15 (a) Show that the equation $x^3 - 5x - 1 = 0$ has a solution between $x = 2$ and $x = 3$ - OCR - GCSE Maths - Question 15 - 2021 - Paper 1

We can use the method of bisection to find the root to one decimal place:

1. Start with the interval $(2, 3)$.
2. Compute the midpoint: $m = \frac{2 + 3}{2} = 2.5$
3. Evaluate $f(2.5)$: $f(2.5) = (2.5)^3 - 5(2.5) - 1 = 15.625 - 12.5 - 1 = 2.125$

Since $f(2) < 0$ and $f(2.5) > 0$, we search in the interval $(2, 2.5)$ next.

4. Compute the new midpoint: $m = \frac{2 + 2.5}{2} = 2.25$
5. Evaluate $f(2.25)$: $f(2.25) = (2.25)^3 - 5(2.25) - 1 = 11.390625 - 11.25 - 1 = -0.859375$

Now, since $f(2.25) < 0$ and $f(2.5) > 0$, we continue in the interval $(2.25, 2.5)$.

6. Compute: $m = \frac{2.25 + 2.5}{2} = 2.375$
7. Evaluate $f(2.375)$: $f(2.375) = (2.375)^3 - 5(2.375) - 1 = 13.419921875 - 11.875 - 1 = 0.544921875$

As $f(2.375) > 0$, we narrow down our interval to $(2.25, 2.375)$:

8. Finally, computing the midpoint gives: $m = \frac{2.25 + 2.375}{2} = 2.3125$
9. Evaluating: $f(2.3125) = (2.3125)^3 - 5(2.3125) - 1 = 12.270462035 - 11.5625 - 1 = -0.292037965$

Thus, we see we can narrow down the range to $(2.3125, 2.375)$ leading to the solution being roughly:

The solution to 1 decimal place is approximately: $\text{x} = 2.3$