Chapter 4.6, Problem 27IP

To determine

To find:the square root to the nearest integer.

Answer to Problem 27IP

Thesquare root to the nearest integeris $\pm 12$ .

Explanation of Solution

Given information :

The expression is $\pm \sqrt{146}$

Calculation :

The first perfect square less than 146 is 144

The first perfect square greater than 146is 169.

  $\begin{array}{l}144<146<169\\ {12}^2<146<{13}^2\\ \sqrt{{12}^2}<146<\sqrt{{13}^2}\\ 12<\sqrt{146}<13\end{array}$

The square root of 146 is between the integers $\pm 12$ and $\pm 13$ .Since 146 is closer to 144 than to 169.

Hence, $\pm 12$ is nearest integer square root to the expression $\pm \sqrt{146}$ .