f(x) = 2x^2 - x - 4 - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 5

Start with the initial value:

$x_0 = 1.35$

Now apply the iteration formula:

1. Calculate $x_1 = \frac{2}{1.35 + 2} = 1.39$

2. Next, $x_2 = \frac{2}{1.39 + 2} \approx 1.394$

3. Finally, $x_3 = \frac{2}{1.394 + 2} \approx 1.392$.

Thus, the values are:

• $x_1 \approx 1.39$
• $x_2 \approx 1.394$
• $x_3 \approx 1.392$.

First, check the function values around the suspected root:

1. Calculate $f(1.391) = 2(1.391)^2 - (1.391) - 4 \approx -0.0003$
2. Calculate $f(1.392) = 2(1.392)^2 - (1.392) - 4 \approx 0.0009$

The sign changes between $1.391$ and $1.392$, indicating a root exists in this interval. Therefore, we can write:

$\alpha \approx 1.392$

To further confirm, one can check the function value at $1.393$:

3. Calculate $f(1.393) \approx 0.0057$, which is positive.

Since the signs change between $1.391$ (negative) and $1.392$ (positive), we conclude that:

$\alpha = 1.392$

to 3 decimal places.