Locating Roots

This lesson covers methods for locating roots of equations — the sign change method, decimal search, and systematic approaches — as required by the Edexcel A-Level Mathematics specification (9MA0). You also need to understand the limitations of these methods.

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The Sign Change Method

The Principle

If f(x) is a continuous function and f(a) and f(b) have opposite signs (one positive, one negative), then there must be at least one root between x = a and x = b.

This is a consequence of the Intermediate Value Theorem.

In mathematical terms: If f(a) × f(b) < 0 (one is positive, one is negative) and f is continuous on [a, b], then there exists at least one value c in (a, b) such that f(c) = 0.

Example

Show that x³ - 3x + 1 = 0 has a root between x = 1 and x = 2.

f(1) = 1 - 3 + 1 = -1 (negative) f(2) = 8 - 6 + 1 = 3 (positive)

Since f(1) < 0 and f(2) > 0, and f is continuous (it is a polynomial), there is a root between x = 1 and x = 2.

Exam Tip: When using the sign change method, you must: (1) evaluate f at both values, (2) state that there is a sign change, and (3) state that f is continuous. Missing any of these steps loses marks.

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Decimal Search

Decimal search is a systematic method to narrow down the location of a root, one decimal place at a time.

Step-by-Step Method

1. Use the sign change method to locate a root between two consecutive integers.
2. Evaluate f(x) at intervals of 0.1 within that interval.
3. Identify the sign change — the root lies between those two values.
4. Evaluate f(x) at intervals of 0.01 within the narrower interval.
5. Continue to the required accuracy.

Worked Example

Find the root of f(x) = x³ + x - 3 between x = 1 and x = 2, correct to 1 decimal place.

Step 1: f(1) = -1 and f(2) = 7. Sign change, so root is in [1, 2].

Step 2: Evaluate at 0.1 intervals:

x	f(x)
1.0	-1
1.1	-0.569
1.2	-0.072
1.3	+0.497

Sign change between x = 1.2 and x = 1.3.

The root is in the interval [1.2, 1.3], so to 1 d.p. the root is x = 1.2 (since we need to check the midpoint to determine rounding).

Step 3 (refinement): Check f(1.25) = 1.953 + 1.25 - 3 = 0.203 > 0

Since f(1.2) < 0 and f(1.25) > 0, the root is in [1.2, 1.25]. Therefore, to 1 d.p., x = 1.2.

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Showing a Root Correct to n Decimal Places

To show that a root α is correct to n decimal places, you must demonstrate a sign change in the interval:

[α - 0.5 × 10⁻ⁿ, α + 0.5 × 10⁻ⁿ]

Example

Show that x = 1.21 is a root of x³ + x - 3 = 0, correct to 2 decimal places.

Check the interval [1.205, 1.215]:

f(1.205) = 1.205³ + 1.205 - 3 = 1.7492 + 1.205 - 3 = -0.0458 f(1.215) = 1.215³ + 1.215 - 3 = 1.7940 + 1.215 - 3 = 0.0090

Since f(1.205) < 0 and f(1.215) > 0, there is a sign change. The root lies in [1.205, 1.215], so the root is x = 1.21 correct to 2 d.p.

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The Bisection Method

The bisection method is a systematic form of the sign change approach:

1. Start with an interval [a, b] where f(a) and f(b) have opposite signs.
2. Find the midpoint: m = (a + b)/2.
3. Evaluate f(m).
4. If f(m) = 0, m is the root. Otherwise, replace whichever endpoint has the same sign as f(m).
5. Repeat until the interval is sufficiently small.

Example

Find a root of f(x) = x² - 3 (i.e., √3) between 1 and 2 using bisection.

Step	a	b	m = (a+b)/2	f(m)	New interval
1	1	2	1.5	-0.75	[1.5, 2]
2	1.5	2	1.75	0.0625	[1.5, 1.75]
3	1.5	1.75	1.625	-0.3594	[1.625, 1.75]
4	1.625	1.75	1.6875	-0.1523	[1.6875, 1.75]
5	1.6875	1.75	1.71875	-0.0459	[1.71875, 1.75]

After 5 steps, the root lies in [1.71875, 1.75]. The root is approximately 1.73 (√3 ≈ 1.7321).

Properties of Bisection

• Always converges (provided there is a sign change)
• Each step halves the interval — very predictable
• Relatively slow compared to Newton-Raphson
• Does not require differentiation

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Limitations of the Sign Change Method

The sign change method fails in several situations:

1. Repeated (Touching) Root

If the curve touches the x-axis but does not cross it (e.g., y = (x - 2)²), there is no sign change. The method cannot detect the root.

2. Two Roots Close Together

If two roots are very close, they may both lie within a single interval, and the function may have the same sign at both endpoints. The sign change method misses them.

3. Discontinuities

If f(x) has a discontinuity (e.g., f(x) = 1/x near x = 0), there may be a sign change without a root. The function jumps from positive to negative (or vice versa) without crossing zero.

4. Very Flat Curves

If f(x) is very close to zero over a wide interval, rounding errors may give incorrect sign information.

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Error Bounds

After applying a root-finding method, the error bound tells you how close your approximation is to the true root.

For Bisection

After n steps starting from interval [a, b], the error is at most: |error| ≤ (b - a) / 2ⁿ

For Decimal Search

After narrowing to an interval of width 10⁻ⁿ, the error is at most 10⁻ⁿ.

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Worked Example: Combining Methods

Show that eˣ = 4x has a root in the interval [0, 1], and find this root correct to 2 decimal places using decimal search.

Step 1: Let f(x) = eˣ - 4x. f(0) = 1 - 0 = 1 > 0 f(1) = e - 4 = 2.718 - 4 = -1.282 < 0

Sign change, so there is a root in [0, 1].

Step 2: Decimal search in [0, 1]:

x	f(x) = eˣ - 4x
0	1
0.1	1.105 - 0.4 = 0.705
0.2	1.221 - 0.8 = 0.421
0.3	1.350 - 1.2 = 0.150
0.4	1.492 - 1.6 = -0.108

Sign change between 0.3 and 0.4.

Step 3: Search [0.3, 0.4]:

x	f(x)
0.35	1.419 - 1.4 = 0.019
0.36	1.433 - 1.44 = -0.007

Sign change between 0.35 and 0.36. Check midpoint: f(0.355) = e^0.355 - 1.42 = 1.426 - 1.42 = 0.006 > 0

Root in [0.355, 0.36], so x = 0.36 to 2 d.p.

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Common Mistakes

1. Not stating f is continuous — The sign change method requires continuity.
2. Checking wrong interval — For "correct to n d.p.", check the interval of width 10⁻ⁿ centred on the answer.
3. Missing roots — Sign change does not guarantee you have found ALL roots. There may be roots where the method does not detect a sign change.
4. Confusing sign change with root — A sign change between f(a) and f(b) means a root exists in (a, b), but it does not tell you the exact value.

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Summary

• Sign change: If f(a) and f(b) have opposite signs and f is continuous, there is a root in (a, b).
• Decimal search: Narrow the interval one decimal place at a time.
• Bisection: Repeatedly halve the interval by checking the sign at the midpoint.
• To show a root is correct to n d.p., demonstrate a sign change in an interval of width 10⁻ⁿ centred on the root.
• Limitations: repeated roots, close roots, discontinuities, flat curves.
• Error bounds: bisection gives |error| ≤ (b - a)/2ⁿ after n steps.

A-Level Deep Dive: Locating Roots

Spec mapping

Edexcel 9MA0 specification section 13 — Numerical methods, sub-strand 13.1 covers locate roots of $f(x) = 0$f(x)=0 by considering changes of sign of $f(x)$f(x) in an interval of $x$x on which $f(x)$f(x) is sufficiently well behaved (refer to the official specification document for exact wording). This is examined in Paper 2 — Pure Mathematics as Year 2 content. It also feeds section 13.2 (iterative methods, including $x_{n+1} = g(x_n)$xn+1​=g(xn​) recurrences) and section 13.3 (Newton-Raphson), and is synoptically linked to section 9 (Differentiation, where stationary points and double roots interact) and section 12 (Integration, occasionally invoked when antiderivatives are not expressible in closed form). The Edexcel formula booklet does not state the sign-change rule — students are expected to know its statement and limitations.

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Worked example with full mark scheme

Question (8 marks): Let $f(x) = x^3 - 5x + 2$f(x)=x3−5x+2.

(a) Show that $f(x) = 0$f(x)=0 has a root $\alpha$α in the interval $[0, 1]$[0,1]. (2)

(b) By using the bisection method twice on the interval $[0, 1]$[0,1], find an interval of width $0.25$0.25 containing $\alpha$α. (4)

(c) Explain, with reference to the value of $f'(x)$f′(x) on this interval, why the bisection method will converge to $\alpha$α and not to a different root. (2)

Solution with mark scheme:

(a) Evaluate at the endpoints:

$f(0) = 0 - 0 + 2 = 2 > 0$f(0)=0−0+2=2>0 $f(1) = 1 - 5 + 2 = -2 < 0$f(1)=1−5+2=−2<0

M1 — evaluating $f$f at both endpoints of $[0, 1]$[0,1].

Since $f$f is a polynomial it is continuous on $[0, 1]$[0,1], and $f(0)$f(0) and $f(1)$f(1) have opposite signs, so by the change-of-sign rule there exists at least one root $\alpha \in (0, 1)$α∈(0,1).

A1 — explicit statement of opposite signs and continuity, leading to existence of a root. Common error: stating "the signs are different so there is a root" without mentioning continuity loses the A1.

(b) Step 1 — first bisection. Midpoint $m_1 = 0.5$m1​=0.5.

$f(0.5) = 0.125 - 2.5 + 2 = -0.375 < 0$f(0.5)=0.125−2.5+2=−0.375<0

Since $f(0) > 0$f(0)>0 and $f(0.5) < 0$f(0.5)<0, the sign change is on $[0, 0.5]$[0,0.5], so the root lies in $[0, 0.5]$[0,0.5].

M1 — correct first bisection with sign comparison.

Step 2 — second bisection. Midpoint $m_2 = 0.25$m2​=0.25.

$f(0.25) = 0.015625 - 1.25 + 2 = 0.765625 > 0$f(0.25)=0.015625−1.25+2=0.765625>0

Since $f(0.25) > 0$f(0.25)>0 and $f(0.5) < 0$f(0.5)<0, the sign change is on $[0.25, 0.5]$[0.25,0.5].

M1 — correct second bisection with sign comparison.

A1 — interval $[0.25, 0.5]$[0.25,0.5], width $0.25$0.25 as required.

A1 — full numerical accuracy preserved (working shown in exact decimals, no premature rounding). Examiners require that $f$f-values are stated, not merely "positive" or "negative" — the sign alone without the value loses the accuracy mark.

(c) Differentiate: $f'(x) = 3x^2 - 5$f′(x)=3x2−5. On $[0.25, 0.5]$[0.25,0.5], $3x^2 \leq 3(0.25) = 0.75$3x2≤3(0.25)=0.75, so $f'(x) \leq 0.75 - 5 = -4.25 < 0$f′(x)≤0.75−5=−4.25<0 throughout.

M1 — computing $f'$f′ and bounding it on the interval.

Since $f'(x) < 0$f′(x)<0 on $[0.25, 0.5]$[0.25,0.5], $f$f is strictly decreasing there, so $f$f is injective on the interval and the sign change identifies a unique root. Bisection therefore converges to $\alpha$α and to no other root.

A1 — explicit "strictly monotonic, hence unique root" argument.

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A function is defined by $g(x) = e^{-x} - x + 1$g(x)=e−x−x+1 for $x \in \mathbb{R}$x∈R.

(a) Show that the equation $g(x) = 0$g(x)=0 has a root $\beta$β in the interval $(1, 2)$(1,2). (2)

(b) Use the bisection method twice, starting from $[1, 2]$[1,2], to find an interval of width $0.25$0.25 containing $\beta$β. Give all $g$g-values to 4 decimal places. (3)

(c) State one limitation of the bisection method that is not relevant here, and explain briefly why it is not relevant. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — evaluating $g(1) = e^{-1} - 1 + 1 = e^{-1} \approx 0.3679 > 0$g(1)=e−1−1+1=e−1≈0.3679>0 and $g(2) = e^{-2} - 2 + 1 = e^{-2} - 1 \approx -0.8647 < 0$g(2)=e−2−2+1=e−2−1≈−0.8647<0.
• A1 (AO2.1) — invoking continuity of $g$g (sum of continuous functions) and opposite signs to conclude existence of a root in $(1, 2)$(1,2).

(b)

• M1 (AO1.1b) — first bisection: $g(1.5) = e^{-1.5} - 0.5 \approx 0.2231 - 0.5 = -0.2769 < 0$g(1.5)=e−1.5−0.5≈0.2231−0.5=−0.2769<0, so root in $[1, 1.5]$[1,1.5].
• M1 (AO1.1b) — second bisection: $g(1.25) = e^{-1.25} - 0.25 \approx 0.2865 - 0.25 = 0.0365 > 0$g(1.25)=e−1.25−0.25≈0.2865−0.25=0.0365>0, so root in $[1.25, 1.5]$[1.25,1.5].
• A1 (AO1.1b) — final interval $[1.25, 1.5]$[1.25,1.5], width $0.25$0.25, with $g$g-values stated to 4 decimal places throughout.

(c)

• B1 (AO3.5a) — for example: "the bisection method can fail at a discontinuity, but $g$g is continuous everywhere, so this limitation does not apply"; or "the method can miss a double root, but on $(1, 2)$(1,2) the sign change is genuine and $g$g is monotonic there, so this is not relevant".

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is an AO1-dominated question with a single AO3 reasoning mark for the limitation discussion. The numerical methods strand of 9MA0 routinely awards an AO3 mark for a "comment on validity" question.

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