A Level Maths 9709 · Paper 1 Pure Mathematics 1 · 1.1 · ~13 min read

Quadratics

Completing the square, the discriminant, quadratic inequalities, line–curve intersections and equations that are quadratics in disguise, with the working Cambridge examiners expect to see.

Completing the squareVertex & rangeDiscriminantSolving quadraticsQuadratic inequalitiesLine meets curveQuadratics in disguise

RSby Riyaz Sir
mentorriyaz.in

Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

Completing the square turns ax² + bx + c into a(x − p)² + q. It gives the vertex (p, q), the axis of symmetry x = p, and the greatest or least value of the expression.
The discriminant b² − 4ac tells you how many real roots there are: two if it is positive, one repeated root if zero, none if negative.
Inequalities: find the critical values, sketch the parabola, then read off the region. Never divide an inequality by something whose sign you don't know.
Line meets curve: substitute to get one quadratic. Its discriminant decides whether the line cuts, touches (tangent) or misses the curve.
Show your working. Cambridge gives no marks for unsupported calculator answers, so factorise or substitute into the formula on paper.

By the end you can

Complete the square for any quadratic, including a ≠ 1, and use it to find a vertex or a range.
Use the discriminant to find conditions on a constant k.
Solve quadratic inequalities and write the answer correctly.
Solve a linear and a quadratic equation simultaneously and decide whether a line is a tangent.
Recognise and solve equations that are quadratic in x², √x, tan x and similar.

1Completing the square

Why it works

x² + bx is the start of a perfect square: (x + b/2)² = x² + bx + (b/2)². So x² + bx = (x + b/2)² − (b/2)².
Halve the x-coefficient, square it, subtract it. Everything else is bookkeeping.
With a ≠ 1, take a out of the x² and x terms first: ax² + bx + c = a[x² + (b/a)x] + c.

ax² + bx + c = a(x − p)² + q

vertex (p, q); axis of symmetry x = p

a > 0: the parabola is ∪-shaped, so q is the minimum value. a < 0: ∩-shaped, so q is the maximum.

The completed square shows the vertex directly. Here the minimum value of the expression is −3, when x = 2.

Worked examples

Worked example

Express 2x² − 12x + 13 in the form a(x + b)² + c, and state the range of f(x) = 2x² − 12x + 13 for x ∈ ℝ.

2x² − 12x + 13 = 2[x² − 6x] + 13 M1
= 2[(x − 3)² − 9] + 13 = 2(x − 3)² − 18 + 13
= 2(x − 3)² − 5 A1 A1 (a = 2, b = −3, c = −5)
(x − 3)² ≥ 0, so f(x) ≥ −5. Range: f(x) ≥ −5 B1

Worked example

Express 5 + 8x − 2x² in the form p − q(x − r)² and hence state the maximum value.

5 + 8x − 2x² = 5 − 2[x² − 4x] M1
= 5 − 2[(x − 2)² − 4] = 13 − 2(x − 2)² A1
the squared term is never negative, so the maximum is 13, when x = 2 B1

Common mistakeNot checking your answer. The June 2024 Paper 1 report says many candidates did not check that their completed square expanded back to the original. Expand it in your head: 2(x − 3)² − 5 = 2x² − 12x + 18 − 5 ✓.

Common mistakeForgetting to multiply the −(b/2)² by a. In 2[(x − 3)² − 9] + 13, the −9 becomes −18, not −9.

2The discriminant

Three cases

b² − 4ac LEARN IT

the expression under the square root in the quadratic formula

Discriminant	Roots of ax² + bx + c = 0	Graph of y = ax² + bx + c
b² − 4ac > 0	two distinct real roots	crosses the x-axis twice
b² − 4ac = 0	one repeated (equal) root	touches the x-axis (vertex on the axis)
b² − 4ac < 0	no real roots	never meets the x-axis
b² − 4ac ≥ 0	real roots (equal or distinct)	meets the x-axis

Shifting the same parabola up changes the discriminant from positive to zero to negative.

Why: the formula x = [−b ± √(b² − 4ac)]/2a has two different values if the square root is of a positive number, one value if it is of zero, and no real value if it is of a negative number.
"Positive for all x" (always above the axis) means a > 0 and b² − 4ac < 0. "Negative for all x" means a < 0 and b² − 4ac < 0.

Conditions on a constant k

Worked example

Find the set of values of k for which kx² + 4x + (k − 3) = 0 has two distinct real roots.

two distinct real roots ⇒ b² − 4ac > 0: 16 − 4k(k − 3) > 0 M1
16 − 4k² + 12k > 0 ⇒ k² − 3k − 4 < 0 (divided by −4, so the sign flips) A1
(k − 4)(k + 1) < 0 ⇒ critical values −1 and 4 M1
a ∪-shaped graph in k is negative between its roots: −1 < k < 4, k ≠ 0 A1

Why k ≠ 0? When k = 0 the equation becomes 4x − 3 = 0, which is linear and has only one root. Examiners reward students who notice this.

Exam tipWhen you divide or multiply an inequality by a negative number, reverse the sign. Safer: move everything so the k² term is positive before factorising.

3Solving quadratic equations

Factorise when the numbers are friendly: check ac and look for two numbers that multiply to ac and add to b.
Formula otherwise FORMULA LIST (MF19): x = [−b ± √(b² − 4ac)]/2a. Write the substitution down, e.g. x = [5 ± √(25 + 24)]/6.
Completing the square when you need exact surd answers or the vertex anyway.

Common mistakeOnly writing the answers from a calculator's equation solver. Cambridge's rubric: "no marks will be given for unsupported answers from a calculator". Show the factorisation or the formula with numbers in.

4Quadratic inequalities

The method

1 · Rearrange so one side is 0 and the x² coefficient is positive.
2 · Critical values: solve the equation = 0.
3 · Sketch a ∪-shaped parabola through the critical values.
4 · Read off: "> 0" means where the graph is above the axis (outside the roots); "< 0" means below (between the roots).

Worked example

Solve 3x² − 5x − 2 > 0.

(3x + 1)(x − 2) = 0 ⇒ critical values x = −⅓ and x = 2 M1
the ∪-shaped graph is above the x-axis outside the roots
x < −⅓ or x > 2 A1

Worked example

Solve x(x + 4) ≤ 12.

x² + 4x − 12 ≤ 0 ⇒ (x + 6)(x − 2) ≤ 0 M1
critical values −6 and 2; the graph is below or on the axis between them
−6 ≤ x ≤ 2 A1

Writing the answer: "x < −⅓ or x > 2" is two separate regions. "−6 ≤ x ≤ 2" is one region. Writing "2 < x < −⅓" is wrong, because no number satisfies it.

Common mistakeUsing ≤ where the question has <. Strict and non-strict inequalities must match the question; the June 2024 report flagged confusion between ≤ and <.

5Where a line meets a curve

Substitute, then use the discriminant

Rearrange the linear equation for one variable and substitute into the quadratic. You get a quadratic in one variable.
Its roots are the x-coordinates of the intersection points. Substitute back into the linear equation to get y (it is simpler and avoids extra roots).

Worked example

Find the points where y = 3 − x meets y = x² − 2x − 3.

x² − 2x − 3 = 3 − x ⇒ x² − x − 6 = 0 M1
(x − 3)(x + 2) = 0 ⇒ x = 3 or x = −2 A1
y = 3 − x gives (3, 0) and (−2, 5) A1

Worked example

The line y = 2x + c is a tangent to y = x² − 4x + 7. Find c and the point of contact.

x² − 4x + 7 = 2x + c ⇒ x² − 6x + (7 − c) = 0 M1
tangent ⇒ one repeated root ⇒ b² − 4ac = 0: 36 − 4(7 − c) = 0 M1
8 + 4c = 0 ⇒ c = −2 A1
x² − 6x + 9 = 0 ⇒ (x − 3)² = 0 ⇒ x = 3, y = 2(3) − 2 = 4: point (3, 4) A1

The line y = x + k cuts the curve twice, touches it once, or misses it, depending on the discriminant.

Discriminant of the combined quadratic	The line…
> 0	meets the curve at two points
= 0	is a tangent (touches at one point)
< 0	does not meet the curve

6Quadratics in disguise

Spot the hidden quadratic

Look for a term that is the square of another: x⁴ = (x²)², x = (√x)², x^2/3 = (x^1/3)², tan²x = (tan x)².
Substitute u for the inner expression, solve the quadratic in u, then go back to x. Check each u is allowed: √x can't be negative, x² can't be negative.

Worked example

Solve x⁴ − 5x² + 4 = 0.

let u = x²: u² − 5u + 4 = 0 M1
(u − 1)(u − 4) = 0 ⇒ u = 1 or u = 4 A1
x² = 1 or x² = 4 ⇒ x = ±1, ±2 A1

Worked example

Solve 2x − 7√x + 3 = 0.

let u = √x: 2u² − 7u + 3 = 0 M1
(2u − 1)(u − 3) = 0 ⇒ u = ½ or u = 3 A1
√x = ½ ⇒ x = ¼; √x = 3 ⇒ x = 9. Both u values are positive, so x = ¼ or x = 9 A1

Worked example

Solve 6 tan²θ − tan θ − 2 = 0 for 0° ≤ θ ≤ 180°.

(3 tan θ − 2)(2 tan θ + 1) = 0 ⇒ tan θ = ⅔ or tan θ = −½ M1 A1
tan θ = ⅔ ⇒ θ = 33.7°
tan θ = −½ ⇒ θ = 180° − 26.6° = 153.4° A1 A1

Common mistakeKeeping an impossible solution. If u = √x gives u = −3, reject it: √x is never negative. In 2024, Paper 1 candidates lost a mark for keeping an extra invalid solution from a quadratic.

Linking questionCompleting the square gives the range of a function (Functions and transformations); the discriminant decides whether a line is a tangent to a circle (Coordinate geometry); and disguised quadratics in sin x and cos x appear in Trigonometry.

Test yourself — cover the answers first

Quick check

What does the completed square a(x − p)² + q tell you about y = ax² + bx + c?
State the condition for ax² + bx + c = 0 to have real roots.
Solve x² − x − 12 < 0.
Express x² + 10x + 3 in completed square form.
Which substitution turns x^2/3 − 3x^1/3 + 2 = 0 into a quadratic?

Exam-style

Find the values of m for which the line y = mx − 1 is a tangent to the curve y = x² + 3.
f(x) = −x² + 6x − 4. (a) Express f(x) as a − (x − b)². (b) Find the range of f. (c) Solve f(x) > 1.
Find the set of values of k for which x² + kx + 9 > 0 for all real x.

Show answers

vertex (p, q); axis x = p; min value q if a > 0, max if a < 0
b² − 4ac ≥ 0
−3 < x < 4
(x + 5)² − 22
u = x^1/3; u = 1 or 2, so x = 1 or 8
x² − mx + 4 = 0; m² − 16 = 0; m = ±4
(a) 5 − (x − 3)²
(b) f(x) ≤ 5
(c) x² − 6x + 5 < 0 ⇒ 1 < x < 5
k² − 36 < 0 ⇒ −6 < k < 6

Practise this chapter

A Level practice papers

Next chapterFunctions and transformations →

Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →

Share with a friend Ask Riyaz Sir