Starting with x0=0x_0 = 0x0=0, use the iteration formula xn+1=2xn2+3\displaystyle x_{n+1} = \frac{2}{x_n^2 + 3}xn+1=xn2+32 once to find x1x_1x1
Use the iteration formula xn+1=2xn2+3\displaystyle x_{n+1} = \frac{2}{x_n^2 + 3}xn+1=xn2+32 two more times to find an estimate for the solution to x3+3x=2x^3 + 3x = 2x3+3x=2
124 exam-style questions on Eduqas GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.