Locate equation roots with graphs and valid sign changes
Turn an equation into f(x) = 0, inspect the graph and bracket a root only where the original function is continuous.
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
- Evaluate a function accurately and read graph intersections.
- Identify polynomial, logarithmic and reciprocal domains.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What do opposite non-zero signs at the endpoints imply for a continuous function?
What you will learn
- Locate approximate roots from axis crossings or intersections of two graphs.
- Use opposite signs to find a root interval while recognising touching roots and false brackets across poles.
Find where the original equation becomes zero
Write the equation as f(x) = 0 without changing its valid inputs. Roots appear where the graph of f meets the x axis. For an equation g(x) = h(x), intersections of their two graphs give the same root inputs as zeros of f(x) = g(x) − h(x). A sketch can locate a plausible interval, but its drawing resolution does not usually justify several decimal places.
For f(x) = x³ − x − 1, f(1) = −1 and f(2) = 5. The polynomial is continuous throughout [1, 2], so it has at least one root strictly between those consecutive integers. Values f(1.3) = −0.103 and f(1.4) = 0.344 refine the positive root's location to (1.3, 1.4). A sign change establishes existence on a continuous interval; uniqueness needs separate information about the function, such as its increasing behaviour there.
A sign change needs continuity, and its absence does not exclude every root
For f(x) = 1/(x − 1), f(0) is negative and f(2) positive, but there is no root: the sign changes across an undefined pole. This interval fails the continuity requirement. Retain all original denominator, square-root and logarithm restrictions before using a sign-change argument. If an endpoint itself gives f(a) = 0, that endpoint is already an exact root rather than a strict interior sign-change bracket.
For f(x) = (x − 1)², the graph touches the axis at x = 1 and remains non-negative. Values at 0 and 2 have the same positive sign despite the root between them. Thus no sign change does not prove no root. Use the graph and the equation together, especially for touching or repeated roots. Numerical sampling over one region also does not by itself establish that every root of the equation has been found.
Pause and explain
Does 1/(x − 1) have a root between 0 and 2 because its endpoint signs differ?
Worked example
Locate a root of x³ − x − 1 = 0 between consecutive integers, then refine the interval using 1.3 and 1.4.
Show the worked solution
- Evaluate f(1) = −1 and f(2) = 5.
- The polynomial is continuous, so the opposite signs locate at least one root in (1, 2).
- Evaluate f(1.3) = −0.103 and f(1.4) = 0.344, giving the narrower bracket (1.3, 1.4).
Answer A root lies in (1, 2), more precisely in (1.3, 1.4).
Try a topic test
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Test this topic →From guided practice to a new situation
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.
Locate both roots of x² − 2 = 0 between consecutive integers using signs.
Give me a hint
Check one negative-input bracket as well as one positive-input bracket.
Compare my reasoning
- At −2 and −1 the values are 2 and −1, so a root lies in (−2, −1).
- At 1 and 2 the values are −1 and 2, so a root lies in (1, 2).
- The polynomial is continuous on both intervals; the two distinct brackets locate the two roots of this quadratic.
Roots in (−2, −1) and (1, 2).
Look for these in your work
- Evaluate signs on both relevant input regions.
- Use continuity to support each root bracket.
For f(x) = (x − 1)², identify the root and explain why testing the signs at 0 and 2 misses it.
Give me a hint
Factor or inspect the original square and describe how its graph meets the axis.
Compare my reasoning
- The function is zero precisely when x − 1 = 0, so x = 1 is a root.
- Both endpoint values are 1, with the same positive sign.
- The upward parabola touches the axis at its repeated root; absence of a sign change does not exclude this contact.
Root x = 1; same-sign endpoint values miss the touching root.
Look for these in your work
- Find the root from the original expression.
- Explain the limitation of sign-change-only searching.
A supplied index model is R(t) = eᵗ + t, with t measured numerically in hours and t ≥ 0. Locate a time when R reaches 3 between consecutive whole hours.
Give me a hint
Use f(t) = eᵗ + t − 3 and check its continuous positive-time interval.
Compare my reasoning
- At t = 0, f(0) = −2.
- At t = 1, f(1) = e − 2 > 0.
- The function is continuous, so a threshold time lies strictly between 0 and 1 hours; this bracket does not yet specify decimal places.
A threshold time lies in 0 < t < 1 hours.
Look for these in your work
- Translate the threshold into a valid zero equation.
- Use a continuous sign-change bracket without claiming unsupported precision.
Common mistakes
- Check continuity throughout the proposed bracket, not only endpoint values.
- A graph locates roots; its visual precision does not certify decimal places.
- Same-sign endpoint values can still surround a touching root.
What can you explain now?
For (x − 1)², does a positive value at both x = 0 and x = 2 prove there is no root between them?
Compare with the explanation
No; x = 1 is a touching root.
The graph meets the axis without crossing it, so there is no sign change across that repeated root.
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Make it stick
Explain how zeros and graph intersections locate equation solutions. Give one sign change across a pole that fails and one touching root without a sign change.
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.