Q. 8.23 - Intermediate Algebra

Question

Simplify: $\left(a\right) \sqrt{100a^2{b}^2} \left(b\right)\sqrt{144p^{12}{q}^{20}} \left(c\right) \sqrt[3]{8x^{30}{y}^{12}}$

Step-by-Step Solution

Answer

The solutions of (a), (b) and (c) are $10\left|ab\right|, 12p^6{q}^{10} and 2x^{10}{y}^4$ respectively.

Part (a) Step 1. Given information

The root: $\sqrt{100a^2{b}^2}$

Part (a) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt{100a^2{b}^2} \\ =\sqrt{{\left(10ab\right)}^2} \sin ce {\left(10ab\right)}^2=100a^2{b}^2 \\ =10\left|ab\right| \end{gathered}$$

Therefore, the solution is $10\left|ab\right|$

Part (b) Step 1. Given information

The root: $\sqrt{144p^{12}{q}^{20}}$

Part (b) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt{144p^{12}{q}^{20}} \\ =\sqrt{{\left(12p^6{q}^{10}\right)}^2} \sin ce {\left(12p^6{q}^{10}\right)}^2=144p^{12}{q}^{20} \\ =12p^6{q}^{10} \end{gathered}$$

Therefore, the solution is $12p^6{q}^{10}$

Part (c) Step 1. Given information

The root: $\sqrt[3]{8x^{30}{y}^{12}}$

Part (c) Step 2. Calculation

We have,

$$\begin{gathered} \sqrt[3]{8x^{30}{y}^{12}} \\ =\sqrt[3]{{\left(2x^{10}{y}^4\right)}^3} \sin ce {\left(2x^{10}{y}^4\right)}^3=8x^{30}{y}^{12} \\ =2x^{10}{y}^4 \end{gathered}$$

Therefore, the solution is $2x^{10}{y}^4$