Question: Use trial and improvement to find an approximate solution to $x = \sqrt{70}$. Give your answer to the nearest integer.
1. **State the problem:** We need to find an approximate value of $x$ such that $x = \sqrt{70}$ using trial and improvement.
2. **Recall the formula:** $x = \sqrt{70}$ means $x^2 = 70$.
3. **Trial and improvement method:** We try integer values of $x$ and check $x^2$ to see which is closest to 70.
4. Try $x=8$: $8^2 = 64$ which is less than 70.
5. Try $x=9$: $9^2 = 81$ which is greater than 70.
6. Since $64 < 70 < 81$, the value of $x$ lies between 8 and 9.
7. To find the nearest integer, compare the differences:
$$70 - 64 = 6$$
$$81 - 70 = 11$$
8. Since 6 is smaller than 11, $x$ is closer to 8.
9. **Final answer:** The approximate value of $\sqrt{70}$ to the nearest integer is **8**.