# Numerical solution of equations

> Edexcel International A-Level Further Mathematics · IAL FMATHS FP1
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp1-numerical-solution-of-equations/

This guide covers the three numerical root-finding methods required for Edexcel IAL FP1 Further Maths, including root location via sign change, step-by-step working, and accuracy justification, aligned to the 2018 specification.

**Prerequisites:** [Roots of P1/P2 elementary and polynomial functions](https://www.owlsprep.com/study/edexcel-ial-pure-math-p1-polynomial-roots/); [Basic differentiation of P1/P2 functions](https://www.owlsprep.com/study/edexcel-ial-pure-math-p2-basic-differentiation/)

## Learning objectives

- Locate roots of f(x)=0 using the sign change test for continuous functions
- Apply interval bisection, linear interpolation, and Newton-Raphson methods to refine root estimates
- Justify answer accuracy using sign change checks on rounding intervals
- Recall unprovided methods for bisection and linear interpolation, and correctly use the given Newton-Raphson formula

## Root Location via Sign Change Test

To solve f(x)=0 numerically, you first confirm a root exists in an interval using the sign change test. For a continuous function, opposite signs of f(a) and f(b) prove at least one root lies in the open interval (a, b).

**Sign Change Test** — For a continuous function f(x), if f(a) < 0 and f(b) > 0 (or vice versa), f(x)=0 has at least one root in (a, b).

*Example:* f(x) = x³ - 3x +1, f(1) = -1, f(2) = 3, so root in (1,2)

**Worked example:** Show that f(x) = x³ - 3x + 1 has a root in the interval (1, 2).

1. Calculate f(1):

   $$f(1) = 1^3 - 3(1) + 1 = -1$$
2. Calculate f(2):

   $$f(2) = 2^3 - 3(2) + 1 = 3$$
3. f(1) is negative, f(2) is positive, and f(x) is a polynomial (continuous for all real x), so a root exists in (1, 2).

> **Exam tip:** Always explicitly state f(x) is continuous when using the sign change test to secure full marks.

## Interval Bisection Method

Interval bisection refines a known root interval by repeatedly splitting it at the midpoint, and retaining the sub-interval that contains the sign change. No formula for this method is provided in the exam, so you must recall the process.

**Interval Bisection** — Iterative method to narrow a root interval (a, b): calculate midpoint c = (a+b)/2, test f(c). If f(a) and f(c) have opposite signs, keep (a, c); else keep (c, b). Repeat until the interval meets your required accuracy.

**Worked example:** Starting with interval (1, 2) for f(x) = x³ - 3x + 1, use interval bisection twice to find a narrower interval containing the root.

1. First iteration: calculate midpoint of (1,2):

   $$c_1 = \frac{1+2}{2} = 1.5$$
2. Calculate f(1.5):

   $$f(1.5) = (1.5)^3 - 3(1.5) + 1 = -0.125$$
3. f(1.5) is negative, f(2) is positive, so retain interval (1.5, 2).
4. Second iteration: calculate midpoint of (1.5,2):

   $$c_2 = \frac{1.5 + 2}{2} = 1.75$$
5. Calculate f(1.75):

   $$f(1.75) = (1.75)^3 - 3(1.75) + 1 \approx 1.109$$
6. f(1.5) is negative, f(1.75) is positive, so the refined interval is (1.5, 1.75).

> **Exam tip:** Write down all intermediate f(x) values explicitly, even if they are simple, to secure method marks if you make an arithmetic error.

## Linear Interpolation Method

Linear interpolation (false position method) is faster than bisection: it estimates the root as the x-intercept of a straight line connecting the points (a, f(a)) and (b, f(b)) of the known root interval. No formula for this method is provided in the exam.

**Linear Interpolation for Root Finding** — For interval (a,b) with opposite sign f(a), f(b), the x-intercept c of the line connecting (a, f(a)) and (b, f(b)) is the next root estimate, derived using similar triangles: \frac{c - a}{b - c} = \frac{|f(a)|}{|f(b)|}.

**Worked example:** Use linear interpolation on the interval (1, 2) for f(x) = x³ - 3x + 1 to find an estimate for the root.

1. We know f(1) = -1, f(2) = 3. Use similar triangles to relate distances:

   $$\frac{c - 1}{2 - c} = \frac{|f(1)|}{|f(2)|} = \frac{1}{3}$$
2. Rearrange to solve for c:

   $$3(c - 1) = 2 - c \implies 3c - 3 = 2 - c \implies 4c = 5 \implies c = 1.25$$
3. Check f(1.25): f(1.25) ≈ -0.797, so the refined interval is (1.25, 2).

> **Exam tip:** Derive the root estimate using similar triangles instead of memorizing the formula to avoid sign errors.

## Newton-Raphson Process

The Newton-Raphson method is the fastest of the three methods, using the derivative of f(x) to refine iterative root estimates. The formula is provided in your exam formula book, and only P1/P2 differentiation rules are required to compute f'(x).

**Newton-Raphson Iteration** — Iterative formula to refine a root estimate x_n to x_{n+1}: x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, where f'(x_n) is the first derivative of f(x) evaluated at x_n.

**Worked example:** Starting with initial estimate x_0 = 1.5 for f(x) = x³ - 3x + 1, use Newton-Raphson twice to find an estimate for the root, and justify it is accurate to 3 decimal places.

1. First compute the derivative of f(x):

   $$f'(x) = 3x^2 - 3$$
2. First iteration (x_0 = 1.5):

   $$f(1.5) = -0.125, f'(1.5) = 3(1.5)^2 - 3 = 3.75$$
3. $$x_1 = 1.5 - \frac{-0.125}{3.75} = 1.5333 \text{ (4 dp)}$$
4. Second iteration (x_1 = 1.5333):

   $$f(1.5333) ≈ 0.0051, f'(1.5333) ≈ 4.0533$$
5. $$x_2 = 1.5333 - \frac{0.0051}{4.0533} ≈ 1.5320 \text{ (4 dp)}$$
6. Justify accuracy to 3 dp: f(1.5315) ≈ -0.0007, f(1.5325) ≈ 0.0034, sign change confirms root is 1.532 to 3 dp.

> **Exam tip:** Double-check your derivative calculation before proceeding with iterations, as an incorrect derivative will invalidate all subsequent estimates.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the sign change test without stating f(x) is continuous
  - Why it fails: Examiners require explicit confirmation of continuity to apply the test; marks are deducted for only stating opposite signs.
  - Correct: Add a line like "f(x) is a polynomial, so continuous for all real x" every time you use the sign change test.
- **Wrong:** Rounding intermediate iterative estimates to the required final accuracy early
  - Why it fails: Early rounding introduces errors that propagate through subsequent iterations, leading to incorrect final values.
  - Correct: Keep at least 4 decimal places for all intermediate estimates, only round the final answer to the required accuracy.
- **Wrong:** Trying to recall pre-written formulas for bisection and linear interpolation
  - Why it fails: Many students misremember formulas and lose marks; no formulas for these methods are provided in the exam.
  - Correct: Practice deriving linear interpolation via similar triangles and calculating bisection midpoints from first principles.
- **Wrong:** Using FP3 differentiation rules (e.g. hyperbolic, inverse trig) for Newton-Raphson
  - Why it fails: Edexcel restricts f(x) to P1/P2 functions, so these derivatives will never appear, and using them is unnecessary.
  - Correct: Only use P1/P2 differentiation rules (power, product, quotient, chain rule for basic trig, exponential, log functions) to compute f'(x).
- **Wrong:** Failing to justify final answer accuracy with a sign change test
  - Why it fails: Examiners require proof the root falls within the rounding interval, not just a matching iterative estimate.
  - Correct: Test the upper and lower bound of the rounding interval (e.g. x ± 0.5×10^(-n) for n decimal places) to confirm a sign change before stating your final answer.

## Cheatsheet

| Method | Process/Formula | Speed | Required Working |
| --- | --- | --- | --- |
| Sign Change Test | Check f(a)·f(b) < 0, confirm f(x) continuous | N/A | f(a), f(b) values, continuity statement |
| Interval Bisection | Midpoint c = (a+b)/2, keep interval with sign change | Slow | All midpoint values, f(c) values, interval after each iteration |
| Linear Interpolation | Derive x-intercept via similar triangles | Medium | Similar triangles derivation, c value, f(c) value, new interval |
| Newton-Raphson | x_{n+1} = x_n - f(x_n)/f'(x_n) (given in formula book) | Fast | Derivative f'(x), all iterate values, sign change for accuracy justification |

## What's next

Now that you have mastered the three numerical root-finding methods for Edexcel IAL FP1, you are ready to practice full exam-style questions combining these methods with sign change accuracy justification. These questions typically make up 8-12 marks of your 75-mark FP1 paper, so regular timed practice is critical to avoid arithmetic errors and ensure you show all required working for full marks. Next, you can move on to other high-weighted FP1 topics including matrices and complex numbers, and practice past paper topic questions to familiarize yourself with Edexcel's marking expectations. Remember to only use a permitted non-CAS scientific calculator for all practice and your actual exam.

- [FP1 Complex Numbers](https://www.owlsprep.com/study/edexcel-ial-further-math-fp1-complex-numbers/)

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