f(x) = x^3 + 3x^2 + 4x - 12 (a) Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{4 - 3x}{3 + x}}, \, x \neq -3 The equation x^3 + 3x^2 + 4x - 12 = 0 has a single root which is between 1 and 2 - Edexcel - A-Level Maths Pure - Question 4 - 2012 - Paper 5

Starting with:

$x_0 = 1$

We calculate:

1. For x_1: $x_1 = \sqrt{\frac{43 - 1}{3 + 1}} = \sqrt{\frac{42}{4}} = \sqrt{10.5} \approx 3.24$

2. For x_2: $x_2 = \sqrt{\frac{43 - 3.24}{3 + 3.24}} = \sqrt{\frac{39.76}{6.24}} \approx 2.23$

3. For x_3: $x_3 = \sqrt{\frac{43 - 2.23}{3 + 2.23}} = \sqrt{\frac{40.77}{5.23}} \approx 2.06$

Thus, we find: $x_1 \approx 3.24, x_2 \approx 2.23, x_3 \approx 2.06$

To confirm that α is approximately 1.272, we evaluate the function:

$f(1.27) = 1.27^3 + 3(1.27)^2 + 4(1.27) - 12 \approx 0.009$ $f(1.28) = 1.28^3 + 3(1.28)^2 + 4(1.28) - 12 \approx -0.001$

Since:

• f(1.27) > 0
• f(1.28) < 0

The root exists in the interval (1.27, 1.28). Further calculations confirm: $f(1.272) \approx 0.00089$ $f(1.271) \approx 0.0008$

Thus, confirming α is approximately 1.272 to 3 decimal places.