Show that the equation $x^2 + 7x - 5 = 0$ has a solution between $x = 0$ and $x = 1$ - Edexcel - GCSE Maths - Question 15 - 2017 - Paper 3

To rearrange the equation, we start with:

$x^2 + 7x - 5 = 0.$

We can move $5$ to the other side:

$x^2 + 7x = 5.$

Next, we factor it as follows:

$x(x + 7) = 5$

Dividing both sides by $(x + 7)$ gives:

$x = \frac{5}{x + 7}.$

Using the iterative formula, we can calculate:

1. For $n = 1$:

   $x_2 = \frac{5}{1 + 7} = \frac{5}{8} = 0.625$

2. For $n = 2$:

   $x_3 = \frac{5}{0.625 + 7} = \frac{5}{7.625} \approx 0.65656$

3. For $n = 3$:

   $x_4 = \frac{5}{0.65656 + 7} \approx \frac{5}{7.65656} \approx 0.65381$

Thus, the third estimate for the solution is approximately $0.65381$.

Substituting $x = 0.65381$ into the equation:

$f(0.65381) = (0.65381)^2 + 7(0.65381) - 5$

Calculating this:

$= 0.4273 + 4.57667 - 5 \approx 0.00397.$

This result suggests that our estimate is quite accurate since it is very close to $0$. Therefore, the solution for $x^2 + 7x - 5 = 0$ is approximately $0.65381$.