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 3 - 2012 - Paper 5

1. Start with x_0 = 1.

2. Compute x_1:

$x_1 = \sqrt{\frac{4 - 1}{3 + 1}} = \sqrt{\frac{3}{4}} = 1.2247 \approx 1.22$

3. Next, compute x_2:

$x_2 = \sqrt{\frac{4 - x_1}{3 + x_1}} = \sqrt{\frac{4 - 1.22}{3 + 1.22}} = \sqrt{\frac{2.78}{4.22}} \approx 1.1482 \approx 1.15$

4. Finally, compute x_3:

$x_3 = \sqrt{\frac{4 - x_2}{3 + x_2}} = \sqrt{\frac{4 - 1.15}{3 + 1.15}} = \sqrt{\frac{2.85}{4.15}} \approx 1.1463 \approx 1.14$

To find α accurately, evaluate f(1.2715) and f(1.2725):

• For f(1.2715): $f(1.2715) = (1.2715)^3 + 3(1.2715)^2 + 4(1.2715) - 12 \approx -0.008027$

• For f(1.2725): $f(1.2725) = (1.2725)^3 + 3(1.2725)^2 + 4(1.2725) - 12 \approx 0.000017$

Since f(1.2715) < 0 and f(1.2725) > 0, the root lies between these values. Repeating this process provides an interval containing 1.272. Hence, we conclude that α = 1.272 to 3 decimal places.