Quadratic Functions, the Discriminant, and Solving Quadratic Equations

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 is, and the one condition that makes it one

A quadratic function is $f(x) = ax^2 + bx + c$ where $a$ , $b$ and $c$ are constants and — this is the load-bearing part — $a \neq 0$ (spec 1.3). The condition is not pedantry. It is the entire reason a whole family of exam questions exists: whenever the coefficient of $x^2$ is itself the unknown you are being asked about, the value that kills it is a case the cannot speak for, and forgetting to handle that case separately is a documented, recurring way to lose the last mark of a question you otherwise did correctly.

The graph of a quadratic is a parabola. If $a > 0$ the arms point upwards and the curve has a minimum; if $a < 0$ the arms point downwards and it has a maximum. That is a testable statement rather than a picture to remember: for large $|x|$ the $ax^2$ term dominates both the others (it grows in proportion to $x^2$ while $bx$ grows only in proportion to $x$ ), so the sign of $a$ decides which way the far ends of the curve go, and a curve with one turning point whose ends both go up must have a minimum.

The roots of the equation $ax^2 + bx + c = 0$ , the zeros of the function $f$ , and the x-intercepts of the curve $y = f(x)$ are three names for the same set of numbers: the x-values where the height of the curve is zero. Keeping that identity in view is what lets a question phrased about a graph ("the curve meets the x-axis at exactly one point") be answered with algebra, and a question phrased about an equation ("has no real roots") be answered with a sketch. The paper switches between the two phrasings freely and expects you not to notice.

A parabola has exactly one turning point and is symmetric about the vertical line through it. Both of those facts are consequences of the completed-square form derived in the next section, not extra facts to store — once the expression is written so that $x$ appears only inside a single square, everything else follows from the two properties of squaring: a square is never negative, and it takes the same value at $t$ and at $-t$ .

Completing the square — the mechanism, then the two complications

Expanding tells you what to do. $(x + q)^2 = x^2 + 2qx + q^2$ : the coefficient of $x$ in a perfect square is always twice the number inside the bracket. So given $x^2 + 6x + \ldots$ , the bracket must contain half of 6, which is 3. That single observation — halve the coefficient of $x$ — is the whole method; everything else is bookkeeping to repair the damage it does.

The damage: $(x+3)^2 = x^2 + 6x + 9$ , which is 9 more than the $x^2 + 6x$ you started with, so you subtract the 9 back. $x^2 + 6x + 1 = (x+3)^2 - 9 + 1 = (x+3)^2 - 8$ . Now $x$ appears exactly once, inside a square, and the whole behaviour of the function is readable: the square is never negative, it is zero only at $x = -3$ , so the minimum value of the expression is $-8$ and it occurs at $x = -3$ . The turning point is $(-3, -8)$ and the line of symmetry is $x = -3$ .

Complication one: the sign flips. $(x + 3)^2$ is zero at $x = -3$ , and $(x - 2)^2$ is zero at $x = +2$ . This costs more marks in this topic than any piece of arithmetic, because it is not arithmetic — nothing goes wrong in the working, the answer is simply read off with the wrong sign at the very end. The check is one substitution: put your claimed turning-point x-value back into the original expression and confirm you get the claimed y-value.

Complication two: a leading coefficient. When $a \neq 1$ , factor $a$ out of the first two terms only, leaving $c$ outside. $2x^2 - 12x + 7 = 2(x^2 - 6x) + 7$ . Now complete the square inside: half of $-6$ is $-3$ , and $(x-3)^2 = x^2 - 6x + 9$ is 9 too big, so $x^2 - 6x = (x-3)^2 - 9$ . Substituting back gives $2\left[(x-3)^2 - 9\right] + 7 = 2(x-3)^2 - 18 + 7 = 2(x-3)^2 - 11$ . The $-9$ was inside the bracket, so it gets multiplied by the 2, giving $-18$ and not $-9$ . Missing that multiplication is the classic slip, and it is the one modelled on the wrong path of the marked solution further down.

Two sanity checks that between them catch almost every error in this procedure, and cost about ten seconds each. First, expand your answer and confirm it returns the original expression — $2(x-3)^2 - 11 = 2x^2 - 12x + 18 - 11 = 2x^2 - 12x + 7$ . Second, evaluate the original at the claimed turning-point x-value and confirm the claimed y-value comes out.

Mechanism

Where $b^2 - 4ac$ actually comes from — it is not a rule, it is a leftover

Run the same procedure on letters instead of numbers, and the discriminant appears whether you were looking for it or not. Start with $ax^2 + bx + c$ and factor $a$ from the first two terms: $a\left(x^2 + \frac{b}{a}x\right) + c$ . Halve the coefficient of $x$ inside the bracket: half of $\frac{b}{a}$ is $\frac{b}{2a}$ , and $\left(x + \frac{b}{2a}\right)^2 = x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}$ , so $x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}$ . Multiply through by the $a$ outside and add the $c$ back: $ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$ . Now combine those last two terms over the common denominator $4a$ : $-\frac{b^2}{4a} + c = \frac{-b^2 + 4ac}{4a} = -\frac{b^2 - 4ac}{4a}$ . So for every quadratic, without exception, $ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}$ . Nobody chose $b^2 - 4ac$ ; it is simply what was left over when the square was completed, and it is the only place the constant $c$ still appears. Read the identity twice and it tells you two separate things. First, the turning point of every quadratic is at $x = -\frac{b}{2a}$ with height $-\frac{b^2 - 4ac}{4a}$ — so for $a > 0$ the turning point lies below the x-axis exactly when $b^2 - 4ac > 0$ , on it exactly when $b^2 - 4ac = 0$ , and above it exactly when $b^2 - 4ac < 0$ . Since an upward parabola's arms both go to $+\infty$ , it crosses the axis twice, touches once, or misses entirely in precisely those three cases: the root count is not a separate rule bolted onto the discriminant, it is a restatement of where the vertex sits. Second, setting the whole thing equal to zero and solving gives $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ . The denominator $4a^2$ is positive whatever $a$ is, so the right-hand side has the same sign as $b^2 - 4ac$ — and a real number squared can never be negative, so a real solution exists if and only if $b^2 - 4ac \geq 0$ . That is the whole theorem, and it is one line of consequence from an identity you derived rather than memorised.

Diagram — The three discriminant cases, drawn as three positions of the same vertex

x-axis: x · y-axis: y = f(x)

C1 · b² − 4ac > 0: An upward parabola (a > 0) whose vertex lies strictly BELOW the x-axis, because the vertex height is −(b² − 4ac)/(4a), which is negative when a is positive and b² − 4ac is positive. Both arms rise to infinity, so the curve is forced to cross the axis once on the way down and once on the way up: two distinct real roots, at x = (−b ± √(b² − 4ac))/(2a). Plotted here for a = 1, b = −4, c = 1: y = x² − 4x + 1, discriminant 16 − 4 = 12, vertex (2, −3), roots at x = 2 ± √3 ≈ 0.27 and 3.73.
C2 · b² − 4ac = 0: The same curve translated upwards until its vertex sits exactly ON the x-axis. It meets the axis at one point and immediately turns away: one repeated root, at x = −b/(2a), which is the line of symmetry itself. The curve touches; it does not cross. Plotted here as the same a = 1, b = −4 as C1 with c raised from 1 to 4: y = x² − 4x + 4 = (x − 2)², discriminant 16 − 16 = 0, vertex (2, 0) sitting exactly on the axis.
C3 · b² − 4ac < 0: Translated further up, so the vertex is strictly ABOVE the axis. The minimum value of the function is positive, so the whole curve is above the axis and never meets it: no real roots. Nothing about the curve's steepness or width changes between the three cases — only the height of the vertex does. Plotted here as the same a = 1, b = −4 as C1 and C2 with c raised further to 7: y = x² − 4x + 7, discriminant 16 − 28 = −12, vertex (2, 3), strictly above the axis.

Vertex (−b/2a, −(b²−4ac)/(4a)): The one point the three cases differ in. Its x-coordinate is the same in all three; only its height moves, and the discriminant is the number that measures that height.
Line of symmetry x = −b/2a: Identical in all three cases, and unaffected by c. Changing c slides the whole curve vertically without moving the line of symmetry — which is exactly why c appears in the discriminant but not in the vertex's x-coordinate.
y-intercept (0, c): The one point you can read straight off the equation with no work at all, since f(0) = c. Mark it on any sketch: examiner reports on curve-sketching questions record otherwise-correct sketches losing marks for leaving it unlabelled.
The a < 0 mirror: For a downward parabola every 'below' above becomes 'above', because the vertex height −(b²−4ac)/(4a) changes sign with a. The discriminant test itself is unchanged: it never referred to which way the curve opens, only to whether the vertex is on the opposite side of the axis from the arms.

Common error: Drawing the b² − 4ac < 0 case as a curve that still just touches the x-axis, and drawing the vertex as a sharp point rather than a smooth turn.

Correct: b² − 4ac < 0 means the curve never meets the axis at all; touching is the b² − 4ac = 0 case. They are two different pictures, not two ways of drawing the same one. A parabola's vertex is also always smooth — an examiner report on a curve-sketching question (a cubic, but the drawing fault is identical) records that 'a few candidates were inclined to feature a cusp at the minimum point' (Oct 2021, Q6(a)).

examiner-report · Oct 2021 · Q6(a)

Mechanism

The quadratic formula is completing the square, done once, in letters

Take the identity derived above and finish the job. From $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ , square-root both sides and keep both signs: $x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2|a|}$ . The modulus can be dropped without changing anything, because the $\pm$ in front already ranges over both signs — writing $2a$ instead of $2|a|$ swaps which branch is which and leaves the pair of solutions identical. Rearranging gives $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . That is the formula: not an independent tool, but the completed square carried out once, symbolically, so that nobody ever has to carry it out symbolically again. This has three consequences that matter on the paper rather than in a textbook. First, the formula can tell you nothing that cannot, because it IS completing the square — but completing the square tells you things the formula cannot, namely where the vertex is, because the formula throws that intermediate line away. If a question asks for a turning point, a minimum value, or the form $a(x+p)^2 + q$ , the formula route has to be abandoned and restarted, while the completing-the-square route has the answer already sitting in its second line. Second, the two routes agree on the discriminant because it is literally the same quantity in both: it is under the square root in the formula, and it is the numerator on the right-hand side just before the square root is taken in the derivation. When the method comparison below solves one equation both ways, the number 169 shows up in both — the same 169, not a coincidence. Third, this is why the mark scheme treats them as interchangeable. The general principles for pure mathematics marking define the method mark for solving a three-term quadratic by three named routes — factorisation with brackets that multiply back to the right leading and constant terms, the formula with values substituted, or completing the square — and any one of them earns it. There is no house method. There is only whichever one is fastest for the question in front of you, and the point of the next block is to make that a decision rather than a habit. One thing the formula is NOT: available on the sheet. The specification states that formulae students are expected to know 'are given below and will not appear in the booklet', and the quadratic formula is on that list. The exam formula booklet's entire Pure Mathematics P1 section is two entries — mensuration (sphere surface area, cone curved surface) and the cosine rule. Nothing else. The formula has to come from memory or from re-deriving it, which is the other reason to be able to complete the square on letters.

Same question, every valid method

Solve $2x^2 - 7x - 15 = 0$ , showing all stages of your working. (VERIDIAN-original question. The instruction is written the way real WMA11 sub-parts write it — 'In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable' is verbatim from the January 2023 paper, attached there to parts testing algebra a calculator would trivially bypass — but this equation is original, not a reproduction of any past-paper question.)

3 valid methods · every one reaches $x = 5$ or $x = -\frac{3}{2}$ · 3 marks available

01
Seek $(mx + p)(nx + q)$ with $mn = 2$ and $|pq| = 15$ . Trying $m = 2$ , $n = 1$ , $p = 3$ , $q = -5$ : $(2x + 3)(x - 5) = 2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15$ . ✓
Method mark for solving a three-term quadratic by factorisation. The general principles for pure marking state the condition explicitly: the brackets must multiply back so that the product of the outer coefficients matches |a| and the product of the constants matches |c|. Here mn = 2 = |a| and |pq| = 15 = |c|, so this pair qualifies. The mark is for producing brackets that pass that test, not for the roots that follow.
M1
02
$2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$
Accuracy mark for the first root. Dependent on the method mark: M0 A1 is impossible under the general marking guidance.
A1
03
$x - 5 = 0 \Rightarrow x = 5$
Accuracy mark for the second root. Both roots are needed; a product of two factors is zero when either factor is zero, and stopping at one of them answers half the question.
A1

Fastest when a factor pair is visible, and the only route that leaves you holding the factorised form — which is precisely what a follow-on part asking you to sketch the curve or state where it meets the axes actually wants. Its one real risk is that hunting for factors that do not exist has no natural stopping point. The discriminant ends that: $b^2 - 4ac = 169 = 13^2$ is a perfect square, so integer factors exist here and are worth looking for. Had it not been a perfect square, no amount of searching would have found them, and one line of arithmetic would have told you to switch routes immediately.

In your own words

In one sentence: why is the quantity under the square root in the quadratic formula the same quantity that decides how many times the curve meets the x-axis?

Where the exam hides the quadratic

Almost no WMA11 question says "here is a quadratic, find its discriminant". The quadratic is one step down, and building it is the step that gets skipped. An examiner report on a real discriminant question puts the failure exactly: candidates "failed to realise that they needed to form a quadratic in $x$ in order to use the discriminant" (Jan 2019, Q9). The technique to practise is therefore not the discriminant itself but the recognition that a discriminant question is present at all.

The most common hiding place is an intersection question. To find where a line $y = mx + d$ meets a curve $y = ax^2 + bx + c$ , set the two expressions equal — at an intersection the heights agree — and collect everything on one side. What you now have is a quadratic whose roots are the x-coordinates of the intersection points, so its discriminant counts them: positive means the line cuts the curve twice, zero means it touches at exactly one point (it is a ), negative means they never meet. Concretely, for $y = 2x + 1$ and $y = x^2 + 4$ : $x^2 + 4 = 2x + 1$ , so $x^2 - 2x + 3 = 0$ , and $b^2 - 4ac = 4 - 12 = -8 < 0$ . They never meet — and you know that without solving anything or drawing anything.

The second hiding place is a substitution. An equation that is not a quadratic in $x$ can be a quadratic in something else: in $3^x$ , or in $x^2$ , or in $\sqrt{x}$ . Two independent series' examiner reports record the same trap here, and it is not the substitution itself — students manage that. It is what happens afterwards: having solved the quadratic in the substituted variable, they stop. One report records that "a significant number of responses did not go any further than finding solutions to the quadratic in p" (Jan 2025, Q4(b)). The variable you solved for is not the variable the question asked about, and the last step is to go back.

The third hiding place is an equation with the unknown in a denominator, cleared by multiplying through — the quadratic appears only after the multiplication. In all three cases the same discipline applies: before writing $b^2 - 4ac$ , be able to point at a single equation of the form $\text{something} = 0$ with exactly three terms, and be able to say which letter is $a$ , which is $b$ and which is $c$ . If you cannot, the discriminant has nothing to attach to yet.

Worked, in full

The full range of $k$ for which $kx^2 + 8kx + 3 = 0$ has no real roots — including the bound the question never states

01
Read the stem first. $k$ is given as a non-zero constant, which is what makes this a quadratic at all and therefore what makes the discriminant applicable — with $a = 0$ there is no $b^2 - 4ac$ to take, only the linear equation $3 = 0$ . Then identify the coefficients: $a = k$ , $b = 8k$ , $c = 3$ . Both $a$ and $b$ contain the unknown, which is the whole difficulty of this question type.
Earns: M1 — attempts $b^2 - 4ac$ with $a = k$ , $b = 8k$ and $c = 3$ . The mark is for the attempt, so a candidate who writes $(8k)^2 - 4(k)(3)$ and then slips arithmetically has already banked it.
02
Evaluate and impose the condition. $(8k)^2 - 4(k)(3) = 64k^2 - 12k$ . "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 $64k^2 - 12k < 0$ .
Earns: dM1 — dependent on the first method mark: sets $b^2 - 4ac < 0$ and simplifies it into a quadratic inequality purely in $k$ . This is still a method mark, not an accuracy one, which matters for partial credit: on the real Jan 2023 anchor this question is modelled on, an examiner report confirms that a candidate who failed to square the parameter's coefficient in full (writing $8k^2$ instead of $64k^2$ , the equivalent slip on this question's own numbers) still 'gained the first two method marks only' — both marks up to and including this line survive an arithmetic slip, because they credit the method of forming and simplifying the inequality, not the specific numbers it produces.
03
Solve it as a quadratic in $k$ — and specifically do not divide by $k$ . Dividing $64k^2 - 12k < 0$ through by $k$ would give $16k - 3 < 0$ , which is wrong twice over: it discards the root $k = 0$ , and it divides an inequality by a quantity whose sign is unknown, so it is not even a legal step. Factorise instead: $4k(16k - 3) < 0$ , with critical values $k = 0$ and $k = \frac{3}{16}$ . Regarded as a function of $k$ , $64k^2 - 12k$ is itself an upward parabola with roots at those two values, so it is negative strictly between them and positive outside — giving the upper limit $k < \frac{3}{16}$ . Check one value to be sure: at $k = \frac{1}{8}$ , $64\left(\frac{1}{64}\right) - \frac{12}{8} = 1 - 1.5 = -0.5 < 0$ . ✓
Earns: A1 — obtains the upper limit $k < \frac{3}{16}$ (not just the value $\frac{3}{16}$ alone), by any of the three credited routes to solving a three-term quadratic (factorisation, formula or completing the square). This is the first mark that actually requires the correct numbers: the same examiner report records that a slip in squaring the parameter's coefficient loses this mark and the one after it, even though the method to this point was sound.
04
State the range: $0 < k < \frac{3}{16}$ . Both bounds are strict, and note where each came from — the upper bound from the discriminant, the lower bound from the fact that $k = 0$ is the other critical value of the same factorised inequality, not a separate argument borrowed from the stem's non-zero condition. Those two facts happen to agree here, but they are logically distinct, and a question can be built where they would not.
Earns: A1 — both bounds present and both strict. This is the mark that disappears in real scripts: not because the algebra is hard, but because the lower bound has no visible prompt in the question. Real WMA11 scripts lose exactly this mark two distinct ways on the equivalent anchor question — dropping the lower bound entirely, or keeping it but stating the boundary as non-strict ( $0 \leq k \leq \frac{3}{16}$ ) when "no real roots" demands strict inequalities throughout.

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 $p$ for which $y = x^2 + (p+4)x + 25$ meets the x-axis exactly once

01
"Meets the x-axis exactly once" means the equation $x^2 + (p+4)x + 25 = 0$ has exactly one root — a repeated root, where the curve touches rather than crosses. That is the $b^2 - 4ac = 0$ case: an equation, not an inequality.
02
Here $a = 1$ , $b = p + 4$ and $c = 25$ , so the condition is $(p+4)^2 - 4(1)(25) = 0$ . Note that $a = 1$ regardless of $p$ , so unlike the worked example above there is no non-zero condition to worry about — the equation is a quadratic whatever $p$ is.

Marked, line by line

$f(x) = 2x^2 - 8x + 11$ . (a) Express $f(x)$ in the form $a(x + b)^2 + c$ , where $a$ , $b$ and $c$ are constants to be found. (3) (b) Hence write down the coordinates of the turning point of the curve with equation $y = f(x)$ . (2) (c) The equation $f(x) = k$ , where $k$ is a constant, has two distinct real roots. Find the range of possible values of $k$ . (3) — VERIDIAN-original question, inspired by the structure of real WMA11 items that pair a completed-square part with a 'hence' part and a discriminant part (spec 1.3, 1.4 and 1.5 in one question). Not a reproduction of any past-paper question, and the per-line mark allocations below are modelled on verified mark-scheme conventions rather than copied from a real scheme.

8 marks available

(a) — 3 marks

01
$a = 2$ : the coefficient of $x^2$ is taken outside as a factor of the first two terms only, giving $f(x) = 2(x^2 - 4x) + 11$ .
Independent accuracy mark. a = 2 can be written down by inspection with no method at all, which is exactly what a B mark is for — the general marking guidance defines B marks as independent accuracy marks where there is no method. It does not depend on any line above or below it and survives intact even if the rest of the part collapses.
B1
02
Half of $-4$ is $-2$ , and $(x-2)^2 = x^2 - 4x + 4$ is 4 too big, so $x^2 - 4x = (x-2)^2 - 4$ and $f(x) = 2\left[(x-2)^2 - 4\right] + 11$ .
Method mark for completing the square: half the coefficient of x inside the bracket, its square subtracted outside. Earned for the correct structure, whatever happens to the arithmetic afterwards — this is the whole difference between an M mark and an A mark.
M1
03
$f(x) = 2(x-2)^2 - 8 + 11 = 2(x-2)^2 + 3$
Accuracy mark, correct answer only, and unavailable without the M mark above. The −4 sat inside the bracket, so multiplying out by the 2 makes it −8, not −4. This is the single most-lost mark in the part.
A1

(b) — 2 marks

101
$(x-2)^2$ is zero at $x = 2$ and strictly positive everywhere else, so $f$ takes its smallest value at $x = 2$ .
Independent mark, follow through their b from part (a). The sign flips: (x − 2)² gives x = +2. Reversing that sign is the most common error in this part, and it is not an arithmetic error — the working is right and only the reading-off is wrong.
B1 ft
102
That smallest value is the leftover constant, $3$ , so the turning point is $(2, 3)$ — a minimum, since $a = 2 > 0$ .
Second independent mark, follow through their c. Naming it as a minimum rather than just a turning point costs one word and pre-answers a follow-up part. Confirm by substitution if there is time: f(2) = 8 − 16 + 11 = 3.
B1 ft

(c) — 3 marks

201
$f(x) = k$ rearranges to $2x^2 - 8x + (11 - k) = 0$ . Two distinct real roots means $b^2 - 4ac > 0$ , strictly: $(-8)^2 - 4(2)(11-k) > 0$ .
Method mark: rearranges to a three-term quadratic equal to zero, with the k carried into the constant, and applies the discriminant. An equally creditable alternative method is available from part (b) — y = k is a horizontal line, so it cuts the curve twice exactly when k is above the minimum value.
M1
202
$64 - 88 + 8k > 0$ , that is $8k - 24 > 0$ .
Correct simplified inequality. Note (−8)² = 64, positive: squaring destroys the sign of b, which is why the sign of b alone can never make a discriminant negative — a useful check on any answer that seems to depend on it.
A1
203
$k > 3$ .
Correct range, strict inequality. It agrees with part (b) exactly: the minimum value of f is 3, so the line y = k meets the curve twice when k is strictly above 3, once when k = 3, and never below it. Two independent routes landing on the same answer is the check worth building the habit of.
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). Note the severity — not a lost accuracy mark but no marks at all, because the method mark is for attempting the discriminant, and $b^2 + 4ac$ is not the discriminant. The derivation is the defence: the $-4ac$ comes from combining $-\frac{b^2}{4a}$ with $+c$ over a common denominator, so the minus sign is structural, not a convention to remember.
coefficient-of-x-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 thing — $36k^2$ , not $6k^2$ . The mis-squared version gives $6k^2 - 20k < 0$ and hence $0 < k < \frac{10}{3}$ , exactly the wrong upper limit the report names, instead of the correct $0 < k < \frac{5}{9}$ . The tell is that the wrong answer is a clean, plausible-looking fraction, so nothing about it looks wrong on the page.
parameter-zero-bound-dropped: The most common error on this question type, confirmed 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 $36k^2 - 20k < 0$ , students divide through by $k$ , which discards the root $k = 0$ and is in any case not a legal operation on an inequality whose divisor has unknown sign. Factorise instead and both critical values survive.
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" excludes the boundary and "two distinct real roots" excludes it too. Only "at least one real root" or "real roots" (unqualified) includes it. Read which of those phrasings the question used before choosing between $<$ and $\leq$ .
discriminant-applied-before-the-quadratic-exists: Confirmed on a question where the quadratic had to be constructed first: "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 same family of error runs in the other direction after a substitution — having formed and solved a quadratic in a substituted variable, students stop: "a significant number of responses did not go any further than finding solutions to the quadratic in p" (Jan 2025, Q4(b)), a trap confirmed independently across two separate series. Both are the same discipline failing at different ends: know which variable you are working in, and know which one the question asked about.
answer-from-the-calculator-with-no-method-shown: WMA11 is a calculator paper with explicit no-calculator sub-parts, and quadratics are exactly where they land. The Jan 2023 paper carries "(Solutions relying on calculator technology are not acceptable.)" on several parts, and one instructs: "In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable." An examiner report on a solve-the-quadratic part records the consequence: "Quite a number used the formula or factorised to solve the quadratic, whilst others did not show a method and found their solution directly from a calculator... this approach could have been penalised" (Jan 2025, Q4(b)). Treat the warning phrase as a cue rather than boilerplate — it marks the parts where the algebra itself is the thing being examined.

Beyond the spec

The specification asks only that you know and use $b^2 - 4ac > 0$ , $= 0$ and $< 0$ (spec 1.4), and a student can score full marks treating the three cases as three memorised outcomes. This is the one-line argument that makes them a single fact instead of three, and it is genuinely useful in the exam for one reason: it makes "discriminant zero means the roots coincide" impossible to get backwards, because it stops being a rule and becomes a triviality. It is not examined at P1 and no question will ask for it.

Write the two roots of $ax^2 + bx + c = 0$ as $\alpha$ and $\beta$ . Then $ax^2 + bx + c = a(x - \alpha)(x - \beta)$ , and expanding the right-hand side and matching coefficients gives $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$ . Now ask how far apart the roots are. For any two numbers, $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ — a one-line identity you can verify by expanding both sides. Substituting: $(\alpha - \beta)^2 = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2}$ . So the discriminant is, up to the positive factor $a^2$ , exactly the squared gap between the roots. Every one of the three cases now reads off a single sentence rather than being remembered. A positive discriminant means a positive gap: two roots at some real distance from each other. A zero discriminant means zero gap: the two roots are at the same place, which is what "repeated root" means, and which is why the curve touches instead of crossing — the two crossing points have slid together. A negative discriminant means the squared gap between the roots is negative, which no real pair of numbers can manage, so there is no real pair. It also explains something the three-case rule never does: why the discriminant is the natural object to have named at all. It is not an arbitrary combination of the coefficients that happens to be useful — it is the coefficients' answer to the question "how far apart are the roots?", asked without solving for them. (The sum and product of the roots appear nowhere in the transcribed P1 unit content — not under 1.3, 1.4 or 1.5, and not anywhere else in the unit — so this is enrichment, not examinable material. Which other unit of the qualification does examine it was not checked in the research pass this lesson was written from, so no claim is made about that here.)

Retrieval — with feedback on every choice

Question 1

2 marks

How many times does the curve $y = -3x^2 + 12x - 12$ meet the x-axis, and why?

Question 2

4 marks

The equation $px^2 + 4px + 3 = 0$ , where $p$ is a non-zero constant, has two distinct real roots. Which is the complete set of possible values of $p$ ?

Question 3

3 marks

Express $3x^2 + 12x + 5$ in the form $a(x + b)^2 + c$ , where $a$ , $b$ and $c$ are constants.

Question 4

2 marks

A quadratic $f(x) = ax^2 + bx + c$ has $a > 0$ , and its minimum point lies strictly below the x-axis. What must be true of $b^2 - 4ac$ ?

Question 5

2 marks

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. Solve $2x^2 - 7x - 15 = 0$ .

Which response is certain to earn the method mark on the question above?

Question 6

1 mark

Part (a) asked for the minimum value of a quadratic. You made an arithmetic slip and wrote 7, when the correct value is 3. Part (b) begins "Hence...". What is the best thing to do?

Reference — not a study method, a lookup

ax² + bx + c = a(x + b/2a)² − (b² − 4ac)/(4a). Vertex: x = −b/2a, height −(b² − 4ac)/(4a).
b² − 4ac: > 0 two distinct roots (crosses); = 0 one repeated root (touches); < 0 none (misses).
x = (−b ± √(b² − 4ac))/(2a) is that same completed square, solved in letters. Not in the booklet — memorise.
Parameters: can it be zero? Is the inequality strict? Never divide an inequality by the unknown.
Method mark, 3-term quadratic: factorisation, formula with values, or completing the square. Any one.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. Every question in this lesson — prequestion, worked chain, chain drill, method comparison, marked solution and MCQ alike — is VERIDIAN-original wording, inspired by confirmed real question types, never a reproduction of a real Pearson question; and 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 12 marks

How many times does the curve $y = -3x^2 + 12x - 12$ meet the x-axis, and why?

Once — it touches the axis without crossing, because $b^2 - 4ac = 12^2 - 4(-3)(-12) = 144 - 144 = 0$
Correct, and the arithmetic is where this one is won: $-4ac = -4 \times (-3) \times (-12)$ is a product of three negative factors, so it is negative, giving $-144$ . Factorising confirms it: $-3x^2 + 12x - 12 = -3(x-2)^2$ , a downward parabola whose maximum sits exactly on the axis at $x = 2$ .
BTwice, because $b^2 - 4ac = 144 + 144 = 288 > 0$
This computes $b^2 + 4ac$ , not $b^2 - 4ac$ — an error a real examiner report records as scoring no marks at all, since the method mark is for attempting the discriminant and this expression is not it. The minus sign is structural: it comes from combining $-\frac{b^2}{4a}$ with $+c$ when the square is completed.
CNever, because $a < 0$ means the curve opens downwards and therefore stays below the axis
The direction of opening tells you nothing on its own about how many times the curve meets the axis — a downward parabola can cross twice, touch once, or miss entirely, exactly as an upward one can. Here it does touch, at $(2, 0)$ . What decides the count is where the vertex sits relative to the axis, which is what the discriminant measures.
DNever, because $b^2 - 4ac = 0$ and a discriminant of zero means there is no root to find
The discriminant value is right and its meaning is inverted. Zero is the boundary case where the two roots coincide into one repeated root — the curve meets the axis at exactly one point. "No real roots" is the strictly negative case.

Traps tested: Discriminant written as b squared plus 4ac · Opening direction substituted for discriminant · Zero discriminant read as no roots

Question 24 marks

The equation $px^2 + 4px + 3 = 0$ , where $p$ is a non-zero constant, has two distinct real roots. Which is the complete set of possible values of $p$ ?

$p < 0$ or $p > \frac{3}{4}$
Correct. $(4p)^2 - 4(p)(3) > 0$ gives $16p^2 - 12p > 0$ , i.e. $4p(4p - 3) > 0$ . As a function of $p$ that is an upward parabola with roots $0$ and $\frac{3}{4}$ , so it is positive OUTSIDE the roots — both branches. Check the negative branch, which is the one that gets dropped: $p = -1$ gives $-x^2 - 4x + 3 = 0$ , discriminant $16 + 12 = 28 > 0$ , two distinct real roots from a downward-opening curve.
B $0 < p < \frac{3}{4}$
This is the region where $16p^2 - 12p$ is NEGATIVE, which is the no-real-roots case, not the two-distinct-roots case. Check $p = \frac{1}{2}$ : the discriminant is $4 - 6 = -2 < 0$ . The two answers are exact complements, which is what makes this distractor tempting rather than obviously wrong.
C $p > \frac{3}{4}$
Only the right-hand branch. This is what dividing $16p^2 - 12p > 0$ through by $p$ produces, and that division is not legal here — the sign of $p$ is unknown, so you do not know whether the inequality should reverse. Factorising instead keeps both critical values and both branches.
D $p \leq 0$ or $p \geq \frac{3}{4}$
The branches are right and the boundaries should not be included. At $p = \frac{3}{4}$ the discriminant is exactly zero, which is one repeated root rather than two distinct ones; and $p = 0$ is excluded by the stem in any case, since the equation is then not a quadratic. Examiner reports on this question type specifically record final answers losing the last mark for including the boundary.

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

Question 33 marks

Express $3x^2 + 12x + 5$ in the form $a(x + b)^2 + c$ , where $a$ , $b$ and $c$ are constants.

$3(x+2)^2 - 7$
Correct. $3(x^2 + 4x) + 5$ ; half of 4 is 2, and $(x+2)^2 = x^2 + 4x + 4$ is 4 too big, so $3\left[(x+2)^2 - 4\right] + 5 = 3(x+2)^2 - 12 + 5 = 3(x+2)^2 - 7$ . Expanding back confirms it: $3x^2 + 12x + 12 - 7 = 3x^2 + 12x + 5$ .
B $3(x+2)^2 + 1$
The $-4$ was subtracted once instead of being multiplied by the 3 outside the bracket: $5 - 4 = 1$ instead of $5 - 12 = -7$ . Expanding gives $3x^2 + 12x + 13$ , not $3x^2 + 12x + 5$ — which is why expanding your own answer catches this every time, in about five seconds.
C $3(x+4)^2 - 43$
The bracket contains the whole coefficient of $x$ instead of half of it. Expanding gives $3x^2 + 24x + 5$ — the constant term is right, which makes this the most dangerous of the three wrong options, but the coefficient of $x$ is double what it should be. A perfect square always has $2q$ as its $x$ -coefficient, so the bracket takes half.
D $(3x+6)^2 - 31$
The leading factor has been taken inside the square. Squaring $3x$ produces $9x^2$ : expanding gives $9x^2 + 36x + 5$ , a different quadratic entirely. The requested form is $a(x+b)^2 + c$ with the $a$ outside the square, which is also the form that lets the turning point be read off directly.

Traps tested: Constant not multiplied by the leading factor · Used b instead of half b · Leading factor kept inside the square

Question 42 marks

A quadratic $f(x) = ax^2 + bx + c$ has $a > 0$ , and its minimum point lies strictly below the x-axis. What must be true of $b^2 - 4ac$ ?

$b^2 - 4ac > 0$
Correct, and it follows from the completed-square identity rather than from a rule: $f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}$ , so the minimum value is $-\frac{b^2 - 4ac}{4a}$ . With $a > 0$ the denominator $4a$ is positive, so that value is negative exactly when $b^2 - 4ac$ is positive. A minimum below the axis on an upward parabola also means the curve must cross twice, which is the same conclusion arrived at from the picture.
B $b^2 - 4ac < 0$
The sign is inverted. There is a minus sign in front of the fraction in the completed-square form, so a POSITIVE discriminant is what pushes the vertex below the axis when $a > 0$ . A negative discriminant lifts the vertex above the axis and the curve never meets it.
C $b^2 - 4ac = 0$ , because the minimum point is where the curve meets the axis
That describes the case where the minimum point sits exactly ON the axis, not strictly below it. The question specifies strictly below, which is a different position of the same vertex and a different sign of the same quantity.
DNothing can be determined without knowing the value of $a$
The sign of $a$ is all that is needed, and the question gives it: $a > 0$ . The minimum value is $-\frac{b^2 - 4ac}{4a}$ , and whether that is negative depends only on the signs of the numerator and denominator, not on their sizes.

Traps tested: Sign of the vertex formula inverted · Minimum confused with the touching case · Overclaims uncertainty

Question 52 marks

In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. Solve $2x^2 - 7x - 15 = 0$ .

Which response is certain to earn the method mark on the question above?

$(2x + 3)(x - 5) = 0$ , therefore $x = -\frac{3}{2}$ or $x = 5$
Correct. The general principles for pure marking name three routes to the method mark for a three-term quadratic — factorisation with brackets that multiply back to the right leading and constant coefficients, the formula with values substituted, or completing the square — and this shows the first of them explicitly. Any of the three would do; what matters is that the method is visible on the page.
B $x = 5$ , $x = -1.5$
Correct answers with no visible method. The general marking guidance is explicit that answers without working may not gain full credit, and the printed instruction on this question removes the usual benefit of the doubt entirely. The values are right and the marks are not safe.
C"By calculator, $x = 5$ and $x = -1.5$ "
This is the response the instruction was written to exclude. An examiner report on a solve-the-quadratic part records that candidates who found their solution directly from a calculator without showing a method could have been penalised — naming the calculator as the method makes explicit exactly what the instruction disallows.
D $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , quoted and left there with no values substituted
Quoting the formula first is genuinely protective — the general principles record the advice that the formula should be quoted first, because otherwise the method mark has to be inferred from correct working and can be lost to any slip in it. But quoting alone is not an attempt at a solution: the method mark is for using the formula with the values of $a$ , $b$ and $c$ from this equation.

Traps tested: Answer without working · Calculator cited as the method · Formula quoted but never used

Question 61 mark

Part (a) asked for the minimum value of a quadratic. You made an arithmetic slip and wrote 7, when the correct value is 3. Part (b) begins "Hence...". What is the best thing to do?

Answer part (b) correctly using your own value of 7, and carry on
Correct. Rule 4 of the general marking guidance states that A marks are correct answer only unless shown, for example as A1 ft, to indicate that previous wrong working is to be followed through — and on a "hence" part the method mark and any ft-marked accuracy marks are available on your own value. A correct method applied correctly to a wrong number is still a correct method, and it is still paid for.
BLeave part (b) blank, since a wrong value in (a) makes everything after it wrong
This is the most expensive response available. It converts a one-mark accuracy slip into the loss of an entire part, and it forfeits marks the scheme was going to award regardless of the earlier error.
CWrite 3 in part (b) without changing part (a), since you now suspect 3 is right
This leaves the script internally inconsistent: part (b) does not follow from part (a), so there is no working for a follow-through mark to attach to, and a value that appears from nowhere has no method behind it. If you believe (a) is wrong, correct (a) itself and let (b) follow from the corrected value.
DCross out the whole question and restart it from part (a), whatever it costs in time
Worth it only if you spot the slip immediately and the paper is going well on time. As a default it is the wrong trade: the follow-through marks in (b) were available without redoing anything, and 75 marks in 90 minutes leaves little room to re-run a question you have already been paid for most of.

Traps tested: Abandons the question after an error · Answer inconsistent with own working · Restarts instead of continuing

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: Oct 2021 · Q6(a) — cited directly in this lesson
Examiner report: Jan 2023 · Q4 — cited directly in this lesson

Pearson's official past-papers portal

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

Quadratic Functions, the Discriminant, and Solving Quadratic Equations

What a quadratic is, and the one condition that makes it one

Completing the square — the mechanism, then the two complications

Where b2−4acb^2 - 4acb2−4ac actually comes from — it is not a rule, it is a leftover

The quadratic formula is completing the square, done once, in letters

Where the exam hides the quadratic

The full range of kkk for which kx2+8kx+3=0kx^2 + 8kx + 3 = 0kx2+8kx+3=0 has no real roots — including the bound the question never states

Complete the chain — the values of ppp for which y=x2+(p+4)x+25y = x^2 + (p+4)x + 25y=x2+(p+4)x+25 meets the x-axis exactly once

Deep dive

Practice this for real

Where $b^2 - 4ac$ actually comes from — it is not a rule, it is a leftover

The full range of $k$ for which $kx^2 + 8kx + 3 = 0$ has no real roots — including the bound the question never states

Complete the chain — the values of $p$ for which $y = x^2 + (p+4)x + 25$ meets the x-axis exactly once