Chapter 4.6, Problem 26IP

To determine

To find:the square root to the nearest integer.

Answer to Problem 26IP

Thesquare root to the nearest integeris $\pm 9$ .

Explanation of Solution

Given information :

The expression is $\pm \sqrt{78}$

Calculation :

The first perfect square less than 78 is 64

The first perfect square greater than 78is 81 .

  $\begin{array}{l}64<78<81\\ 8^2<78<9^2\\ \sqrt{8^2}<78<\sqrt{9^2}\\ 8<\sqrt{78}<9\end{array}$

The square root of 78 is between the integers $\pm 8$ and $\pm 9$ .Since 78 is closer to 81 than to 64.

Hence, $\pm 9$ is nearest integer square root to the expression $\pm \sqrt{78}$ .