Chapter 4.6, Problem 6GP

To determine

To find:the square root to the nearest integer.

Answer to Problem 6GP

Thesquare root to the nearest integeris $\pm 6$ .

Explanation of Solution

Given information :

The expression is $\pm \sqrt{39}$

Calculation :

The first perfect square less than 39 is 36

The first perfect square greater than 39 is 49.

  $\begin{array}{l}36<39<49\\ 6^2<39<7^2\\ \sqrt{6^2}<39<\sqrt{7^2}\\ 6<\sqrt{39}<7\end{array}$

The square root of 39 is between the integers $\pm 6$ and $\pm 7$ .Since 48 is closer to 36 than to 49.

Hence, $\pm 6$ is nearest integer square root to the expression $\pm \sqrt{39}$ .