7 The equation \(x ^ { 2 } = x ^ { 3 } + x - 3\) has a single solution, \(x = \alpha\)
7
- By considering a suitable change of sign, show that \(\alpha\) lies between 1.5 and 1.6
[0pt]
[2 marks]
7 - Show that the equation \(x ^ { 2 } = x ^ { 3 } + x - 3\) can be rearranged into the form
$$x ^ { 2 } = x - 1 + \frac { 3 } { x }$$
7
- Use the iterative formula
$$x _ { n + 1 } = \sqrt { x _ { n } - 1 + \frac { 3 } { x _ { n } } }$$
with \(x _ { 1 } = 1.5\), to find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to four decimal places.
7 - Hence, deduce an interval of width 0.001 in which \(\alpha\) lies.