To find the two consecutive integers between which \(\sqrt{28}\) lies, we can start by estimating the square root.
First, we note the perfect squares nearest to 28:
- \(5^2 = 25\)
- \(6^2 = 36\)
Since \(25 < 28 < 36\), we know that: \[ 5 < \sqrt{28} < 6 \]
To confirm this, we can calculate the approximate value of \(\sqrt{28}\). Simplifying \(\sqrt{28}\): \[ \sqrt{28} = \sqrt{4 \cdot 7} = \sqrt{4} \cdot \sqrt{7} = 2\sqrt{7} \]
Next, we need to estimate \(\sqrt{7}\). We know:
- \(2^2 = 4\)
- \(3^2 = 9\)
Thus, \(2 < \sqrt{7} < 3\), which leads to: \[ 4 < 2\sqrt{7} < 6 \] This means: \[ 2 < \sqrt{7} < 3 \] and consequently: \[ 4 < 2\sqrt{7} < 6 \]
Thus, \(\sqrt{28}\) is indeed between 5 and 6.
Therefore, the square root of 28 lies between the two consecutive integers 5 and 6.