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AS Level · Further Pure Mathematics 1 WFM01

Numerical Solutions of Equations

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Edexcel IAL · Further Pure 1

Numerical Solutions
of Equations

When an equation is too hard to solve algebraically, we use clever numerical tricks — sign changes, bisection, interpolation, and tangent lines — to pin down a root to any level of accuracy we need.

Roots & Intervals Interval Bisection Linear Interpolation Newton–Raphson

Contents

  • Roots in Intervals & Sign-Change Test
  • When the Sign-Change Test Fails
  • Interval Bisection
  • Linear Interpolation
  • Newton–Raphson Method
  • When Newton–Raphson Fails
  • What to Memorise
  • Concepts Checklist
  • Exam Tips
Topic 1
Roots in Intervals & the Sign-Change Test

What exactly is a root?

A root of the equation f(x) = 0 is simply a value of x that makes the equation true. If x = α is a root, then f(α) = 0 — the graph of y = f(x) crosses the x-axis there.

Some equations — like x² − 4 = 0 — can be solved exactly. But many real equations (especially cubics, quartics, or anything mixing polynomials with trig or log) have no neat algebraic solution. That's when we approximate numerically.

Analogy: Think of it like zooming in on a map. You can't walk to the exact coordinate of a buried treasure, but you can get close enough to dig. Each numerical method is a different way of zooming in.

What is an interval?

An interval a < x < b is a range of x-values that we believe contains a root, even though we don't know the root precisely. Bracket notation: (a, b) means strict inequality; [a, b] includes the endpoints.

The Sign-Change & Continuity Test

This is the foundational tool for all three methods. Here's the logic:

If f(a) is negative and f(b) is positive (or vice versa), the graph must have crossed the x-axis somewhere between a and b — that crossing is a root. But this only works if f(x) is continuous (no gaps, no jumps, no asymptotes) in the interval.

The Test
f(a) · f(b) < 0
If f(a) and f(b) have opposite signs AND f(x) is continuous on (a, b), then a root must lie in the interval (a, b).
Critical: Always state continuity in your conclusion. "f(x) has a sign change in [a, b] and f(x) is continuous in the interval, so a root must lie in [a, b]." Missing this loses marks.

Proving a root to a given accuracy

If asked to show that a root is x = 1.39 to 2 d.p., use the bounds of that rounded value:

  • Lower bound: 1.385 (anything below rounds down to 1.38, not 1.39)
  • Upper bound: 1.395 (anything above rounds up to 1.40)
  • Apply the sign-change test to (1.385, 1.395)
  • Conclude: "All values in this interval round to 1.39 to 2 d.p."
Worked Example
Given f(x) = x³ − 5x + 2, show that a root lies in the interval (1, 2).
  1. Evaluate at each endpoint: f(1) = 1 − 5 + 2 = −2 (negative)
  2. Evaluate: f(2) = 8 − 10 + 2 = 0... wait, that is the root. Let's adjust: f(x) = x³ − 5x + 1: f(1) = −3 < 0, f(2) = 8−10+1 = −1 < 0. Try f(3) = 27−15+1 = 13 > 0, so root is in (2, 3).
  3. Opposite signs confirmed: f(2) < 0, f(3) > 0.
  4. Check continuity: f(x) = x³ − 5x + 1 is a polynomial — continuous everywhere.
  5. Conclusion:
Practice Question 1
Topic 1b
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Also in the full note
  • The core idea
  • How to use it — step by step
  • The Gradient Method (most efficient)
  • The Formula
  • How to apply it
  • When Does Newton–Raphson Fail?
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More Further Pure Mathematics 1 topics

Operations with Complex Numbers
Introduction to Complex Numbers
Coordinate Systems
Matrix Algebra
Solving Equations with Complex Roots
Roots of Quadratic Equations
Roots of Quadratic Equations
Introduction to Matrices
Roots in Intervals
The Parabola