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Notes/Mathematics/Paper 2/Numerical Solution of Equations
CAIEAS Level9709§2.6

Numerical Solution of Equations

For equations no amount of algebra will solve: trap the root between two numbers, then iterate until the digits stop changing.

26 min read 2 sub-topics
68
question parts
2021–2025 · 37 papers
5 marks
per paper
≈ 9% of the paper
2.2/3
avg difficulty
moderate
#6
most examined
of 6 topics by marks

Some equations simply cannot be solved exactly. x3−2x−3=0x^3 - 2x - 3 = 0 has a perfectly real root near 1.91.9, but no rearrangement will ever produce it, and ex=4−x\mathrm{e}^x = 4 - x is worse. When algebra runs out, you find the root numerically: locate it roughly, then improve the estimate until it stops moving.

This is the smallest topic on Paper 2 — about 5 of the 50 marks — and the least demanding, at 2.2/32.2/3. It is also the most mechanical: the recorded techniques read like a checklist ("use a sign change to locate the root", "evaluate the function at interval endpoints", "choose a starting value from a known interval", "round each iterate to the required accuracy"). Almost every mark is for doing the standard thing and showing that you did.

Which is exactly where the marks go missing. These questions are marked on your working, not your answer: an unsupported "x=1.893x = 1.893" scores very little.

Before you start you should be able to

  • Substitute numbers into a function accurately, in radian mode where relevant

  • Rearrange an equation to make a chosen term the subject

  • Read intersections off a sketch (see Coordinate Geometry)

By the end of this page you can

  • Show that a root lies in a given interval by demonstrating a change of sign

  • Find an interval containing a root yourself, by trial

  • Use a graph of two curves to say how many roots an equation has and roughly where

  • Verify that a given iterative formula rearranges to the original equation

  • Apply xn+1=F(xn)x_{n+1} = \mathrm{F}(x_n) from a starting value and reach a root to a stated accuracy

  • Explain what convergence looks like, and that an iteration may fail to converge

01

Locating a root

Syllabus requirement · §2.6

“

locate approximately a root of an equation, by means of graphical considerations and/or searching for a sign change, e.g. finding a pair of consecutive integers between which a root lies.

”

If a continuous curve is below the axis at one point and above it at another, it must have crossed somewhere in between. That is the whole of the sign-change method, and it is enough to trap a root between two consecutive integers.

xy1232-2f(1) = −4f(2) = +1rootThe sign changesbetween 1 and 2,and the curve iscontinuous —so it must crosssomewhere between.Show BOTH valuesand say “sign change”.

f(1) is negative and f(2) is positive, so the curve crosses the axis between 1 and 2. Both values have to be shown — the conclusion is not accepted on its own.

Showing that a root lies between a and b
  1. 1

    Rearrange to the form f(x)=0\mathrm{f}(x) = 0 if it is not already. ex=4−x\mathrm{e}^x = 4 - x becomes ex+x−4=0\mathrm{e}^x + x - 4 = 0.

    The sign-change argument is about one function crossing zero, so everything must be on one side first.

  2. 2

    Evaluate f\mathrm{f} at both endpoints and write both values down, signs included.

  3. 3

    State the conclusion in full: "there is a change of sign, and f\mathrm{f} is continuous, so there is a root between aa and bb."

    The mark is for the sentence as much as the arithmetic. “Sign change” alone is usually accepted; nothing at all is not.

Worked example

2 marks

Show that the equation x3−2x−3=0x^3 - 2x - 3 = 0 has a root between x=1x = 1 and x=2x = 2.

Show full working
  1. 1

    Let f(x)=x3−2x−3\mathrm{f}(x) = x^3 - 2x - 3 — it is already in the right form.

  2. 2

    f(1)=1−2−3=−4\mathrm{f}(1) = 1 - 2 - 3 = -4.

  3. 3

    f(2)=8−4−3=+1\mathrm{f}(2) = 8 - 4 - 3 = +1.

  4. 4

    There is a change of sign between x=1x = 1 and x=2x = 2, and f\mathrm{f} is continuous, so a root lies between them.

Answer

f(1)=−4<0\mathrm{f}(1) = -4 < 0 and f(2)=1>0\mathrm{f}(2) = 1 > 0; the sign change means a root lies in 1<x<21 < x < 2.

Two marks, two values, one sentence. Never write just "yes there is a root" — the examiner is marking the evidence.

The graphical route

An equation like ex=4−x\mathrm{e}^x = 4 - x can also be handled by sketching y=exy = \mathrm{e}^x and y=4−xy = 4 - x on the same axes: the roots are the intersections. That instantly tells you how many roots there are, which a sign-change search cannot. Questions often ask for this first and the sign change second.

Practise locating rootsReal past-paper questions · Locating roots by sign change or graphical methods

The rest of this note

Checking your access…

Can you do all of these?

  • Rearrange ex=4−x\mathrm{e}^x = 4 - x into the form f(x)=0\mathrm{f}(x) = 0

  • Show a root of x3−2x−3=0x^3 - 2x - 3 = 0 lies between 11 and 22, with both values and a conclusion

  • Use a sketch of two curves to say how many roots an equation has

  • Verify that x=2x+33x = \sqrt[3]{2x+3} rearranges to x3−2x−3=0x^3 - 2x - 3 = 0

  • Run an iteration from a given x1x_1 and write out every value

  • Say when to stop iterating, and round to the accuracy asked for

  • Explain that a different rearrangement might not converge

Now do the questions
68 real Paper 2 parts from 2021–2025, sorted by difficulty, with mark schemes
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