?

Check which letter the question actually asked for (almost always x, never a region's label like R), sketch or use a number line to fix the region, and carry the original inequality's strictness through to the boundary.
Correct, and a direct response to the two real, quoted findings this lesson is built around: candidates who found the right boundary value still lost marks by writing the answer in y or in a region's label (Jan 2019, Q4), and a separate report found that skipping a sketch was where a necessary condition went missing on a reciprocal-inequality question (Oct 2021, Q3(i)). All three habits address a documented failure mode, not a generic reminder.
BSkip the sketch when time is short, since the algebra alone always gives the right region.
This is the exact habit a real examiner report names as the cause of lost marks: "it was rare to see a sketch graph, which would have helped students to see that x > 0 was required" (Oct 2021, Q3(i), on a single reciprocal inequality, 3/x > 4). The algebra is not wrong to trust — the point is that a purely symbolic approach is where a necessary condition is easiest to lose track of, and a sketch is the check that catches it.
CState the final answer using whichever letter labels the shaded region on the diagram, since that is what the picture is asking about.
This is the second named trap this lesson is built around, confirmed verbatim: "others stated inequalities involving R, which gained no credit" (Jan 2019, Q4). A region's label captions a picture; it is never a coordinate, and it cannot appear in an inequality about x.
DGive the boundary values only, without stating whether the inequality is strict, since the strictness can be inferred from the picture.
A mark scheme's accuracy marks are correct-answer-only: an interval with the wrong strictness at even one end is not the same answer as the correct one, whatever the accompanying sketch shows. The final written line has to carry the strict/non-strict distinction itself, not leave it to be inferred.

Traps tested: Sketch skipped a sign case goes unnoticed · Region label used as the answer variable · Boundary strict vs non strict mismatched

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2019 · Q4 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

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Up next

Perpendicularity Proofs, Rectangle/Point-Construction Problems, and Gradient-Pythagoras Combinations

A perpendicularity proof has two halves, and the exam only pays for both. The half everyone remembers is the calculation — multiply two gradients, or square two sides — and the half that quietly loses marks on real scripts is the sentence after it, the one that actually states the two lines are perpendicular. This lesson is built around that exact gap, confirmed on the richest verified example in the whole WMA11 research pass, plus the two traps that sit either side of it: an arithmetic slip in the length route that looks completely innocent on the page, and a 'find the fourth vertex' question where the obvious method is measurably the less reliable one.

55 min

Quadratic inequalities, interpreted and represented graphically

What a quadratic inequality is actually asking — and why the graph answers it directly

Two steps, every time: find the boundary, then read the region off the shape

Why 'outside' and 'between' aren't two rules to memorise — it's one sign pattern

Representing the answer: a number line, or a shaded region with a solid or dashed boundary

Solve x2−2x−15≤0x^2 - 2x - 15 \leq 0x2−2x−15≤0 and 3x−1>53x - 1 > 53x−1>5 together — the interval a number line catches and pure algebra can lose

Complete the chain — solve x2−4x−5≥0x^2 - 4x - 5 \geq 0x2−4x−5≥0 and 2x+3<112x + 3 < 112x+3<11 together, including the branch that disappears entirely

Deep dive

Practice this for real

Solve $x^2 - 2x - 15 \leq 0$ and $3x - 1 > 5$ together — the interval a number line catches and pure algebra can lose

Complete the chain — solve $x^2 - 4x - 5 \geq 0$ and $2x + 3 < 11$ together, including the branch that disappears entirely