Show that the equation 2x3−x2−k=02x^3 - x^2 - k = 02x3−x2−k=0, where 1<k<121 < k < 121<k<12, has a solution between x=1x = 1x=1 and x=2x = 2x=2.
Show that the equation 2x3−x2−k=02x^3 - x^2 - k = 02x3−x2−k=0 can be rearranged to give: x=k2x−1\displaystyle x = \sqrt{\frac{k}{2x - 1}}x=2x−1k
Starting with x0=1x_0 = 1x0=1, use the iteration formula xn+1=k2xn−1\displaystyle x_{n+1} = \sqrt{\frac{k}{2x_n - 1}}xn+1=2xn−1k twice to find, in terms of kkk, an estimate for the solution to 2x3−x2−k=02x^3 - x^2 - k = 02x3−x2−k=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.