Refine root brackets and justify the requested rounding
Retain the interval containing a root, narrow it with new sign checks and distinguish interval width from a rounding guarantee.
Section 2.6 · Numerical solution of equations
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 graph and algebra knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.6 and 3.6 include locating roots graphically or by sign changes, notation for convergent approximation sequences, and using a given iteration or rearrangement to a prescribed accuracy. An iteration may fail to converge. Knowledge of the general condition for convergence is not required here. Check the original equation, domains and requested rounding. Written practice is self-checked.
Before you begin
- Locate a root using valid sign changes.
- Round decimal numbers and evaluate a function without premature rounding.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For opposite signs at a and b, how is midpoint refinement used?
What you will learn
- Refine a continuous sign-change bracket, using midpoint tests or chosen decimal inputs.
- Use an interval lying within one rounding band to justify an answer to prescribed decimal places.
Keep the half that retains the sign change
Once a continuous function has opposite signs at a and b, test the midpoint m = (a + b)/2. If f(m) = 0, you have found an exact root. Otherwise, keep whichever of [a, m] and [m, b] has opposite endpoint signs. Repeating this bracket-refinement method halves the interval width each time. The midpoint is a candidate approximation, not an exact root unless its original function value is zero.
For x² − 2 on [1, 2], midpoint 1.5 gives 0.25, so keep [1, 1.5]. Next midpoint 1.25 gives −0.4375, so keep [1.25, 1.5]. The new endpoint evidence preserves a root while narrowing its location. After n halvings the width is (b − a)/2ⁿ, and the midpoint's absolute error is at most half that width for a root within the bracket. This is a support method for reliable root location; use any particular iteration the question explicitly gives when requested.
Check a rounding band as well as a narrow width
To justify a rounded answer, make the entire root bracket lie inside the band that rounds to that answer. For the positive root of x² − 2, f(1.4142) = −0.00003836 and f(1.4143) = 0.00024449. The continuous sign change locates a root between them, and every input there rounds to 1.414 to three decimal places. Reporting 1.414 is a rounding claim; it does not mean f(1.414) equals zero.
A bracket narrower than 0.001 can still cross a three-decimal rounding boundary. For example, (1.4144, 1.4146) crosses 1.4145, so its width alone does not establish whether the rounded value is 1.414 or 1.415. Decimal places and significant figures are different instructions: 0.6823 rounds to 0.682 at three decimal places or three significant figures, but these notions do not always coincide for other-sized numbers. State the requested precision and retain extra digits during calculations.
Pause and explain
What extra check turns a narrow root bracket into a justified rounded answer?
Worked example
Use the bracket (1.4142, 1.4143) to justify the positive root of x² − 2 = 0 to three decimal places.
Show the worked solution
- Evaluate opposite signs −0.00003836 and 0.00024449 at the two endpoints.
- The polynomial is continuous, so a root lies strictly between those inputs.
- The whole bracket lies inside the rounding band [1.4135, 1.4145), so the root rounds to 1.414.
Answer Positive root approximately 1.414 to three decimal places, justified by the sign-change bracket.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For f(x) = x² − 2, start with [1, 2] and perform two midpoint refinements. Give the resulting bracket.
Give me a hint
Evaluate the midpoint sign and replace only the endpoint with the same sign.
Compare my reasoning
- The first midpoint 1.5 gives 0.25, so retain [1, 1.5].
- The second midpoint 1.25 gives −0.4375, so retain [1.25, 1.5].
- The two endpoint signs still differ and the final width is 1/4.
After two refinements: [1.25, 1.5], width 1/4.
Look for these in your work
- Retain a valid continuous sign-change interval at both stages.
- Use the new endpoints when calculating the next midpoint.
A root of x³ − x − 1 lies between 1.3247 and 1.3248. Check the signs and report the root to three decimal places, explaining the rounding.
Give me a hint
Both endpoint values lie on the same side of the three-decimal rounding threshold 1.3245.
Compare my reasoning
- The function values are −0.000076579777 and 0.000349908992.
- Continuity locates a root inside this bracket.
- Every value in the bracket lies in [1.3245, 1.3255), giving rounded answer 1.325.
Root approximately 1.325 to three decimal places.
Look for these in your work
- Verify the bracket using original polynomial values.
- Use the entire rounding band rather than round just one trial input.
For a continuous function, suppose a root lies in (1.4144, 1.4146). A student reports 1.414 to three decimal places because the interval is less than 0.001 wide. Is that justified, and what further evidence is needed?
Give me a hint
Compare the bracket with the rounding threshold 1.4145.
Compare my reasoning
- The stated interval has width 0.0002 but crosses 1.4145.
- Some possible roots round to 1.414 and others to 1.415, so the report is not justified by this bracket alone.
- Refine with valid sign checks until the root bracket lies wholly within one rounding band, or establish an exact boundary root if that case occurs.
Not justified; refine the bracket to one side of 1.4145, or establish the exact boundary case.
Look for these in your work
- Distinguish a width-based error statement from a rounding decision.
- Specify evidence that actually resolves the requested precision.
Common mistakes
- Retain opposite signs and the original function's domain at every refinement.
- A midpoint is an approximation until checked as an exact root.
- A narrow interval that crosses a rounding boundary does not settle the rounded answer.
What can you explain now?
Does a root bracket (1.4144, 1.4146) by itself establish a unique three-decimal rounded answer?
Compare with the explanation
No; it crosses the rounding boundary 1.4145.
Values on its two sides round differently even though the bracket is narrower than 0.001.
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Make it stick
Refine a sign-change bracket twice. Explain how a midpoint error bound and a whole-band rounding check support different parts of a numerical answer.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.