Q. 251 - Intermediate Algebra

Question

In the following exercise , simplify 

$$\begin{gathered} \left(i\right)\frac{\sqrt{108p^5{q}^2}}{\sqrt{3p^3{q}^6}} \\ \left(ii\right)\frac{\sqrt[3]{-16a^4{b}^{-2}}}{\sqrt[3]{2a^{-2}b}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The solution  is $\frac{6p}{q^2}$.

Part (b) The solution  is $-\frac{2a^2}{b}$

Step 1. Given information

The given equations are

$$\begin{gathered} \left(i\right)\frac{\sqrt{108p^5{q}^2}}{\sqrt{3p^3{q}^6}} \\ \left(ii\right)\frac{\sqrt[3]{-16a^4{b}^{-2}}}{\sqrt[3]{2a^{-2}b}} \end{gathered}$$

Step 2. Calculation Part(a)

$$\begin{gathered} \left(i\right)\frac{\sqrt{108p^5{q}^2}}{\sqrt{3p^3{q}^6}}=\frac{\sqrt{108p^5{q}^2}}{\sqrt{3p^3{q}^6}} [\mathrm{Write} \mathrm{the} \mathrm{expression}] \\ =\frac{\sqrt{108p^5{q}^2}}{\sqrt{3p^3{q}^6}}[\mathrm{Rewrite} \mathrm{using} \mathrm{quotient} \mathrm{property} \\ =\frac{\sqrt{36p^2}}{\sqrt{q^4}} [\mathrm{Cancel} \mathrm{out} \mathrm{common} \mathrm{property}] \\ =\frac{6p}{q^2}[\mathrm{Simplify}] \\ \mathrm{Hence} \mathrm{the} \mathrm{simplified} \mathrm{form} \mathrm{is}\frac{6p}{q^2}. \end{gathered}$$

Step 3. Calculation Part(b)

$$\begin{gathered} \left(ii\right)\frac{\sqrt[3]{-16a^4{b}^{-2}}}{\sqrt[3]{2a^{-2}b}}=\frac{\sqrt[3]{-16a^4{b}^{-2}}}{\sqrt[3]{2a^{-2}b}}\mathrm{Write} \mathrm{the} \mathrm{expressio}n \\ =\frac{\sqrt[3]{-16a^4{b}^{-2}}}{\sqrt[3]{2a^{-2}b}} \mathrm{Rewrite} \mathrm{using} \mathrm{quotient} \mathrm{form} \\ =\frac{\sqrt[3]{8a^6}}{\sqrt[3]{b^3}} \mathrm{Cancel} \mathrm{out} \mathrm{common} \mathrm{factors} \\ =-\frac{2a^2}{b} \\ \mathrm{Hence} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{expression} \mathrm{is} -\frac{2a^2}{b} \end{gathered}$$