Use the Newton-Raphson method, starting from $x_0=1$ , to find a root of $x^3+3x-7=0$ , correct to 2 d.p.

Model answer

1.41

Also accepted: 1.406, 1.41026

Explanation: With

f(x)=x^3+3x-7

and

f'(x)=3x^2+3

x_1=1.5

x_2\approx1.41026

x_3\approx1.40630

, effectively converged. Root

\approx1.41

(2 d.p.).

Derived from ZIMSEC Pure Mathematics Paper 2, November 2025, Q10

Get the full paper, not just one question

Danho has every sitting for this paper, with your progress tracked question by question, offline.