The Discriminant with a Parameter — "No Real Roots" Style

~45 min · WMA11 · 1.4

WMA11 · 1.4 · 45 min

A parameter turns one calculation into two. Once a letter sits inside $a$ , $b$ or $c$ , $b^2 - 4ac$ stops being a number you evaluate and becomes an expression you solve — and the paper has three separate ways of catching a candidate who treats it as the same old evaluation: a squaring slip that produces an equally plausible wrong number, a value of the parameter that quietly stops the equation being a quadratic at all, and a boundary that the question's own wording either includes or excludes depending on a single word.

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 actually changes when a letter replaces a number

Every question in the previous lesson had $a$ , $b$ and $c$ as specific numbers, so $b^2 - 4ac$ evaluated to a specific number too — compute it, read off which of the three cases applies, done. The moment one of $a$ , $b$ or $c$ is itself a letter — conventionally $k$ , $p$ , $m$ , $t$ or similar — $b^2 - 4ac$ stops being a number and becomes an EXPRESSION in that letter. Applying the discriminant condition now means solving an equation or inequality FOR the letter, not evaluating a number. The mechanism is identical; what you are doing to it is not.

This has a concrete consequence worth stating before anything else. You are no longer asking "what is $b^2 - 4ac$ "; you are asking "for which values of the parameter is $b^2 - 4ac$ positive, zero, or negative." That is a genuinely different question — its answer is a range or a pair of values, not a single number — and it is why this topic sits downstream of solving quadratic inequalities (spec 1.7–1.9) almost as much as it sits downstream of the discriminant itself: the last step of nearly every question here is solving a quadratic inequality, just with the letters relabelled from $x$ to whatever the parameter is called.

One more shift matters, and it is easy to miss because it isn't part of the discriminant calculation at all. When the parameter appears in the coefficient of $x^2$ — the leading coefficient $a$ — there is a value of the parameter that makes $a = 0$ . At that value the equation is not a quadratic; it collapses to something linear, and the discriminant has nothing to say about a linear equation. That value has to be handled as its own case, outside whatever the discriminant inequality produces — usually by excluding it from the final answer. Questions state this explicitly (" $k$ is a non-zero constant") for exactly this reason: it is not otherwise implied by anything else on the page.

Where each named trap actually lives

Three failure modes are confirmed on real scripts for this exact question type — all three from the one series that tests it in full, Jan 2023 Q4 — and each one lives at a different line of the working, which is worth knowing individually rather than filing under one generic "be careful with parameters" warning, because each has its own specific tell and its own specific fix.

The sign of $b^2 - 4ac$ itself. The discriminant is $b^2 - 4ac$ , not $b^2 + 4ac$ — an error a real script can make while doing everything else correctly, and one that scores nothing at all, because the method mark is for attempting the discriminant and $b^2 + 4ac$ is not it. Where the parameter sits inside $b$ , squaring $b$ means squaring the WHOLE coefficient: $(6k)^2$ is $36k^2$ , not $6k^2$ . Both errors produce a clean, plausible-looking wrong number, which is exactly what makes them dangerous — nothing about the final line looks wrong on the page.

The parameter's own constraint. If the stem states the parameter is non-zero, that is not throat-clearing — it is telling you that $a = 0$ would stop the equation being a quadratic at all, and the resulting excluded value very often coincides with one of the two numbers the discriminant inequality produces anyway. The natural way to lose it is dividing an inequality through by the parameter to tidy it up, which both silently discards the root at parameter-equals-zero and is not even a legal move: dividing an inequality by a quantity of unknown sign can flip its direction, and the parameter's sign is exactly what is being decided. Factorise instead of dividing, and every critical value stays on the page.

Strict versus non-strict at the boundary. "No real roots" and "two distinct real roots" both exclude $b^2 - 4ac = 0$ , because a repeated root is still a real root. "At least one real root", an unqualified "real roots", and — as the prequestion above showed — "does not cross the x-axis" all include it. The question's own wording decides between $<$ / $>$ and $\leq$ / $\geq$ ; decide before writing the final line, not after.

A fourth thing has to happen before any of the three above can even begin: recognising that a discriminant question is present at all. Not every question says "find the discriminant." Some hide a three-term quadratic one algebraic step below the surface — behind a fraction that has to be cleared, or a line-meets-curve equation that has to be set up first. That recognition step is covered on its own later in this lesson, because failing to spot it is a documented trap in its own right, separate from the three above.

Mechanism

The discriminant of the discriminant — solving "no real roots" as its own quadratic inequality

Once $a$ , $b$ and $c$ are substituted in — possibly with the parameter inside one, two, or all three of them — $b^2 - 4ac$ becomes an algebraic expression in that parameter, and the condition supplied by the question's wording ( $>0$ , $=0$ , $<0$ , $\geq 0$ , $\leq 0$ ) turns it into an equation or inequality to solve. When both $a$ and $b$ contain the parameter — the shape that produces every trap in this lesson at once, and the shape both worked examples below use — that expression is itself a QUADRATIC in the parameter: expand it out and there is a squared term, because $b$ was squared to get there. Solving a quadratic INEQUALITY needs one step beyond solving a quadratic EQUATION. The first part — finding the critical values — is exactly the same three-route problem as solving any three-term quadratic: factorise, use the formula, or , and the general principles for pure marking credit all three equally, whichever variable is actually being solved for. But a second question then has to be answered, one a plain equation never asks: which side of those critical values — or the region between them — actually satisfies the inequality. The safest way to answer it is the reasoning this course has already used once, for a quadratic in $x$ : the expression, regarded as a function of the parameter, is itself an upward or downward parabola (upward here, always — a squared term coming from $b^2$ can never carry a negative coefficient), so it is positive outside its own roots and negative between them, and one substituted test value in each region confirms which. What must never happen, at any point in this process, is dividing the inequality through by the parameter to simplify it first — that move discards whichever root sits at parameter-equals-zero and is not valid in general, since dividing an inequality by a quantity of unknown sign can silently flip its direction. Every piece of this paragraph is the exact machinery the previous lesson used to find where a quadratic in $x$ crosses its axis, applied one level up: a quadratic inequality in the parameter, solved by finding where a second, related quadratic crosses ITS axis.

Diagram — D(p) = 144p² − 48p, sketched as its own parabola in p

x-axis: p · y-axis: D(p) = b² − 4ac

D(p) = 144p² − 48p: An upward parabola in p — the coefficient of p² is 144, positive — crossing its own horizontal axis at p = 0 and p = 1/3, with a minimum at p = 1/6 where D = −4. That vertex location is not a separate fact to memorise: any parabola is symmetric about its vertex, so the vertex sits exactly midway between two roots it has — the average of 0 and 1/3 is 1/6, which is why the minimum falls there. This is the SAME reasoning the previous lesson used for a quadratic in x, applied one level up: an upward parabola's vertex tells you where the curve sits relative to zero, and its two crossing points bound the region where it is negative. Here, that region is the answer to the original question.

p = 0: One of the two critical values of D(p) — and also the value the stem's own 'p is a non-zero constant' condition already excludes on its own. The two arguments agree here, which is why it is tempting to think only one of them needs checking; the discriminant would exclude p = 0 even if the stem never mentioned it, and the stem would exclude it even if the discriminant produced a different critical value there.
p = 1/3: The other critical value, where D(p) returns to zero — the upper boundary of the no-real-roots region.
Shaded region: 0 < p < 1/3: Where D(p) is strictly negative — the values of p for which the ORIGINAL equation 3px² + 12px + 4 = 0 has no real roots. Both edges are open circles, not filled: the boundary itself is the one-repeated-root case, which belongs to a different question ('exactly one real root'), not this one.

Common error: Marking p = 0 and p = 1/3 as solid, included boundary points on the sketch — as though the shaded 'no real roots' region were 0 ≤ p ≤ 1/3.

Correct: Both boundary points are excluded. The condition is D(p) < 0, strict, because at D(p) = 0 the original equation has one repeated real root — a real root, so it belongs to the boundary, not to either open region beside it. A real script on the equivalent numerical question lost its final mark for writing exactly this closed version of the boundary.

examiner-report · Jan 2023 · Q4

Worked, in full

The full range of $p$ for which $3px^2 + 12px + 4 = 0$ has no real roots, given that $p$ is a non-zero constant

01
Read the stem first. " $p$ is a non-zero constant" is not throat-clearing — it is what makes the equation a quadratic at all. Identify the coefficients: $a = 3p$ , $b = 12p$ , $c = 4$ . The parameter multiplies both the leading coefficient and the middle one, which is exactly the shape that produces the full trap set.
Earns: M1 — attempts $b^2 - 4ac$ with $a = 3p$ , $b = 12p$ and $c = 4$ . The mark is for the attempt: the real mark scheme for this exact question type (Jan 2023 Q4) explicitly condones a candidate who does not square the numeric part of $b$ at this stage — e.g. writing $12p^2$ in place of $(12p)^2 = 144p^2$ — provided $p$ itself is genuinely squared somewhere. A slip here still banks this mark, though not the two accuracy marks that come later.
02
Evaluate and impose the condition. $(12p)^2 - 4(3p)(4) = 144p^2 - 48p$ . "No real roots" is the $b^2 - 4ac < 0$ case, strictly: at exactly zero there is one repeated real root, which is a real root, so the boundary belongs to the other case. The condition is $144p^2 - 48p < 0$ .
Earns: No independent mark yet. In the real 4-mark scheme for this exact question type, forming this strict inequality correctly and solving it to a critical value (next stage) are ONE dependent method mark, not two separate ones — there is no mark-scheme line credited for 'the correct inequality' on its own. Getting the whole-coefficient squaring and the strict inequality right here is what makes the dM1 below, and both A marks after it, reachable at all.
03
Solve it as a quadratic in $p$ — and specifically do not divide by $p$ . Dividing $144p^2 - 48p < 0$ through by $p$ would give $144p - 48 < 0$ , i.e. $p < \frac{1}{3}$ , which silently discards the root $p = 0$ and is in any case not a legal step: dividing an inequality by a quantity of unknown sign can flip its direction, and $p$ 's sign is exactly what is being decided. Factorise instead: $48p(3p - 1) < 0$ , with critical values $p = 0$ and $p = \frac{1}{3}$ .
Earns: dM1 — dependent on the M1 above, and this single mark is what the real mark scheme awards for BOTH setting the discriminant expression against zero AND solving the resulting quadratic-in- $p$ down to a non-zero critical value — stage 2 and this stage together earn it, not stage 2 alone. Any of the three general-principles routes (factorisation, formula, completing the square) credits this mark equally, whichever variable is actually being solved for.
04
Decide which side of the critical values satisfies the inequality. Regarded as a function of $p$ , $144p^2 - 48p$ is itself an upward parabola with roots at $0$ and $\frac{1}{3}$ , so it is negative strictly between them and positive outside. Check one value to be sure: at $p = 0.1$ , $144(0.01) - 48(0.1) = 1.44 - 4.8 = -3.36 < 0$ . ✓
Earns: A1 — for identifying the UPPER of the two critical values ( $\frac{1}{3}$ ) as the upper limit of the answer, not merely for having a value that solves the quadratic. The real mark scheme phrases this mark as being for 'an upper limit... not just the value,' and — at this stage only — condones a non-strict form of it; only the final line below fixes the boundary as strict.
05
State the range: $0 < p < \frac{1}{3}$ . Both bounds are strict. Notice where each one actually comes from: the upper bound is a genuine consequence of the discriminant on its own; the lower bound coincides with the stem's own non-zero condition on $p$ , but it is ALSO independently produced by the discriminant here — the two arguments happen to land on the same number, which will not always be true (the beyond-spec block later in this lesson works out exactly when it is and isn't), so both still have to be checked separately, every time.
Earns: A1 — the complete range, both bounds present and both strict. This is the second and final accuracy mark in the real 4-mark structure (M1, dM1, A1, A1) — and the one that disappears most often in real scripts: not because the algebra is hard, but because the lower bound looks, at a glance, like it was already handled by the word 'non-zero' in the stem.

Source — Examiner report, Jan 2023

"many candidates ignored the solution, k = 0, so the most common error seen was the absence of the lower bound 0 < k"

Complete it yourself

Complete the chain — the values of $t$ for which $x^2 + tx + (t + 3) = 0$ has two distinct real roots

01
"Two distinct real roots" means $b^2 - 4ac > 0$ , strictly — the discriminant must be positive, not merely non-negative.
02
Here $a = 1$ — fixed, with no letter attached, so unlike the worked example above there is no non-zero-parameter condition to worry about this time — $b = t$ and $c = t + 3$ . The condition is $t^2 - 4(t + 3) > 0$ .

Marked, line by line

The equation $x + \dfrac{4}{x} = c$ , where $c$ is a constant and $x \neq 0$ , is to be solved for $x$ . (a) Show that the equation can be written as $x^2 - cx + 4 = 0$ . (1) (b) Given that the equation has no real solutions for $x$ , find the set of possible values of $c$ . (3) (c) Find the value(s) of $c$ for which the equation has exactly one real solution for $x$ , and state that solution in each case. (3) — VERIDIAN-original question, inspired by the confirmed real finding that a discriminant-with-parameter question can require forming the quadratic first, from an equation that does not initially look like one (Jan 2019 Q9's documented trap, cited in the trap-taxonomy below) — not a reproduction of any past-paper question, and the coefficients are original.

7 marks available

(a) — 1 mark

“Show that” — the answer is already printed above

Matching the printed result isn’t the same as deriving it — real examiner reports describe scripts that adjust flawed working just to still land on it. Write your own full working below before checking it against the mark scheme.

(b) — 3 marks

101
No real solutions for $x$ means the quadratic from part (a) has no real roots: $b^2 - 4ac < 0$ with $a = 1$ , $b = -c$ , $c = 4$ , i.e. $(-c)^2 - 4(1)(4) < 0$ .
Method mark for forming the discriminant of the QUADRATIC derived in part (a), not of the original fractional equation — the discriminant only has something to say once a genuine three-term quadratic is actually on the page.
M1
102
$c^2 - 16 < 0$
Correct simplified inequality. $(-c)^2 = c^2$ : the sign on $b$ disappears under squaring, which is worth noticing on its own — it is also why the sign of $b$ alone can never, by itself, make a discriminant negative.
A1
103
$-4 < c < 4$
Correct range, both bounds strict. $c^2 < 16$ is not the same statement as $c < 4$ : a squared quantity below 16 means $c$ is within 4 of zero on EITHER side, so the lower bound $c > -4$ is exactly as real a condition as the upper one, not an afterthought attached to it.
A1

201
Exactly one real solution for $x$ is the repeated-root case: $c^2 - 16 = 0$ .
Method mark: sets the discriminant of the part (a) quadratic equal to zero. This is the boundary of the region found in part (b) — the two parts are the same calculation, evaluated at $<$ and at $=$ respectively.
M1
202
$c = 4$ or $c = -4$
Both values required — an equation of the form $c^2 = 16$ always has two solutions, and these are exactly the two boundary values excluded in part (b), not a coincidence.
A1
203
$c = 4$ : $x^2 - 4x + 4 = (x - 2)^2 = 0 \Rightarrow x = 2$ . $c = -4$ : $x^2 + 4x + 4 = (x + 2)^2 = 0 \Rightarrow x = -2$ . Check against the ORIGINAL equation: $2 + \frac{4}{2} = 4$ ✓ and $-2 + \frac{4}{-2} = -4$ ✓.
Correct solution stated for both cases, each checked against the original fractional equation rather than only the derived quadratic — confirming a value introduced by multiplying through by $x$ is not spurious is precisely the discipline part (a) exists to test.
A1

Named traps

discriminant-written-as-b-squared-plus-4ac: Confirmed directly on a real discriminant-with-parameter question: "a few used an incorrect expression b² + 4ac, gaining no marks" (Jan 2023, Q4). This is not a one-mark slip — the method mark itself is for attempting $b^2 - 4ac$ , so a candidate who evaluates $b^2 + 4ac$ scores nothing on the line, however careful the rest of the working is. The sign is not a convention to remember: it falls out of combining $-\frac{b^2}{4a}$ with $+c$ when the square is completed (see the previous lesson's derivation), which is why getting it wrong here usually means the derivation was never actually understood, only the three-case outcome.
coefficient-not-squared-in-full: Confirmed on the same question, where the coefficient of $x$ was $6k$ : "a small number did not square the 6, obtaining an upper limit of 10/3" (Jan 2023, Q4). Squaring $6k$ means squaring the whole product, $36k^2$ , not $6k^2$ . This lesson's own worked example shows the identical mechanism with different numbers: mis-squaring $12p$ to $12p^2$ instead of $144p^2$ produces the equally clean-looking wrong bound $p < 4$ in place of the correct $p < \frac{1}{3}$ — the tell is always that the wrong answer looks exactly as plausible as the right one, so checking the squaring is not optional the moment a parameter sits inside the $b$ term.
parameter-zero-bound-dropped: The most common error confirmed on this question, verbatim: "many candidates ignored the solution, k = 0, so the most common error seen was the absence of the lower bound 0 < k" (Jan 2023, Q4). The mechanism is almost always division: faced with an inequality like $144p^2 - 48p < 0$ , it is tempting to divide through by $p$ , which silently discards the root $p = 0$ and is in any case not a legal move on an inequality whose divisor has unknown sign. Factorise instead of dividing, and both critical values survive automatically.
boundary-included-when-the-inequality-is-strict: Confirmed on the same question: "some gave their final answer as 0 ≤ k ≤ 5/9, losing the final mark" (Jan 2023, Q4). The distinction is exact, not stylistic: at $b^2 - 4ac = 0$ there is one repeated real root, which is a real root, so "no real roots" and "two distinct real roots" both exclude that boundary — while "at least one real root", "real roots" unqualified, and "does not cross the x-axis" all include it. The marked-solution question above is built to show a boundary that is genuinely INCLUDED, in part (c) — reading which case is which matters in both directions, not only the strict one.
discriminant-applied-before-the-quadratic-exists: Confirmed on a question where the quadratic first had to be built: "This question proved challenging for some candidates, who failed to realise that they needed to form a quadratic in x in order to use the discriminant" (Jan 2019, Q9). The trap is not the discriminant technique itself — it is not recognising that a discriminant question is present at all. A term in $\frac{1}{x}$ cleared by multiplying through, a substitution, or a line set equal to a curve can all hide a three-term quadratic one algebraic step below the surface; the marked-solution question above is built around exactly this pattern, and part (a) exists specifically to make that hidden step visible before the discriminant is ever taken.
not-every-term-multiplied-through: Confirmed on the real question this lesson's marked-solution is modelled on, Jan 2019 Q9 (the equation $\frac{3}{x} + 5 = -2x + c$ , "has no real roots, find the range of possible values of c"): "Examiners occasionally saw responses in which all terms except c were multiplied by x. This led to a discriminant which was linear in c and limited further progress could be made." The real mark scheme is explicit that this is a named risk, not a hypothetical one — it even states its own method mark "may be implied by later work" and explicitly "condone[s] one term not being multiplied by x for this mark but all four terms must be on one side," meaning the attempt still banks that one mark even when a term is missed, but every mark after it is lost, because an expression missing one multiplication is not actually the three-term quadratic the discriminant technique needs. This is the exact mechanism the marked-solution's part (a) drills: every term on both sides has to be multiplied by $x$ — not only the fractional one — or the "quadratic" that results is not genuinely quadratic in the variable being eliminated.
adjusted-working-to-fit-the-printed-answerfudged-reverse-fit: Part (a) above is an "ag" (answer given) mark — the target line $x^2 - cx + 4 = 0$ is printed on the page before any working is written. Confirmed on a different WMA paper's own "show that" question, Jan 2021 Q10: "The given answer here persuaded many candidates to 'adjust' their working following obvious mistakes." This is a different KIND of wrong from every other trap on this list: the final line can be completely, word-for-word correct and still earn nothing, because the mark is for a genuine forward derivation, not for a page that ends in the right place. The lesson's own warrant-check gate on part (a) exists specifically to make this distinction concrete rather than abstract — writing your own working BEFORE comparing it against the model line is what separates deriving $x^2 - cx + 4 = 0$ from reverse-engineering a plausible path to it.

In your own words

In one sentence: why does the critical-value equation for the parameter — here, $144p^2 - 48p = 0$ — always have $p = 0$ as one of its two solutions, whenever the parameter multiplies BOTH $a$ and $b$ but $c$ is a fixed number with no parameter in it at all?

Beyond the spec

The spec asks only that you know and use $b^2 - 4ac > 0$ , $= 0$ and $< 0$ (spec 1.4); a student can score full marks on every question in this lesson by mechanically forming the discriminant and solving. This is the one-line reason the $p = 0$ boundary is so often present in exactly this question shape — and so often the one that gets dropped — and it is genuinely useful for spotting the trap coming, not required to pass.

Suppose the parameter multiplies both the leading coefficient and the coefficient of $x$ — $a = mp$ and $b = np$ for some fixed numbers $m$ , $n$ — while $c$ is a fixed number with no $p$ in it at all, exactly the shape of the worked-chain example above. Then $b^2 - 4ac = (np)^2 - 4(mp)(c) = n^2p^2 - 4mcp = p(n^2p - 4mc)$ . Every term in that expression carries a factor of $p$ — $b^2$ contributes $p^2$ , which always has $p$ as a factor, and $4ac$ contributes exactly one factor of $p$ — so $p = 0$ is not a coincidence produced by these particular numbers; it is forced by the SHAPE of the question, every single time $a$ and $b$ share a parameter this way and $c$ does not. That is also, structurally, why $p = 0$ is so easy to lose: it is not one of the two 'interesting' numbers produced by factorising the bracket, it is the trivial factor sitting in front of it — the one that looks, at a glance, as though the word 'non-zero' in the stem had already taken care of it. (The moment $c$ itself also contains the parameter — as it does in $x^2 + tx + (t+3) = 0$ , used in the chain-drill above, where $a = 1$ carries no parameter at all — this factorisation does not apply, $p = 0$ stops being forced, and the two critical values can be any pair of numbers at all. Compare the worked-chain's $0$ and $\frac{1}{3}$ against the chain-drill's $-2$ and $6$ : that contrast is exactly the difference this argument predicts.)

Retrieval — with feedback on every choice

Question 1

4 marks

The equation $qx^2 - 10qx + 9 = 0$ , where $q$ is a non-zero constant, has two distinct real roots. Which is the complete set of possible values of $q$ ?

Question 2

3 marks

The equation $5rx^2 + 20rx + 2 = 0$ , where $r$ is a non-zero constant, has no real roots. A student's working reads: ' $(20r)^2 - 4(5r)(2) = 20r^2 - 40r < 0$ , so $0 < r < 2$ .' What is the actual correct range?

Question 3

4 marks

The equation $2mx^2 - 4mx + 3 = 0$ , where $m$ is a non-zero constant, has at least one real root. Which is the complete set of possible values of $m$ ?

Question 4

3 marks

For which values of the constant $m$ does the equation $x - \dfrac{6}{x} = m$ have no real solutions for $x$ (with $x \neq 0$ )?

Question 5

2 marks

In which of these does using the discriminant require first checking that the stated constant cannot take a specific 'forbidden' value, or the equation stops being a quadratic?

Reference — not a study method, a lookup

Parameter in a, b or c turns b²−4ac from a number into an expression — solve FOR it, don't just evaluate it.
Squaring a term with the parameter inside it means squaring the WHOLE coefficient: (6k)² = 36k², not 6k².
Parameter multiplies a? Check it can't be zero — separately from the discriminant, even if the stem never restates it.
'No real roots' / 'two distinct roots' exclude b²−4ac=0. 'At least one root' / 'does not cross' include it.
Never divide an inequality by the parameter — factorise instead, or a root vanishes and the sign may flip unseen.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson's own WMA11 material was independently verified against the primary Pearson document by the WMA11 research pass this lesson was written from, drawing on exactly two WMA11 series — January 2023 Q4 and January 2019 Q9 — and no claim about THIS paper's own past series extends past what those two series' mark schemes and examiner reports actually state. One further quotation, on the part (a) warrant-check gate, is sourced from a sibling paper instead of WMA11 itself: WMA14 January 2021 Q10, cited because it is the clearest confirmed instance of the 'adjusted working to fit a printed answer' pattern found in the research pass, and every WMA paper shares the same 'ag'/'cso' mark-scheme discipline this quote illustrates — it is flagged as cross-paper in that block's own trap-taxonomy entry, not presented as a WMA11 finding. Every question in this lesson — prequestion, worked chain, chain drill, marked solution and MCQ alike — is VERIDIAN-original wording, inspired by those two confirmed real question types, never a reproduction of a real Pearson question; because the questions are original, the per-line mark allocations attached to them are modelled on verified mark-scheme conventions (what M, A and B marks mean, when follow-through applies, which routes earn the method mark for a three-term quadratic) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 14 marks

The equation $qx^2 - 10qx + 9 = 0$ , where $q$ is a non-zero constant, has two distinct real roots. Which is the complete set of possible values of $q$ ?

$q < 0$ or $q > \frac{9}{25}$
Correct. $(-10q)^2 - 4(q)(9) > 0$ gives $100q^2 - 36q > 0$ , i.e. $4q(25q - 9) > 0$ . As a function of $q$ this is an upward parabola with roots $0$ and $\frac{9}{25}$ , so it is positive OUTSIDE them. Check the negative branch, the one most often dropped: $q = -1$ gives $-x^2 + 10x + 9 = 0$ , i.e. $x^2 - 10x - 9 = 0$ , discriminant $100 + 36 = 136 > 0$ — two distinct real roots, confirmed.
B $0 < q < \frac{9}{25}$
This is the region where $100q^2 - 36q$ is NEGATIVE — the no-real-roots case, not two distinct roots. Check $q = 0.1$ : the discriminant is $1 - 3.6 = -2.6 < 0$ , confirming this range is the wrong one.
C $q > \frac{9}{25}$
Only the right-hand branch. This is what dividing $100q^2 - 36q > 0$ through by $q$ produces, and that division is not a legal step here — the sign of $q$ is unknown, so the inequality could flip. Factorising instead keeps both critical values and both branches.
D $q \leq 0$ or $q \geq \frac{9}{25}$
The branches are right; the boundaries should not be included. At $q = \frac{9}{25}$ the discriminant is exactly zero — one repeated root, not two distinct ones — and $q = 0$ is excluded by the stem in any case, since the equation is then not a quadratic.

Traps tested: Inequality side reversed · Negative branch dropped · Boundary included when the inequality is strict

Question 23 marks

$0 < r < \frac{1}{10}$
Correct — and the student's own working shows exactly where it went wrong: $(20r)^2$ is $400r^2$ , not $20r^2$ . The 20 was never squared, only the letter was — the same documented error as a real series's version of this question, where the coefficient was 6 instead of 20. Redo it correctly: $400r^2 - 40r < 0 \Rightarrow 40r(10r - 1) < 0 \Rightarrow 0 < r < \frac{1}{10}$ .
B $0 < r < 2$
This is the student's own wrong answer, reached by squaring only the letter $r$ and leaving the 20 unsquared. It looks exactly as plausible as the correct range — a clean bound with a small integer — which is the entire reason this error survives to the final line of a real script undetected.
C $0 \leq r \leq \frac{1}{10}$
The critical values are corrected, but the boundaries should not be included. At $r = \frac{1}{10}$ the discriminant is exactly zero — one repeated root, which is a real root, so the boundary belongs to a different case than 'no real roots'.
D $r < 0$ or $r > \frac{1}{10}$
This is the sign of the (correctly computed) inequality reversed — the region OUTSIDE the critical values, which is where the discriminant is POSITIVE, i.e. two distinct real roots, not the no-real-roots case the question asks for.

Traps tested: Coefficient not squared in full · Boundary included when the inequality is strict · Inequality side reversed

Question 34 marks

The equation $2mx^2 - 4mx + 3 = 0$ , where $m$ is a non-zero constant, has at least one real root. Which is the complete set of possible values of $m$ ?

$m < 0$ or $m \geq \frac{3}{2}$
Correct, and it is worth noticing this range is not symmetric in how it treats its two boundaries. $(-4m)^2 - 4(2m)(3) \geq 0$ gives $16m^2 - 24m \geq 0$ , i.e. $8m(2m - 3) \geq 0$ , true for $m \leq 0$ or $m \geq \frac{3}{2}$ — but $m = 0$ has to be cut from the lower branch anyway, because the stem already states $m$ is non-zero and at $m = 0$ there is no quadratic at all. $m = \frac{3}{2}$ , by contrast, survives: check it directly — $3x^2 - 6x + 3 = 0 \Rightarrow x^2 - 2x + 1 = (x-1)^2 = 0$ , a genuine repeated real root, which counts as 'at least one'.
B $m \leq 0$ or $m \geq \frac{3}{2}$
The inequality itself is solved correctly, but $m = 0$ has to be removed separately: at $m = 0$ the equation is $3 = 0$ , which is not a quadratic and has no roots of any kind, real or otherwise, so it cannot belong in an answer about how many real roots an equation has.
C $m < 0$ or $m > \frac{3}{2}$
This drops $m = \frac{3}{2}$ , but it shouldn't be dropped here: the question says 'at least one real root', which is the NON-strict case $b^2 - 4ac \geq 0$ , and a repeated root at the boundary is a genuine real root. Excluding it treats every boundary as though the inequality were strict, when this one specifically is not.
D $0 < m < \frac{3}{2}$
This is the region BETWEEN the critical values, where the discriminant is negative — the no-real-roots case, the opposite of what the question asks for. It is also the one range of the four that violates the stem's own $m \neq 0$ condition in the wrong direction, by including small positive values that were never in question.

Traps tested: Parameter zero bound dropped · Boundary included when the inequality is strict · Inequality side reversed

Question 43 marks

For which values of the constant $m$ does the equation $x - \dfrac{6}{x} = m$ have no real solutions for $x$ (with $x \neq 0$ )?

No such value of $m$ exists — the equation has real solutions for every value of $m$
Correct, and worth pausing on. Multiplying through by $x$ gives $x^2 - mx - 6 = 0$ , so $b^2 - 4ac = m^2 - 4(1)(-6) = m^2 + 24$ . Since $m^2 \geq 0$ for every real $m$ , this sum is at least $24$ — strictly positive, always, whatever $m$ is. The discriminant condition here doesn't produce a range of excluded values; it proves there isn't one. A parameter question does not have to have a non-trivial answer, and assuming one must exist just because the question is phrased as though it might is itself a trap.
B $-2\sqrt{6} < m < 2\sqrt{6}$
This is the range produced by wrongly writing the discriminant as $m^2 - 24$ instead of $m^2 + 24$ — dropping a sign on $-4ac$ when $c = -6$ is itself negative, so $-4ac = -4(1)(-6) = +24$ , not $-24$ . Two negative signs combine to a positive one, and it is easy to apply only the first of the two.
C $m > 2\sqrt{6}$ or $m < -2\sqrt{6}$
This answers the wrong question twice over: it uses the incorrectly-signed discriminant $m^2 - 24$ from option B, and then takes the wrong side of it besides. The correctly-signed discriminant, $m^2 + 24$ , is never negative for any real $m$ , so no such excluded range exists at all.
DIt cannot be determined without a specific value of $m$
It can be determined for every value of $m$ at once: the discriminant reduces to the single expression $m^2 + 24$ regardless of which particular $m$ is chosen, and that expression's sign is settled for all real $m$ simultaneously — which is exactly what makes the question answerable without testing individual values one at a time.

Traps tested: Sign of negative c mishandled · Overclaims uncertainty

Question 52 marks

In which of these does using the discriminant require first checking that the stated constant cannot take a specific 'forbidden' value, or the equation stops being a quadratic?

$kx^2 + 3x - 7 = 0$ , where $k$ is a constant
Correct — $k$ is the coefficient of $x^2$ here. At $k = 0$ this becomes $3x - 7 = 0$ : linear, one root, no discriminant at all. Whenever the parameter sits in $a$ , this check is required, independently of whatever the discriminant condition itself produces.
B $x^2 + kx - 7 = 0$ , where $k$ is a constant
No such check is needed here: $k$ only affects $b$ , and the coefficient of $x^2$ is fixed at $1$ regardless of $k$ 's value, so the equation is a genuine quadratic for every real $k$ , including $k = 0$ .
C $x^2 + 3x - k = 0$ , where $k$ is a constant
No such check is needed: $k$ only affects $c$ , the constant term, and $a = 1$ is again fixed. Every value of $k$ still gives a genuine quadratic with a well-defined discriminant.
D $7x^2 + kx - 3 = 0$ , where $k$ is a constant
No such check is needed: the coefficient of $x^2$ is fixed at $7$ , unaffected by $k$ . Only when the PARAMETER ITSELF is, or multiplies, the coefficient of $x^2$ does a forbidden value need excluding — being present somewhere in the equation is not the same as being present in $a$ .

Traps tested: Unnecessary exclusion applied

Practice this for real

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Examiner report: Jan 2023 · Q4 — cited directly in this lesson

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

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The hidden quadratic — substitution from an indices/exponential equation

An equation that never once shows you an x^2 can still be a quadratic wearing a disguise. Two independently-verified WMA11 series — Jan 2023 and Jan 2025 — record the same failure on the same question type: candidates spot the substitution, do the algebra, solve the quadratic in the new letter correctly, and hand that in as the final answer, never returning to the variable the question actually asked about. Finding the substitution is never the hard part. Remembering that it was a substitution, and not the answer, is.

45 min