Q. 8.24 - Intermediate Algebra

Question

Simplify: $\left(a\right) \sqrt{225m^2{n}^2} \left(b\right)\sqrt{169x^{10}{y}^{14}} \left(c\right)\sqrt[3]{27w^{36}{z}^{15}}$

Step-by-Step Solution

Answer

The solutions of (a), (b) and (c) are $15\left|mn\right|, 13\left|x^5{y}^7\right| and 3w^{12}{z}^5$ respectively.

Part (a) Step 1. Given information

The root: $\sqrt{225m^2{n}^2}$

Part (a) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt{225m^2{n}^2} \\ =\sqrt{{\left(15mn\right)}^2} \sin ce {\left(15mn\right)}^2=225m^2{n}^2 \\ =15\left|mn\right| \end{gathered}$$

Therefore, the solution is $15\left|mn\right|$

Part (b) Step 1. Given information

The root: $\sqrt{169x^{10}{y}^{14}}$

Part (b) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt{169x^{10}{y}^{14}} \\ =\sqrt{{\left(13x^5{y}^7\right)}^2} \sin ce {\left(13x^5{y}^7\right)}^2=169x^{10}{y}^{14} \\ =13\left|x^5{y}^7\right| \end{gathered}$$

Therefore, the solution is $13\left|x^5{y}^7\right|$

Part (c) Step 1. Given information

The root: $\sqrt[3]{27w^{36}{z}^{15}}$

Part (c) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt[3]{27w^{36}{z}^{15}} \\ =\sqrt[3]{{\left(3w^{12}{z}^5\right)}^3} \sin ce {\left(3w^{12}{z}^5\right)}^3=27w^{36}{z}^{15} \\ =3w^{12}{z}^5 \end{gathered}$$

Therefore, the solution is $3w^{12}{z}^5$