---
title: "Secant: x^5 - 3x^3 - 1"
subject: "Numerical Method"
chapter: 2
key: "semester_4/numerical_method:secant-x5-3x3-1"
source: https://www.bctengineer.com/q/semester_4/numerical_method/secant-x5-3x3-1
site: BCT Engineering
---

# Secant: x^5 - 3x^3 - 1

Numerical Method · Chapter 2: Solutions of Nonlinear Equations

- Asked in: 2068 Bhadra, 8 marks
- Video: [Secand method in 5 minutes | solution of algebraic equation](https://www.youtube.com/watch?v=6S2SOhWZ0bU) — Smart Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/secant-x5-3x3-1
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-2

## Question

Find a real root of x^5 - 3x^3 - 1 = 0 correct up to four decimal places using the Secant Method.

## Short answer

Use secant method with x0=1.5, x1=2 on f(x)=x^5-3x^3-1. Iterates converge to 1.7823, so the real root is approximately 1.7823.

## Answer

**Secant formula:** x(n+1) = x(n) − f(x(n))·[x(n) − x(n−1)] / [f(x(n)) − f(x(n−1))].

#### Step 1: Starting values
f(x) = x⁵ − 3x³ − 1
f(1.5) = 7.5938 − 10.1250 − 1 = −3.5313 < 0 and f(2) = 32 − 24 − 1 = 7 > 0, so a root lies in (1.5, 2).
Take x0 = 1.5 and x1 = 2.

#### Step 2: Iterations
| n | x(n-1) | x(n) | f(x(n-1)) | f(x(n)) | x(n+1) |
|---|---|---|---|---|---|
| 1 | 1.5000 | 2.0000 | -3.5313 | 7.0000 | 1.6677 |
| 2 | 2.0000 | 1.6677 | 7.0000 | -2.0153 | 1.7419 |
| 3 | 1.6677 | 1.7419 | -2.0153 | -0.8182 | 1.7927 |
| 4 | 1.7419 | 1.7927 | -0.8182 | 0.2323 | 1.7815 |
| 5 | 1.7927 | 1.7815 | 0.2323 | -0.0176 | 1.7823 |
| 6 | 1.7815 | 1.7823 | -0.0176 | -0.0003 | 1.7823 |

The last two iterates agree to four decimals.

**Answer: root ≈ 1.7823**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Root in (1.5, 2) because:
   A. f(1.5) < 0 and f(2) = 7 > 0
   B. f(2) < 0
   C. f(1.5) = 0
   D. both negative
   Answer: A. Sign change. (Key point: Bracket)

2. Each new x uses:
   A. f'(x)
   B. Bracket halving
   C. The two previous points, no derivative
   D. Fixed g(x)
   Answer: C. Secant method. (Key point: Method)

3. The root is:
   A. 1.6677
   B. 2.0000
   C. 1.7927
   D. 1.7823
   Answer: D. Last iterates agree at 1.7823. (Key point: Result)

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