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

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.

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

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

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.

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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