Chapter 1.3, Problem 1P

To determine

To find: the simplest form of an expression.

Answer to Problem 1P

  $\left(a\right)8,\left(b\right)5,\left(c\right)\frac{1}{6}\text{ and }\left(d\right)\frac{9}{11}$ .

Explanation of Solution

Given:

  $\left(a\right)\sqrt{64},\left(b\right)\sqrt{25},\left(c\right)\sqrt{\frac{1}{36}}\text{and }\left(d\right)\sqrt{\frac{81}{121}}$ .

Concept used:

In general, an expression is in simplest form when it is easiest to use.

Common ways to help it simplify is:

Combining like terms.

Factoring.

Expanding the opposite of factoring.

Clear out fractions by multiplying.

Recognizing the patterns and it has seen before, like the difference of squares etc.

Calculation:

Option $\left(a\right)$ can be simplified as:

  $\begin{array}{l}\left(a\right):\sqrt{64}.\\ \Rightarrow \sqrt{64}=\sqrt{8^2}.\\ \Rightarrow {\left(8^2\right)}^{\frac{1}{2}}=8.\end{array}$

Option $\left(b\right)$ can be simplified as:

  $\begin{array}{l}\sqrt{25}.\\ \Rightarrow \sqrt{5^2}.\\ \Rightarrow {\left(5^2\right)}^{\frac{1}{2}}.\end{array}$

Option $\left(c\right)$ can simplified as:

  $\begin{array}{l}\sqrt{\frac{1}{36}}.\\ \Rightarrow \sqrt{\frac{1}{6^2}}={\left(\frac{1}{6^2}\right)}^{\frac{1}{2}}.\\ \Rightarrow \frac{1}{6}.\end{array}$

Option $\left(d\right)$ can simplified as:

  $\begin{array}{l}\sqrt{\frac{81}{121}}.\\ \Rightarrow \sqrt{\frac{9^2}{{11}^2}}=\sqrt{{\left(\frac{9}{11}\right)}^2}.\\ \Rightarrow {\left(\frac{9}{11}\right)}^{2\times \frac{1}{2}}=\frac{9}{11}.\end{array}$

Hence, $\left(a\right)8,\left(b\right)5,\left(c\right)\frac{1}{6}\text{ and }\left(d\right)\frac{9}{11}$ .