Question

Approximate square root of, 96, end square root
96
​

square root of, 96, end square root
96
​
must lie between the whole numbers
9
9 and
10
10 because 9, squared, equals, start box, 81, end box9
2
=
81
​
and 10, squared, equals, start box, 100, end box10
2
=
100
​
, and 96 lies between these values.
Drag square root of, 96, end square root
96
​
based on your estimate above:
Real
√96
__
_
To one decimal place, square root of, 96, end square root
96
​
must lie between
9.7
9.7 and
9.8
9.8 because 9, point, 7, squared, equals, start box, 94, point, 0, 9, end box9.7
2
=
94.09
​
and 9, point, 8, squared, equals, start box, 96, point, 0, 4, end box9.8
2
=
96.04
​
, and 96 lies between these values.
Drag square root of, 96, end square root
96
​
based on your estimate above:
Real
√96
__
_
To two decimal places, square root of, 96, end square root
96
​
must lie between
and

Answer 1

To find the square root of 96 to two decimal places, we will narrow it down from the previous estimates.

From our previous findings:
- \(\sqrt{96} \approx 9.7\) (since \(9.7^2 = 94.09\))
- \(\sqrt{96} \approx 9.8\) (since \(9.8^2 = 96.04\))

Now let's check more decimal places between 9.7 and 9.8:

1. **Estimate \(9.75\):**
\[
9.75^2 = 95.0625
\]
This is less than 96.

2. **Estimate \(9.76\):**
\[
9.76^2 = 95.2976
\]
This is still less than 96.

3. **Estimate \(9.77\):**
\[
9.77^2 = 95.8729
\]
This is still less than 96.

4. **Estimate \(9.78\):**
\[
9.78^2 = 96.0484
\]
This is greater than 96.

So, we find that:
- \(9.77^2 = 95.8729 < 96 < 9.78^2 = 96.0484\)

Thus, to two decimal places, \(\sqrt{96}\) lies between **9.77 and 9.78**.

In summary, the range is:
\[
\text{To two decimal places, } \sqrt{96} \text{ must lie between } 9.77 \text{ and } 9.78.
\]