Q. 253 - Intermediate Algebra

Question

In the following exercise , simplify

$$\begin{gathered} \left(i\right)\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}} \\ \left(ii\right)\frac{\sqrt{16x^4{y}^{-2}}}{\sqrt{-54x^{-2}{y}^4}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The solution  is $\frac{8m^4}{3n^4}$.

Part (b) The solution  is $\frac{-2x^2}{3y^2}$

Step 1. Given information

The given equations are $$\begin{gathered} \left(i\right)\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}} \\ \left(ii\right)\frac{\sqrt{16x^4{y}^{-2}}}{\sqrt{-54x^{-2}{y}^4}} \end{gathered}$$

Step 2. Calculation Part(a).

$$\begin{gathered} \left(i\right)\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}= \frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}} [\mathrm{Write} \mathrm{the} \mathrm{expression}] \\ =\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}[\mathrm{Rewrite} \mathrm{using} \mathrm{quotient} \mathrm{property} \\ =\frac{\sqrt{64m^8}}{\sqrt{9n^8}} [\mathrm{Cancel} \mathrm{out} \mathrm{common} \mathrm{property}] \\ =\frac{\sqrt{8m^8}}{\sqrt{3n^8}} [\mathrm{Simplify}] \\ \mathrm{Hence} \mathrm{the} \mathrm{simplified} \mathrm{form} \mathrm{is} \frac{\sqrt{8m^8}}{\sqrt{3n^8}} \end{gathered}$$

Step 3. Calculation Part(b)

$$\begin{gathered} \left(ii\right)\frac{\sqrt{16x^4{y}^{-2}}}{\sqrt{-54x^{-2}{y}^4}}=\frac{\sqrt{16x^4{y}^{-2}}}{\sqrt{-54x^{-2}{y}^4}} \mathrm{Write} \mathrm{the} \mathrm{expression} \\ =\frac{\sqrt{16x^4{y}^{-2}}}{\sqrt{-54x^{-2}{y}^4}} \mathrm{Rewrite} \mathrm{using} \mathrm{quotient} \mathrm{property} \\ =\frac{\sqrt{8x^4}}{\sqrt{-27y^4}} \mathrm{Cancel} \mathrm{out} \mathrm{common} \mathrm{factors} \\ =\frac{-2x^2}{3y^2} \mathrm{Simplify} \\ \mathrm{The} \mathrm{simplified} \mathrm{form} \mathrm{is} \frac{-2x^2}{3y^2} . \end{gathered}$$