Question

to estimate the square root of a non - perfect square.
perfect squares
estimate the square roots by finding the perfect squares that are less and greater than the number. find the square roots of the perfect squares for your estimate. then, circle the integer that the estimate is closest to. the first one is done for you.
\begin{tabular}{|c|c|c|}hline&\text{consecutive perfect squares}&\text{in between the integers}\hline$sqrt{10}$&9 and 16&3 and 4\hline$sqrt{77}$&&\hline$sqrt{6}$&&\hline$sqrt{105}$&&\hline$sqrt{34}$&&\hline$sqrt{50}$&&\hline$sqrt{143}$&&\hline$sqrt{60}$&&\hline$sqrt{2}$&&\hline$sqrt{85}$&&\hline$sqrt{47}$&&\hlineend{tabular}

Explanation:

Step1: Recall perfect - square values

We know that \(n^2\) gives perfect - squares. For example, \(1^2 = 1\), \(2^2=4\), \(3^2 = 9\), \(4^2=16\), \(5^2 = 25\), \(6^2=36\), \(7^2 = 49\), \(8^2=64\), \(9^2 = 81\), \(10^2=100\), \(11^2 = 121\), \(12^2=144\).

Step2: Find consecutive perfect - squares for \(\sqrt{77}\)

Since \(8^2=64\) and \(9^2 = 81\), the consecutive perfect squares are 64 and 81. And \(\sqrt{77}\) is between 8 and 9.

Step3: Find consecutive perfect - squares for \(\sqrt{6}\)

Since \(2^2 = 4\) and \(3^2=9\), the consecutive perfect squares are 4 and 9. And \(\sqrt{6}\) is between 2 and 3.

Step4: Find consecutive perfect - squares for \(\sqrt{105}\)

Since \(10^2 = 100\) and \(11^2=121\), the consecutive perfect squares are 100 and 121. And \(\sqrt{105}\) is between 10 and 11.

Step5: Find consecutive perfect - squares for \(\sqrt{34}\)

Since \(5^2 = 25\) and \(6^2=36\), the consecutive perfect squares are 25 and 36. And \(\sqrt{34}\) is between 5 and 6.

Step6: Find consecutive perfect - squares for \(\sqrt{50}\)

Since \(7^2 = 49\) and \(8^2=64\), the consecutive perfect squares are 49 and 64. And \(\sqrt{50}\) is between 7 and 8.

Step7: Find consecutive perfect - squares for \(\sqrt{143}\)

Since \(11^2 = 121\) and \(12^2=144\), the consecutive perfect squares are 121 and 144. And \(\sqrt{143}\) is between 11 and 12.

Step8: Find consecutive perfect - squares for \(\sqrt{60}\)

Since \(7^2 = 49\) and \(8^2=64\), the consecutive perfect squares are 49 and 64. And \(\sqrt{60}\) is between 7 and 8.

Step9: Find consecutive perfect - squares for \(\sqrt{2}\)

Since \(1^2 = 1\) and \(2^2=4\), the consecutive perfect squares are 1 and 4. And \(\sqrt{2}\) is between 1 and 2.

Step10: Find consecutive perfect - squares for \(\sqrt{85}\)

Since \(9^2 = 81\) and \(10^2=100\), the consecutive perfect squares are 81 and 100. And \(\sqrt{85}\) is between 9 and 10.

Step11: Find consecutive perfect - squares for \(\sqrt{47}\)

Since \(6^2 = 36\) and \(7^2=49\), the consecutive perfect squares are 36 and 49. And \(\sqrt{47}\) is between 6 and 7.

Answer:

	Consecutive Perfect Squares	In Between the Integers
\(\sqrt{6}\)	4 and 9	2 and 3
\(\sqrt{105}\)	100 and 121	10 and 11
\(\sqrt{34}\)	25 and 36	5 and 6
\(\sqrt{50}\)	49 and 64	7 and 8
\(\sqrt{143}\)	121 and 144	11 and 12
\(\sqrt{60}\)	49 and 64	7 and 8
\(\sqrt{2}\)	1 and 4	1 and 2
\(\sqrt{85}\)	81 and 100	9 and 10
\(\sqrt{47}\)	36 and 49	6 and 7