Show that the equation 3x3−x2−4=03x^3 - x^2 - 4 = 03x3−x2−4=0 has a solution between x=1x = 1x=1 and x=2x = 2x=2.
Show that the equation 3x3−x2−4=03x^3 - x^2 - 4 = 03x3−x2−4=0 can be rearranged to give: x=43x−1\displaystyle x = \sqrt{\frac{4}{3x - 1}}x=3x−14
Starting with x0=1x_0 = 1x0=1, use the iteration formula xn+1=43xn−1\displaystyle x_{n+1} = \sqrt{\frac{4}{3x_n - 1}}xn+1=3xn−14 twice to find an estimate for the solution to 3x3−x2−4=03x^3 - x^2 - 4 = 03x3−x2−4=0
Give your answer in exact form.
124 exam-style questions on OCR GCSE Maths Iteration. Each one has a worked solution and a mark scheme showing where the marks go.