I1PureCore

Locating Roots

Not every equation can be solved with a formula — $x^{3} - x - 1 = 0$ has no nice factorisation. But you do not need an exact algebraic answer to be CONFIDENT a root exists somewhere in a given interval, and to narrow down exactly where.

28 min Video by TLMaths Numerical Methods

The big picture

Numerical methods exist for exactly the equations algebra cannot crack. This first technique, the $change-of-sign method$ , relies on a simple, powerful idea: if a continuous function is negative at one point and positive at another, it must have crossed zero somewhere in between. This single observation is the starting point for every numerical method in the chapter — locating roots is always step one before refining them.

A-Level Maths I1-01: Locating Roots: Introducing the Change of Sign MethodWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Understand that a continuous function changing sign over an interval implies a root within it
Use the change-of-sign method to show a root lies in a given interval
Narrow down an interval to a required degree of accuracy by repeated halving
Recognise situations where the change-of-sign method can fail

The change-of-sign method

For a continuous function $f (x)$ , if $f (a) < 0$ and $f (b) > 0$ (or vice versa), the graph must cross the $x$ -axis at least once between $x = a$ and $x = b$ — so a root of $f (x) = 0$ lies somewhere in that interval.

f (1) = 1 - 1 - 1 = - 1

(negative).

f (2) = 8 - 2 - 1 = 5

(positive).

3Since

f (1) < 0

and

f (2) > 0

, and

f

is continuous (a polynomial), there is a root between

x = 1

and

x = 2

AnswerA root of

f (x) = 0

lies in the interval

(1, 2)

Tip — State explicitly that $f$ is CONTINUOUS on the interval (true for any polynomial) — this is the condition that makes the sign-change argument valid.

Narrowing the interval

Once a root is trapped in an interval, testing the MIDPOINT and checking its sign halves the interval size, trapping the root even more precisely. Repeating this process (interval bisection) can locate a root to any required degree of accuracy.

1Test the midpoint

x = 1.5

f (1.5) = 3.375 - 1.5 - 1 = 0.875

(positive).

2Since

f (1) < 0

and

f (1.5) > 0

, the root lies in the smaller interval

(1, 1.5)

AnswerThe root lies in

(1, 1.5)

Why this works

Repeating this halving process is exactly how a calculator (or computer) can find a root to as many decimal places as needed — each repetition roughly halves the size of the interval containing the true root.

When the method fails

The change-of-sign method can miss roots, or wrongly suggest one, in two situations: if the function has an EVEN number of roots within the interval (the sign returns to where it started, showing no NET change), or if the function is discontinuous (undefined) somewhere within the interval.

f (1) = (- 1) (- 3) = 3

(positive).

f (5) = (3) (1) = 3

(positive).

2No sign change is detected — yet the function actually has TWO roots (

x = 2

and

x = 4

) within this interval.

AnswerThe method misses BOTH roots because the function returns to the same sign, showing no net change — a sketch or extra checking is needed to catch an even number of roots.

Tip — A sketch (or checking a few extra points within the interval) always strengthens a change-of-sign argument, especially over a wide interval where multiple roots could hide.

Think like an examiner

Always state the function is CONTINUOUS on the interval you are using — this justifies the conclusion via the Intermediate Value Theorem.
Show BOTH function values (with their signs) explicitly, and state the conclusion in words ("since $f (a)$ is negative and $f (b)$ is positive, there is a root between...").
Be alert to the possibility of an even number of roots hiding within a wide interval — a lack of sign change does not guarantee no root at all.

Common misconceptions

Concluding a root exists without checking the function is continuous.

The change-of-sign argument relies on continuity — for any A-Level function (polynomials, trig, exponentials on their domain) this is safe to state, but it must be mentioned.

Assuming no sign change means no root exists in the interval.

An interval could contain an EVEN number of roots, causing the function to return to the same sign — a lack of sign change is not conclusive proof of no root.

Stopping after one halving step when a specific accuracy is required.

Keep halving (bisecting) the interval repeatedly until the required number of decimal places of accuracy is achieved.

Change-of-sign method

f (a) < 0, f (b) > 0 \Rightarrow root in (a, b)

For continuous f, a sign change guarantees a root.

Stretch yourself

Show that $f (x) = ln x - 4 + x$ has a root between $x = 2$ and $x = 3$ , then narrow the interval to width 0.5.

Hint — Evaluate $f$ at $x = 2$ and $x = 3$ to confirm the sign change, then test the midpoint $x = 2.5$ .

Questions students ask

Why does a sign change guarantee a root?

For a continuous function, if it is negative at one point and positive at another, its graph must cross the x-axis somewhere between them — this is a consequence of the Intermediate Value Theorem.

Can the change-of-sign method ever be wrong?

It never gives a false positive (if it detects a sign change, a root really does exist), but it can give a false negative — an interval containing an even number of roots may show no overall sign change.

How do I narrow down a root to more decimal places?

Repeatedly test the midpoint of the current interval and check which half still shows a sign change — each repetition roughly halves the width of the interval trapping the root.

Key takeaways

If $f (a)$ and $f (b)$ have opposite signs and $f$ is continuous, a root of $f (x) = 0$ lies in $(a, b)$ .
Testing the midpoint and checking its sign narrows the interval, locating the root more precisely.
The method can fail (miss roots) when there is an even number of roots in the interval, or where $f$ is discontinuous.
Always state the continuity condition explicitly as part of your argument.

How this fits the course

Build on

Differential Equations

Iteration

Leads to

Iteration

Test yourself

Ready to lock in Locating Roots? Pick a mode and earn XP & Dobloons.