f(x) = x³ + 2x² - 3x - 11 (a) Show that f(x) = 0 can be rearranged as x = \sqrt{\frac{3x + 11}{x + 2}}; x ≠ -2 - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 2

Using the iterative formula:

$x_{n+1} = \frac{3x_n + 11}{x_n + 2}$

1. For x₁ = 0: $x_2 = \frac{(3(0) + 11)}{(0 + 2)} = \frac{11}{2} = 5.5$

2. For x₂ = 5.5: $x_3 = \frac{(3(5.5) + 11)}{(5.5 + 2)} = \frac{(16.5 + 11)}{7.5} = \frac{27.5}{7.5} \approx 3.6667$

3. For x₃ = 3.6667: $x_4 = \frac{(3(3.6667) + 11)}{(3.6667 + 2)} = \frac{(11.0001)}{5.6667} \approx 1.9422$

After iterating these values we find:

• x₂ ≈ 5.500,
• x₃ ≈ 3.667,
• x₄ ≈ 1.942.

Let f(x) = x³ + 2x² - 3x - 11.

To find root a, we can evaluate f at several points to find a sign change:

• f(2.056) ≈ -0.0013781637 (negative)
• f(2.057) ≈ 0.0041410494 (positive)
• f(2.058) ≈ 0.0056324894 (positive)

The sign change between 2.056 and 2.057 indicates there is a root in this interval. Since both values are correct to 3 decimal places, we find that:

• a ≈ 2.057 is the value correct to 3 decimal places.