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Quadratic Functions

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Cambridge IGCSE Additional Maths

Quadratic Functions

💡 The Big Idea: A quadratic equation is any equation with an x² term in it, and because squaring can "hide" a sign, every quadratic naturally wants to give you two answers — your job is to find both, using whichever tool (factorising, the formula, or completing the square) is fastest for the numbers you're given.

Summary — What's in this Chapter

Solving by factorising — split the quadratic into two brackets and set each to zero.
The quadratic formula — a "works every time" formula, especially good with large or ugly numbers.
Completing the square — rewrites the quadratic so x appears only once; reveals the turning point.
Choosing the right method — a decision-making skill examiners specifically test.
Hidden quadratics — equations in disguise (x⁴, √x, 2ˣ, tan x) that are secretly quadratic in structure.
The discriminant (b² − 4ac) — tells you how many real roots a quadratic has, without solving it.
Quadratic graphs — the parabola shape, roots, y-intercept, and turning point.
Quadratic inequalities — solving ax² + bx + c > 0 (or <, ≤, ≥) using a sketch.
Inequalities on graphs — shading a region that satisfies several inequalities (linear + quadratic) at once.

1. Solving Quadratics by Factorising

The Core Idea

Think of a quadratic equation like (x − 3)(x + 7) = 0 as two numbers being multiplied together to make zero. The only way two numbers multiply to give zero is if at least one of them IS zero. There's no other way — you can't multiply two non-zero numbers and land on zero. So if A × B = 0, then either A = 0 or B = 0 (or both).

This is the entire trick behind factorising. Once your quadratic is written as two brackets multiplied together and set equal to zero, you just solve each bracket separately.

ax² + bx + c = 0 → factorise → (bracket 1)(bracket 2) = 0 Rearrange so zero is on one side (ideally with a positive x² term) before you factorise.

Worked Example — Step by Step

Solve (2x − 3)(3x + 5) = 0

Step 1 — Set the first bracket to zero: 2x − 3 = 0 → 2x = 3 → x = 3/2

Step 2 — Set the second bracket to zero: 3x + 5 = 0 → 3x = −5 → x = −5/3

Step 3 — Write both solutions together: x = 3/2 or x = −5/3

⚠️ The classic mistake — don't lose a solution! To solve x(x − 4) = 0, it's tempting to divide both sides by x, leaving x − 4 = 0, giving only x = 4. But this silently deletes the x = 0 solution. Dividing by an unknown that could be zero is illegal in algebra — always treat x as its own bracket instead. Correct answer: x = 0 or x = 4.

Practice Question 1

Solve x(5x − 1) = 0.

Practice Question 2

Solve (8x + 7)(2x − 3) = 0.

2. The Quadratic Formula

Why It Always Works

Not every quadratic factorises nicely — sometimes the numbers are just too ugly or "random". The quadratic formula is the universal backup plan: plug in a, b and c from ax² + bx + c = 0, and it spits out both solutions, guaranteed, every single time (as long as real solutions exist).

x = ( −b ± √(b² − 4ac) ) / 2a In plain words: take negative b, plus-or-minus the square root of "b squared minus 4ac", all divided by 2a.

🛡️ Safety habit: bracket your negatives Whenever you substitute a negative number into the formula, put brackets around it first. If b = −5, write −(−5), not just −−5. This single habit prevents most sign-error mistakes in this topic.

Worked Example — Step by Step

Solve 2x² − 5x + 2 = 0 using the formula

Step 1 — Identify: a = 2, b = −5, c = 2

Step 2 — Substitute (bracket negatives): x = ( −(−5) ± √((−5)² − 4×2×2) ) / (2×2)

Step 3 — Simplify inside the root: x = (5 ± √(25 − 16)) / 4 = (5 ± √9) / 4 = (5 ± 3) / 4

Step 4 — Split + and −: x = 8/4 = 2, or x = 2/4 = 1/2

Answer: x = 2 or x = 1/2

Practice Question

Use the quadratic formula, without a calculator, to find the exact solutions of 12x² − 17x + 6 = 0.

3. Completing the Square

The Core Idea

Completing the square rewrites y = ax² + bx + c as y = a(x + p)² + q. The magic of this form is that x only appears once. That single change unlocks two superpowers: you can solve the equation using ±√ (no factorising needed), and you can instantly read off the turning point of the graph.

When a = 1: p = b/2 and q = c − p² When a ≠ 1, factor a out of the x² and x terms first, then apply the same rule inside the bracket.

Worked Example — Solving by Completing the Square

Solve 2x² − 8x − 24 = 0 by completing the square

Step 1 — Divide everything by 2 (the coefficient of x²): x² − 4x − 12 = 0

Step 2 — Halve the middle number (−4 → −2) and rewrite: (x − 2)² − (−2)² − 12 = 0

Step 3 — Simplify: (x − 2)² − 4 − 12 = 0 → (x − 2)² − 16 = 0

Step 4 — Make x the subject: (x − 2)² = 16 → x − 2 = ±√16 = ±4

Step 5 — Solve both cases: x = 4 + 2 = 6, or x = −4 + 2 = −2

Answer: x = 6 or x = −2

💡 Why bother, if factorising is faster? Completing the square isn't just an alternative solving method — it's the fastest way to find a parabola's turning point without calculus. Once in the form a(x + p)² + q, the turning point is always at (−p, q). This works because a squared term can never be negative, so its smallest possible value is 0 — which happens exactly when x = −p.

Practice Question

Write 3x² − 12x + k in the form a(x + p)² + k + q, where a, p and q are constants, and k is an unknown constant.

4. Which Method Should I Use?

This is a skill examiners deliberately test — knowing WHICH tool fits the job, not just knowing all three tools individually.

Situation	Best Method
Question explicitly says "solve by factorisation"	Factorising
Two-term quadratics (e.g. x² − 4x = 0, or x² − 9 = 0)	Factorising (common factor, or difference of two squares)
Coefficients are large or "ugly", and factorising is hard to spot	Quadratic Formula
Answer needs to be rounded (e.g. "to 2 d.p." or "3 s.f.")	Quadratic Formula
Answer needs to be exact, involving surds	Quadratic Formula (surds appear naturally)
Question explicitly says "complete the square" or "hence solve"	Completing the Square
You need the turning point of the graph	Completing the Square
You genuinely can't decide	Quadratic Formula — it always works!

5. Hidden Quadratic Equations

Spotting the Disguise

A "hidden" quadratic has the exact same structure as a normal quadratic — a(something)² + b(something) + c = 0 — just wearing a costume. The "something" could be x², √x, 2ˣ, or even tan(3x). If you can see a repeating expression where one version is the square of the other, it's a hidden quadratic.

x⁴ − 3x² − 4 = 0 is a quadratic in x² (because x⁴ = (x²)²)
x − 3√x − 4 = 0 is a quadratic in √x (because (√x)² = x)
4ˣ − 3(2ˣ) − 4 = 0 is a quadratic in 2ˣ (because 4ˣ = (2²)ˣ = (2ˣ)²)
3tan²(3x) + 4tan(3x) − 6 = 0 is a quadratic in tan(3x)

Substitute u = "the something" → solve the simple quadratic in u → replace u back → solve for x

Worked Example

Solve x − √x − 6 = 0

Step 1 — Let u = √x, so u² = x. Rewrite: u² − u − 6 = 0

Step 2 — Factorise: (u − 3)(u + 2) = 0 → u = 3 or u = −2

Step 3 — Replace u with √x: √x = 3 → x = 9. But √x = −2 has no solution, because a square root can never be negative.

Answer: x = 9 (always check by substituting back in!)

✅ Golden checking habit With hidden quadratics, one of your two u-solutions often turns out to be impossible once you substitute back (like √x = −2 above, or x² = −1). Always check both branches — don't assume both u-values lead to valid x-values.

Practice Question

Solve x⁸ − 17x⁴ + 16 = 0.

6. Discriminants

What It Tells You Without Solving Anything

Look back at the quadratic formula — specifically the bit under the square root sign: b² − 4ac. This is called the discriminant, sometimes written as Δ (capital delta). It's incredibly useful because you can calculate it in seconds, and it tells you exactly how many real solutions a quadratic has — before you even try to solve it.

Δ = b² − 4ac

Discriminant value	What it means	Graph behaviour
b² − 4ac > 0	Two distinct real roots	Crosses the x-axis twice
b² − 4ac = 0	One repeated (equal) root	Touches the x-axis once
b² − 4ac < 0	No real roots	Never touches the x-axis

Line Meets Curve: Tangent Test

The discriminant also tells you how many times a straight line crosses a curve. Set the line equal to the curve, rearrange into ax² + bx + c = 0 form, then check the discriminant. If it equals exactly 0, the line just grazes the curve — it's a tangent.

Worked Example: Show that y = 3x − 7 is tangent to y = (x + 3)(x − 2)

Step 1 — Set equal: 3x − 7 = (x + 3)(x − 2)

Step 2 — Expand and rearrange: 3x − 7 = x² + x − 6 → x² − 2x + 1 = 0

Step 3 — Discriminant: Δ = (−2)² − 4(1)(1) = 4 − 4 = 0

Since Δ = 0, the line touches the curve at exactly one point — it's a tangent. ✓

Practice Question

Find, in terms of k, the discriminant of 3x² + 2kx − k = kx² − 4kx + 2.

7. Quadratic Graphs

The Parabola Shape

Every quadratic graph makes a curved shape called a parabola. There are only two possible orientations, and the sign of a (the coefficient of x²) tells you instantly which one you'll get:

a > 0 (positive) → "∪-shape" (like a smile) → has a minimum point
a < 0 (negative) → "∩-shape" (like a frown) → has a maximum point

🗺️ Four features to find before sketching 1) Overall shape (sign of a) · 2) x-intercepts / roots (solve = 0, use discriminant to check how many exist) · 3) y-intercept (set x = 0, it's always (0, c)) · 4) Turning point (complete the square, or spot it if there's only one root).

Worked Example — Full Sketch

Sketch y = x² − 5x + 6

Shape: a = 1 > 0 → ∪-shape

y-intercept: c = 6 → crosses y-axis at (0, 6)

Roots: factorise → (x − 2)(x − 3) = 0 → x = 2 or x = 3, so crosses x-axis at (2, 0) and (3, 0)

The turning point sits neatly between the two roots, at the minimum of the ∪-shape.

Using the Graph to Find the Range

The turning point tells you the range of a quadratic function — the set of all possible output (y) values. If the graph has a minimum point at (xₘᵢₙ, yₘᵢₙ), then the function can never produce a y-value below yₘᵢₙ, so the range is f(x) ≥ yₘᵢₙ. Similarly for a maximum point, the range is f(x) ≤ yₘₐₓ.

Practice Question

Sketch y = f(x) where f(x) = 2x² − 4x − 6. Find the turning point and write down the range of f(x).

8. Quadratic Inequalities

Solving with a Sketch

A quadratic inequality looks like a quadratic equation but with <, >, ≤ or ≥ instead of =. The reliable way to solve these — every single time — is to sketch the graph and read off which region of x satisfies the inequality. Trying to reason it out purely algebraically is a common source of errors.

Step 1: Rearrange to ax² + bx + c ⋚ 0 (positive x² term) Step 2: Solve ax² + bx + c = 0 to find the roots x₁ and x₂ (x₁ ≤ x₂) Step 3: Sketch the ∪-shaped graph, labelling the roots Step 4: Read off the region — "> 0" means above the x-axis (x < x₁ or x > x₂); "< 0" means below the x-axis (x₁ < x < x₂)

Worked Example

Find the set of values for which 3x² + 2x − 6 > x² + 4x − 2

Step 1 — Rearrange: 3x² + 2x − 6 − x² − 4x + 2 > 0 → 2x² − 2x − 4 > 0

Step 2 — Simplify by dividing by 2: x² − x − 2 > 0

Step 3 — Factorise: (x − 2)(x + 1) > 0, so roots are x = −1 and x = 2

Step 4 — Sketch a ∪-shape through −1 and 2. We want where the graph is ABOVE the x-axis (since > 0).

Answer: x < −1 OR x > 2

⚠️ The negative-coefficient trap Never multiply or divide an inequality by a negative number without flipping the sign (< becomes >, and so on). It's much safer to rearrange the whole inequality first so the x² term is positive, avoiding the flip altogether. Also — quadratic inequality answers are almost never written with "AND" when the solution is two separate ranges; check carefully whether your answer should use "or" (two separate regions) or a single squeezed range like x₁ < x < x₂.

Practice Question

Find the values of k such that (k + 1)x² − 4x + (k − 2) = 0 has no real roots. (Hint: use the discriminant to set up a quadratic inequality in k, then solve it.)

9. Inequalities on Graphs (Regions)

Shading a Region

This is a different (but related) skill: instead of one variable, these inequalities involve two — x and y. Several inequalities are drawn together on the same set of axes, and the solution isn't a set of x-values, it's a whole 2D area, usually labelled R.

Step 1: Find key points for each line/curve Step 2: Draw each — dotted line for strict inequalities (< or >), solid line for weak inequalities (≤ or ≥) Step 3: Shade the UNWANTED side of each line/curve Step 4: Whatever's left unshaded is your answer region, R

🎯 Quick test for which side to shade If you're not sure which side of a line satisfies an inequality, plug in the origin (0, 0) — it's the easiest point to test. If (0, 0) satisfies the inequality, then the whole side of the line containing the origin is the "wanted" side.

👀 Read the question carefully Some exam questions ask you to shade the wanted region directly, rather than shading the unwanted parts and leaving R blank. Always check exactly which convention the question is using before you start shading — it changes what your final diagram should look like.

What to Memorise

Concept	Formula / Rule
Quadratic Formula	x = (−b ± √(b² − 4ac)) / 2a
Completing the Square (a = 1)	p = b/2, q = c − p² → y = (x + p)² + q
Discriminant	Δ = b² − 4ac
Δ > 0	Two distinct real roots — graph crosses x-axis twice
Δ = 0	One repeated root — graph touches x-axis once (tangent case)
Δ < 0	No real roots — graph never meets x-axis
y-intercept of ax² + bx + c	Always (0, c)
Turning point from a(x + p)² + q	Always at (−p, q)
Shape rule	a > 0 → ∪ (minimum); a < 0 → ∩ (maximum)
Range (minimum point)	f(x) ≥ yₘᵢₙ
Range (maximum point)	f(x) ≤ yₘₐₓ
Hidden quadratic method	Substitute u = "the repeating expression", solve for u, then convert back to x
Solving ax²+bx+c > 0	x < x₁ or x > x₂ (outside the roots)
Solving ax²+bx+c < 0	x₁ < x < x₂ (between the roots)

Concepts Checklist

Tick each one off only once you could explain it to someone else without looking at your notes.

I can solve a quadratic by factorising, including two-term quadratics like x(x − 4) = 0
I know why dividing by x can lose a solution, and I avoid doing it
I can substitute correctly into the quadratic formula, bracketing negative numbers
I can complete the square when a = 1 and when a ≠ 1
I can use completing the square to solve an equation using ±√
I can choose the most sensible method (factorise / formula / complete the square) for a given question
I can recognise and solve hidden quadratics using substitution (u = ...)
I check both u-solutions actually lead to valid x-solutions
I can calculate the discriminant and interpret what it means (0, 1, or 2 roots)
I can use the discriminant to test whether a line is tangent to a curve
I can sketch a quadratic graph labelling y-intercept, roots, and turning point
I can state the range of a quadratic function from its graph
I can solve quadratic inequalities using a sketch, avoiding the negative-flip trap
I can shade a region on a graph satisfying several inequalities at once

Exam Tips — Common Mistakes & Mark-Scheme Traps

🚫 Losing the x = 0 solution Dividing both sides of x(x − 4) = 0 by x throws away a valid answer. Always solve "bracket = 0" for every factor, including bare x.

🚫 Forgetting ± when square-rooting (x + 5)² = 16 has TWO solutions: x + 5 = +4 and x + 5 = −4. Skipping the ± is one of the most common lost marks in this whole chapter.

✅ Match large coefficients to a factor e.g. solutions x = −2/3 and x = 1/2 from 12x² + 2x − 4 = 0 do NOT simply give (3x+2)(2x−1) — check by expanding, and multiply by the missing constant if needed.

✅ "Correct to 2 d.p." is a strong hint Whenever a question asks for a rounded answer, it's telling you factorising probably won't work cleanly — reach for the quadratic formula.

🚫 Expanding back out after completing the square When solving via completing the square, keep the squared bracket intact until the very last step (±√) — expanding it early throws away the whole point of the method.

🚫 Forgetting to flip the inequality Dividing an inequality by a negative number flips the direction of the sign. Best practice: rearrange first so the x² coefficient is positive, avoiding the need to divide by a negative at all.

✅ Always sketch for inequality questions Both quadratic inequalities (one variable) and region-shading questions (two variables) become far less error-prone once you sketch first and reason from the picture.

✅ Calculators can check, not replace, working Many calculators solve quadratics directly — use this to verify your answer, but always show full algebraic working, since that's where the marks are.

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Exam Tips — Common Mistakes & Mark-Scheme Traps

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