Learn9709Paper 2 · Pure Mathematics 2Numerical solution of equationsLocating roots: sketches and sign changes

Numerical solution of equations · Subtopic 1 of 3

Locating Roots: Sketches and Sign Changes

Most equations have no exact solution. Before you can approximate a root you need to know how many there are and roughly where: sketch two graphs to count them, then trap each one with a change of sign.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Rewrite an equation as two simpler graphs and use a sketch to count its roots.
Show that a root lies in an interval by a change of sign, with a full written conclusion.
Confirm that a root is correct to a given number of decimal places by testing the bounds.
Recognise when a change of sign is misleading: asymptotes, touching roots and pairs of roots.

Pause and think

There is no formula for the solutions of $x^5+x=1$ . Without solving it, how could you tell that it has a solution between $0$ and $1$ ?

Section 2 of 14

Core idea

Start hereEquations nobody can solve exactly

$x=\cos x$ has a solution (draw the graphs, they cross), but no algebra will ever produce it as a formula. Yet a calculator gives $0.739085\ldots$ How? By trapping the root in smaller and smaller intervals. The first tool is simple: if a smooth curve goes from below the axis to above it, it must cross in between.

The big idea

If $f$ is continuous on $[a,b]$ and $f(a)$ , $f(b)$ have opposite signs, then $f(x)=0$ has a root between $a$ and $b$ . Watch out for asymptotes (sign change, no root) and touching roots (root, no sign change).

1. The sign-change rule

y = x³ − 2x − 5 changes sign between 2 and 2.1

f(2)<0

and

f(2.1)>0

, and the curve is unbroken, so the root lies between

2

and

2.1

(it is

2.0946\ldots

f(a)\cdot f(b)<0\ \text{and}\ f\text{ continuous on }[a,b]\ \Rightarrow\ \exists\,\alpha\in(a,b):\ f(\alpha)=0

continuous: the graph can be drawn without lifting the pen: no gaps, jumps or asymptotes.
$\exists$: "there exists".
$\alpha$: the root.

Worked example $x^3-2x-5=0$ has a root in $(2,2.1)$

Show it.

Goal: A sign change.
Strategy: Evaluate $f$ at both ends.

$f(2)=8-4-5=-1<0$
Left end.
$f(2.1)=9.261-4.2-5=0.061>0$
Right end.
Sign change and $f$ (a polynomial) is continuous, so a root lies in $(2,2.1)$ .

AnswerShown.

✓ Check

f(2.1)

is much smaller in size than

f(2)

, so the root is near

2.1

(it's

2.0946

) ✓.

Worked example $x=\cos x$

Show there is a root between $0.7$ and $0.8$ .

Goal: Sign change.
Strategy: Rearrange to $f(x)=0$ first.

$f(x)=x-\cos x$
Radians.
$f(0.7)=-0.065,\quad f(0.8)=0.103$
Evaluate.

AnswerSign change ⇒ root in

(0.7,0.8)

✓ CheckThe graphs below cross once ✓.

y = x and y = cos x cross once

The graphs cross once, so

x=\cos x

has exactly one root, close to

0.739

Key takeawayCounting roots: sketch two graphs Rewrite

f(x)=0

g(x)=h(x)

with two familiar graphs; count crossings.

2. When the rule fails

Two traps: an asymptote and a touching root

Left: a sign change between

1.5

and

2.5

, but no root, just a break in the graph. Right: a root at

4.5

, but the curve touches and bounces back, so the sign never changes.

Asymptote: $\frac1{x-1}$ changes sign across $x=1$ but has no root.
Touching root: $(x-2)^2$ has a root but no sign change.
Two close roots can hide each other.

Worked exampleProving accuracy

Show the root of $x^3-2x-5=0$ is $2.09$ to 2 d.p.

Goal: Bound the root.
Strategy: Test the rounding boundaries $2.085$ and $2.095$ .

$f(2.085)=-0.106<0$
Lower bound.
$f(2.095)=0.0050>0$
Upper bound.

AnswerRoot in

(2.085,2.095)

, so

2.09

to 2 d.p.

✓ CheckAny number in that interval rounds to

2.09

✓.

Pause and think

Show $e^x=3-x$ has a root between 0.7 and 0.8.

Summary

Continuous + sign change ⇒ root.
Sketch to count roots.
Bounds at rounding limits.

Continuous: No breaks in the graph.

HistoryProving the obvious. That an unbroken curve going from below the axis to above it must cross the axis seems obvious. Bernard Bolzano, a priest and mathematician in Prague, was one of the first to prove it carefully, in 1817. It is now called the intermediate value theorem, and it is the reason every sign-change argument works.

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