Question: The table below can be used to find an approximate solution to
x = \sqrt{93}
a) Copy and complete the table. What exact values should replace A,
B and C?
b) Write down the approximate solution to x = \sqrt{93} to the nearest
integer.
x x^2 Comment
8 64 Too low
9 A
10 B
9.5 C
1. **State the problem:** We want to approximate $x = \sqrt{93}$ using a table of values for $x$ and $x^2$.
2. **Complete the table:**
- For $x=9$, calculate $x^2 = 9^2 = 81$. So, $A = 81$.
- For $x=10$, calculate $x^2 = 10^2 = 100$. So, $B = 100$.
- For $x=9.5$, calculate $x^2 = (9.5)^2 = 9.5 \times 9.5 = 90.25$. So, $C = 90.25$.
3. **Interpret the table:**
- $8^2 = 64$ is too low compared to 93.
- $9^2 = 81$ is still less than 93.
- $9.5^2 = 90.25$ is closer but still less than 93.
- $10^2 = 100$ is greater than 93.
4. **Approximate $\sqrt{93}$:** Since $9.5^2 = 90.25$ is less than 93 and $10^2 = 100$ is greater than 93, $\sqrt{93}$ lies between 9.5 and 10.
5. **Nearest integer:** The closest integer to $\sqrt{93}$ is $10$ because 93 is closer to 100 than to 81.
**Final answers:**
- $A = 81$
- $B = 100$
- $C = 90.25$
- Approximate solution to $\sqrt{93}$ to the nearest integer is $10$.