Show that the equation 3x3−x2−8=03x^3 - x^2 - 8 = 03x3−x2−8=0 has a solution between x=1x = 1x=1 and x=2x = 2x=2.
Show that the equation 3x3−x2−8=03x^3 - x^2 - 8 = 03x3−x2−8=0 can be rearranged to give: x=83x−1\displaystyle x = \sqrt{\frac{8}{3x - 1}}x=3x−18
Starting with x0=1x_0 = 1x0=1, use the iteration formula xn+1=83xn−1\displaystyle x_{n+1} = \sqrt{\frac{8}{3x_n - 1}}xn+1=3xn−18 twice to find an estimate for the solution to 3x3−x2−8=03x^3 - x^2 - 8 = 03x3−x2−8=0
124 exam-style questions on Eduqas GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.