B3PureCore

Quadratic Functions

Q: How do I know whether to factorise or use the formula?

Try factorising first; if two whole numbers multiply to ac and add to b, it factorises and that is quickest. If they do not exist, or the question asks for decimals or surds, use the formula, which always works.

Q: Why can’t I just divide both sides by x?

Because x might be zero, and dividing by it deletes the root x = 0. Always move everything to one side and factor the x out instead.

Q: What does it mean if the formula gives a negative under the square root?

It means b^2 - 4ac < 0, so there are no real roots — the parabola never crosses the x-axis. The discriminant lesson explains this in full.

Q: What is a “quadratic in disguise”?

An equation that becomes a quadratic after a substitution, such as x^4 - 5x^2 + 4 = 0 with u = x^2, or an equation in e^x or x. Spotting the hidden u^2 + pattern lets you solve it with the usual tools.

A quadratic is any expression with an $x^{2}$ term as its highest power. Solving one means finding where its curve crosses the $x$ -axis — and you have three tools for the job: factorising, the formula, and completing the square.

30 min Video by TLMaths Algebra and Functions

The big picture

Quadratics are the first genuinely curved functions you study, and they are everywhere: the path of a thrown ball, the area of a rectangle with a fixed perimeter, the profit at a given price. Their U-shaped (parabola) graph turns once, which is why they can have two solutions, one, or none. Almost every later topic — the discriminant, completing the square, quadratic inequalities, even disguised quadratics in trigonometry and logarithms — is built on being fluent here. The goal of this lesson is not just to solve a quadratic, but to know instantly which of the three methods will be fastest.

A-Level Maths B3-15: Three Ways to Solve a Quadratic EquationWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Recognise a quadratic and connect its roots to where the graph meets the $x$ -axis
Solve a quadratic by factorising, using the zero-product idea
Solve any quadratic with the quadratic formula
Choose the most efficient method for a given equation

What a quadratic is, and what a root means

A quadratic function has the form $f (x) = a x^{2} + b x + c$ , where $a \neq = 0$ (if $a$ were 0 there would be no $x^{2}$ term and it would just be a straight line). Its graph is a $parabola$ — a symmetric U-shape (or ∩-shape if $a < 0$ ).

The $roots$ (or solutions) of $a x^{2} + b x + c = 0$ are the $x$ -values that make the function zero. Graphically, these are exactly the points where the curve crosses the $x$ -axis. A parabola can cross twice, touch once, or miss entirely — which is why a quadratic has two, one, or no real roots.

Why this works

Setting $f (x) = 0$ and asking “where does the curve hit the $x$ -axis?” are the same question. Keeping the graph in mind stops you accepting an impossible answer, like a single crossing where the algebra should give two.

Method 1 — factorising

Factorising rewrites the quadratic as a product of two brackets. It relies on one powerful fact: if two things multiply to give zero, at least one of them must be zero. So once $(x - p) (x - q) = 0$ , either $x = p$ or $x = q$ .

This is the fastest method $when it works$ — but not every quadratic factorises with whole numbers, which is where the other methods come in.

(x - p) (x - q) = 0 \Rightarrow x = p or x = q

The zero-product principle — the engine of factorising.

1Find two numbers that multiply to

+ 12

and add to

- 7

: they are

- 3

and

- 4

2Factorise:

(x - 3) (x - 4) = 0

3Set each bracket to zero:

x - 3 = 0

x - 4 = 0

Answer

x = 3

x = 4

Tip — Never “divide through by $x$ ” to solve a quadratic — you would throw away the root $x = 0$ . Instead factor the $x$ out: $x^{2} - 5 x = 0$ becomes $x (x - 5) = 0$ , giving $x = 0$ or $x = 5$ .

Method 2 — the quadratic formula

When factorising is awkward or impossible, the quadratic formula solves $a x^{2} + b x + c = 0$ every time. It is derived by completing the square on the general quadratic (the next lesson), so it carries all the same information.

The $\pm$ is what produces the two roots — one from adding the square root, one from subtracting it. Getting the signs of $a$ , $b$ and $c$ right (including their own minus signs) is where care pays off.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Works for every quadratic — memorise it exactly.

1Identify

a = 3

b = - 5

c = - 1

2Substitute:

x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 3 ) ( - 1 )}{2 ( 3 )} = \frac{5 \pm 25 + 12}{6}

= \frac{5 \pm 37}{6}

4Evaluate:

\frac{5 + 6.083}{6} = 1.85

and

\frac{5 - 6.083}{6} = - 0.180

Answer

x = 1.85

x = - 0.180

(3 s.f.)

Why this works

The part under the root, $b^{2} - 4 a c$ , is the $discriminant$ . Its sign alone tells you how many roots exist before you finish the calculation — the subject of a lesson all its own.

Choosing your method

Try to $factorise first$ — if it factorises cleanly it is quickest, and it gives exact answers. If it resists factorising (the numbers do not fall out, or the question asks for surds or decimals), go straight to the $formula$ . Save $completing the square$ for when you also need the turning point or an exact surd form.

A quick check: if the question says “give your answer to 2 decimal places” or “in the form $p \pm q r$ ”, the quadratic almost certainly does not factorise, so reach for the formula immediately.

Think like an examiner

Read the instruction: “to 2 d.p.” signals the formula; “in exact/surd form” signals the formula or completing the square; a clean question often factorises.
When using the formula, write $a$ , $b$ , $c$ (with signs) before substituting — this earns a method mark and prevents sign slips.
A quadratic set to “ $= 0$ ” is essential before any method. If it is not, rearrange first: everything on one side, zero on the other.
Give $both$ roots unless the context (e.g. a length) rules one out — and if it does, state why you reject it.

Common misconceptions

Dividing

x^{2} = 5 x

x

to get

x = 5

Move everything to one side:

x^{2} - 5 x = 0 \Rightarrow x (x - 5) = 0

, so

x = 0

x = 5

. Dividing by

x

loses the

x = 0

root.

Mishandling signs in the formula, e.g. taking

b = 5

when the equation has

- 5 x

Use

b = - 5

. Substitute the coefficients exactly as they appear, including their signs.

Forgetting the

\pm

and giving only one root.

The

\pm

gives two roots; report both unless the context excludes one.

Applying the formula to an equation not yet equal to zero.

Rearrange to

a x^{2} + b x + c = 0

first, or the coefficients will be wrong.

Solving a quadratic

a x^{2} + b x + c = 0

Standard form (

a \neq = 0

(x - p) (x - q) = 0 \Rightarrow x = p, q

Zero-product (factorising).

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The quadratic formula.

Stretch yourself

Solve $x^{4} - 5 x^{2} + 4 = 0$ . (Hint: it is secretly a quadratic.)

Hint — Let $u = x^{2}$ . The equation becomes a quadratic in $u$ . Solve for $u$ , then work back to $x$ — and expect up to four values.

Questions students ask

How do I know whether to factorise or use the formula?

Try factorising first; if two whole numbers multiply to

a c

and add to

b

, it factorises and that is quickest. If they do not exist, or the question asks for decimals or surds, use the formula, which always works.

Why can’t I just divide both sides by

x

Because

x

might be zero, and dividing by it deletes the root

x = 0

. Always move everything to one side and factor the

x

out instead.

What does it mean if the formula gives a negative under the square root?

It means

b^{2} - 4 a c < 0

, so there are no real roots — the parabola never crosses the

x

-axis. The discriminant lesson explains this in full.

What is a “quadratic in disguise”?

An equation that becomes a quadratic after a substitution, such as

x^{4} - 5 x^{2} + 4 = 0

with

u = x^{2}

, or an equation in

e^{x}

sin x

. Spotting the hidden

u^{2} + \dots

pattern lets you solve it with the usual tools.

Key takeaways

A quadratic is $a x^{2} + b x + c$ with $a \neq = 0$ ; its roots are where the parabola meets the $x$ -axis.
Factorising uses the zero-product idea — fastest when it works.
The quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ solves any quadratic.
Rearrange to “ $= 0$ ” first, and never divide by $x$ .
Substitution turns disguised quadratics (in $x^{2}$ , $e^{x}$ , $sin x$ ) into ordinary ones.

How this fits the course

Build on

Index Laws
Factorising (GCSE)soon

Leads to

Completing the Square

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