Q. 252 - Intermediate Algebra

Question

In the following exercise , simplify 

$$\begin{gathered} \left(i\right)\frac{\sqrt{98r{s}^{10}}}{\sqrt{2r^3{s}^4}} \\ \left(ii\right)\frac{\sqrt{-375y^4{z}^{-2}}}{\sqrt{3y^{-2}{z}^4}} \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) The solution  is $\frac{7s^3}{r}$.

Part (b) The solution  is $-\frac{5y^2}{z^2}$

Step 1. Given information

The given equations 

$$\begin{gathered} \left(i\right)\frac{\sqrt{98r{s}^{10}}}{\sqrt{2r^3{s}^4}} \\ \left(ii\right)\frac{\sqrt{-375y^4{z}^{-2}}}{\sqrt{3y^{-2}{z}^4}} \end{gathered}$$

Step 2. Calculation Part(a)

$$\begin{gathered} \left(i\right)\frac{\sqrt{98r{s}^{10}}}{\sqrt{2r^3{s}^4}}=\frac{\sqrt{98r{s}^{10}}}{\sqrt{2r^3{s}^4}} \mathrm{Write} \mathrm{the} \mathrm{expression} \\ =\frac{\sqrt{98r{s}^{10}}}{\sqrt{2r^3{s}^4}} \mathrm{Rewrite} \mathrm{using} \mathrm{quotient} \mathrm{property} \\ =\frac{\sqrt{49s^6}}{\sqrt{r^3}} \mathrm{Cancel} \mathrm{out} \mathrm{common} \mathrm{factors} \\ =\frac{7s^3}{r} \mathrm{Simplify} \\ \mathrm{Therefore} , \mathrm{The} \mathrm{simplified} \mathrm{form} \mathrm{is} \frac{7s^3}{r}. \end{gathered}$$

Step 3. Calculation Part(b)

$$\begin{gathered} \left(ii\right)\frac{\sqrt{-375y^4{z}^{-2}}}{\sqrt{3y^{-2}{z}^4}}=\frac{\sqrt{-375y^4{z}^{-2}}}{\sqrt{3y^{-2}{z}^4}} \mathrm{Write} \mathrm{the} \mathrm{expression} \\ =\frac{\sqrt{-375y^4{z}^{-2}}}{\sqrt{3y^{-2}{z}^4}} \mathrm{Rewrite} \mathrm{using} \mathrm{quotient} \mathrm{form} \\ =\frac{\sqrt[3]{-125y^6} }{z^6}\mathrm{Cancel} \mathrm{out} \mathrm{common} \mathrm{factor} \\ =-\frac{5y^2}{z^2} \\ T\mathrm{hus} \mathrm{the} \mathrm{simplified} \mathrm{form} \mathrm{is} -\frac{5y^2}{z^2} \end{gathered}$$