Locating and approximating roots

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Most equations met in practice, such as x³-2x-5 = 0 or e^(-x) = x, cannot be solved exactly. Numerical methods find their roots to any required accuracy. This lesson locates a root with a sign change, refines it with Newton–Raphson and with fixed-point iteration, explains geometrically why each method can fail, and shows how to confirm that a quoted root really is correct to the stated number of decimal places.

A sign change locates a root

If f is continuous on [a,b] and f(a), f(b) have opposite signs, then f(x) = 0 has at least one root between a and b. Continuity matters: f(x) = 1/(x-1) changes sign between 0 and 2 but has no root, because its graph breaks at x = 1. And no sign change does not mean no root: (x-1)² touches zero at x = 1 without changing sign.

Throughout this lesson, f(x) = x³-2x-5. Since f(2) = -1 and f(3) = 16, and polynomials are continuous, there is a root α in (2,3). A sketch or graphing calculator shows it is the only real root.

Linear interpolation replaces the curve by the chord between (2,-1) and (3,16):

x₁ = 2 + (0-(-1))/(16-(-1)) × (3-2) = 2 + 1/17 ≈ 2.0588.

Then f(x₁) ≈ -0.391, so α is in (x₁,3) and the chord can be redrawn. Because the curve bends upwards here, every chord estimate stays on the left of α, and convergence is slow.

Newton–Raphson follows the tangent

The tangent at xₙ crosses the x-axis at

xₙ₊₁ = xₙ-f(xₙ)/(f'(xₙ)).

f(xₙ): −1.00
f′(xₙ): 10.0
Next estimate xₙ₊₁: 2.10

Predict where the tangent will meet the axis, then change the current estimate. The fixed curve is f(x) = x³ − 2x − 5. The filled point is (xₙ, f(xₙ)); the open point is the next estimate. The slider stays where f′(x) is positive.

Use the Newton–Raphson method with x₀ = 2 to find the root of x³-2x-5 = 0 correct to six decimal places, and justify the accuracy.

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Set up the iteration
Method
f'(x) = 3x²-2.
Reason
The tangent’s gradient is needed at each estimate.
Working
Answer
xₙ₊₁ = xₙ-(xₙ³-2xₙ-5)/(3xₙ²-2).
Iterate
Method
Keep full calculator precision between steps.
Reason
Rounding early can stop the digits settling.
Working
Answer

n xₙ
0 2
1 2.1
2 2.094568121
3 2.094551482
4 2.094551482
Confirm with a sign change
Method
Test the bounds 2.0945505 and 2.0945515.
Reason
Settled digits suggest the answer; a sign change across the rounding interval proves it.
Working
Answer
f(2.0945505) < 0 and f(2.0945515) > 0, so α = 2.094551 to six decimal places.

n	xₙ
0	2
1	2.1
2	2.094568121
3	2.094551482
4	2.094551482

Quick check

From x₁ onwards, the number of correct digits roughly doubles each step, much faster than interpolation. The sign-change check is what turns “the digits stopped changing” into a justified accuracy.

Fixed-point iteration needs a shallow gradient

Rearranging f(x) = 0 as x = F(x) gives the iteration xₙ₊₁ = F(xₙ). If it converges, it converges to a root, because the limit satisfies α = F(α). Near the root, each error is multiplied by about F'(α), so the iteration converges from close enough starting values when |F'(α)| < 1 and moves away when |F'(α)| > 1.

A student rearranges x³-2x-5 = 0 as x = (x³-5)/2 and iterates from x₀ = 2.1, getting 2.1305, 2.3352, 3.8671, 26.4, …, and concludes the equation has no root near 2.1. Explain the error, and give a rearrangement that works.

Write your answer, then compare it with the solution

The error and a convergent rearrangement

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Test the student's rearrangement
Method
F(x) = (1/2)(x³-5), so F'(x) = (3/2)x².
Reason
At α ≈ 2.0946, |F'(α)| ≈ 6.58 > 1, so errors are multiplied by more than 6 each step.
Working
Answer
The iteration diverges even though the root exists; divergence says nothing about whether a root is there.
Rearrange differently
Method
Use x = cube root of (2x + 5).
Reason
F'(x) = (2/3)(2x + 5)^(-2/3) ≈ 0.152 at the root.
Working
Answer
From x₀ = 2: 2.080084, 2.092351, 2.094217, 2.094501, 2.094544, … converging steadily to 2.094551.

Common mistake

Both rearrangements have the same root, so the equation is not the problem; the choice of F is. Small |F'(α)| gives fast convergence, and a negative F'(α) makes the estimates alternate around the root.

Explain what goes wrong in each case.

Newton–Raphson for x³-3x + 3 = 0 started at x₀ = 1.
Linear interpolation for tan x = 0 on the interval [1,2].
Newton–Raphson for x³-2x + 2 = 0 started at x₀ = 0.

What breaks in each case

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f'(1) = 3-3 = 0: the tangent at x₀ is horizontal and never meets the x-axis, so x₁ is undefined.
tan 1 ≈ 1.557 and tan 2 ≈ -2.185 have opposite signs, but tan x is not continuous on [1,2]: it has an asymptote at x = π/2. The “root” found is false.
x₁ = 0-2/(-2) = 1, then x₂ = 1-1/1 = 0: the estimates cycle between 0 and 1 for ever, and never approach the real root near -1.769.

Common mistakes

Concluding that a sign change proves a root without checking that the function is continuous on the interval.
Concluding that no sign change means no root: a curve can touch the axis, or cross it twice, within one interval.
In Newton–Raphson, using f'(xₙ)/(f(xₙ)) instead of f(xₙ)/(f'(xₙ)).
Starting Newton–Raphson near a stationary point, where the tangent sends the next estimate far away.
Stating a root to k decimal places because two iterates agree, without a sign-change check across the rounding interval.

Before you move on

The equation x³-2x-5 = 0 has a root near x = 2. Use the Newton–Raphson method with x₀ = 2 to find x₁, then find the root correct to 4 decimal places.

x₁

Root, 4 decimal places

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With f(x) = x³-2x-5 and f'(x) = 3x²-2,

x₁ = 2-(-1)/10 = 2.1,; x₂ ≈ 2.094568,; x₃ ≈ 2.094551.

To confirm 2.0946, check the signs at the ends of its rounding interval: f(2.09455) < 0 and f(2.09465) > 0, so the root lies between them and rounds to 2.0946.

The trapezium rule, Simpson’s rule and Euler’s method are taught in Numerical integration and Euler methods.

Syllabus and review details

GCE A-Level H2 Further Mathematics 2027 · 2027

Last reviewed: 2026-11-06

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Guided steps

Reading settings

A sign change locates a root

Newton–Raphson follows the tangent

A root to six decimal places

Set up the iteration

Iterate

Confirm with a sign change

Fixed-point iteration needs a shallow gradient

Two rearrangements of one equation

Test the student's rearrangement

Rearrange differently

Why each method fails

Common mistakes

Before you move on

Newton’s own example

Lesson done

A sign change locates a root

Newton–Raphson follows the tangent

A root to six decimal places

Fixed-point iteration needs a shallow gradient

Two rearrangements of one equation

Why each method fails

Common mistakes

Before you move on

Newton’s own example

Questions