9.7 Higher roots  (Page 2/8)

Solution

We use the absolute value to be sure to get the positive root.

ⓐ
$\begin{array}{ccc}& & \sqrt{x^2}\hfill \\ \text{Since}{\left(x\right)}^2=x^2\text{and we want the positive root.}\hfill & & \left|x\right|\hfill \end{array}$

ⓑ
$\begin{array}{ccc}& & \sqrt[3]{n^3}\hfill \\ \begin{array}{c}\text{Since}{\left(n\right)}^3=n^3.\text{It is an odd root so there is}\hfill \\ \text{no need for an absolute value sign.}\hfill \end{array}\hfill & & n\hfill \end{array}$

ⓒ
$\begin{array}{ccc}& & \sqrt[4]{p^4}\hfill \\ \text{Since}{\left(p\right)}^4=p^4\text{and we want the positive root.}\hfill & & \left|p\right|\hfill \end{array}$

ⓓ
$\begin{array}{ccc}& & \sqrt[5]{y^5}\hfill \\ \begin{array}{c}\text{Since}{\left(y\right)}^5=y^5.\text{It is an odd root so there}\hfill \\ \text{is no need for an absolute value sign.}\hfill \end{array}\hfill & & y\hfill \end{array}$

Solution

1. ⓐ
   $\begin{array}{ccc}& & \sqrt[3]{y^{18}}\hfill \\ \text{Since}{\left(y^6\right)}^3=y^{18}.\hfill & & \sqrt[3]{{\left(y^6\right)}^3}\hfill \\ & & y^6\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{ccc}& & \sqrt[4]{z^8}\hfill \\ \text{Since}{\left(z^2\right)}^4=z^8.\hfill & & \sqrt[4]{{\left(z^2\right)}^4}\hfill \\ \text{Since}{z}^2\text{is positive, we do not need an absolute value sign.}\hfill & & z^2\hfill \end{array}$

Solution

ⓐ
$\begin{array}{ccc}& & \sqrt[3]{64p^6}\hfill \\ \text{Rewrite}64p^6\text{as}{\left(4p^2\right)}^3.\hfill & & \sqrt[3]{{\left(4p^2\right)}^3}\hfill \\ \text{Take the cube root.}\hfill & & 4p^2\hfill \end{array}$

ⓑ
$\begin{array}{ccc}& & \sqrt[4]{16q^{12}}\hfill \\ \text{Rewrite the radicand as a fourth power.}\hfill & & \sqrt[4]{{\left(2q^3\right)}^4}\hfill \\ \text{Take the fourth root.}\hfill & & 2\left|q^3\right|\hfill \end{array}$

Solution

1. ⓐ
   $\begin{array}{ccc}& & \sqrt[3]{x^4}\hfill \\ \begin{array}{c}\text{Rewrite the radicand as a product using the}\hfill \\ \text{largest perfect cube factor.}\hfill \end{array}\hfill & & \sqrt[3]{x^3·x}\hfill \\ \text{Rewrite the radical as the product of two radicals.}\hfill & & \sqrt[3]{x^3}·\sqrt[3]{x}\hfill \\ \text{Simplify.}\hfill & & x\sqrt[3]{x}\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{ccc}& & \sqrt[4]{x^7}\hfill \\ \begin{array}{c}\text{Rewrite the radicand as a product using the}\hfill \\ \text{greatest perfect fourth power factor.}\hfill \end{array}\hfill & & \sqrt[4]{x^4·x^3}\hfill \\ \text{Rewrite the radical as the product of two radicals.}\hfill & & \sqrt[4]{x^4}·\sqrt[4]{x^3}\hfill \\ \text{Simplify.}\hfill & & \left|x\right|\sqrt[4]{x^3}\hfill \end{array}$

Solution

1. ⓐ
   $\begin{array}{ccc}& & \sqrt[3]{16}\hfill \\ & & \sqrt[3]{2^4}\hfill \\ \begin{array}{c}\text{Rewrite the radicand as a product using the}\hfill \\ \text{greatest perfect cube factor.}\hfill \end{array}\hfill & & \sqrt[3]{2^3·2}\hfill \\ \text{Rewrite the radical as the product of two radicals.}\hfill & & \sqrt[3]{2^3}·\sqrt[3]{2}\hfill \\ \text{Simplify.}\hfill & & 2\sqrt[3]{2}\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{ccc}& & \sqrt[4]{243}\hfill \\ & & \sqrt[4]{3^5}\hfill \\ \begin{array}{c}\text{Rewrite the radicand as a product using the}\hfill \\ \text{greatest perfect fourth power factor.}\hfill \end{array}\hfill & & \sqrt[4]{3^4·3}\hfill \\ \text{Rewrite the radical as the product of two radicals.}\hfill & & \sqrt[4]{3^4}·\sqrt[4]{3}\hfill \\ \text{Simplify.}\hfill & & 3\sqrt[4]{3}\hfill \end{array}$

Solution

1. ⓐ
   $\begin{array}{ccc}& & \sqrt[3]{24x^7}\hfill \\ \begin{array}{c}\text{Rewrite the radicand as a product using}\hfill \\ \text{perfect cube factors.}\hfill \end{array}\hfill & & \sqrt[3]{2^3{x}^6·3x}\hfill \\ \text{Rewrite the radical as the product of two radicals.}\hfill & & \sqrt[3]{2^3{x}^6}·\sqrt[3]{3x}\hfill \\ \text{Rewrite the first radicand as}{\left(2x^2\right)}^3.\hfill & & \sqrt[3]{{\left(2x^2\right)}^3}·\sqrt[3]{3x}\hfill \\ \text{Simplify.}\hfill & & 2x^2\sqrt[3]{3x}\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{ccc}& & \sqrt[4]{80y^{14}}\hfill \\ \\ \\ \begin{array}{c}\text{Rewrite the radicand as a product using}\text{perfect fourth power factors.}\hfill \end{array}\hfill & & \sqrt[4]{2^4{y}^{12}·5y^2}\hfill \\ \\ \\ \text{Rewrite the radical as the product of two radicals.}\hfill & & \sqrt[4]{2^4{y}^{12}}·\sqrt[4]{5y^2}\hfill \\ \\ \\ \text{Rewrite the first radicand as}{\left(2y^3\right)}^4.\hfill & & \sqrt[4]{{\left(2y^3\right)}^4}·\sqrt[4]{5y^2}\hfill \\ \\ \\ \text{Simplify.}\hfill & & 2\left|y^3\right|\sqrt[4]{5y^2}\hfill \end{array}$

Solution

1. ⓐ
   $\begin{array}{ccc}& & \sqrt[3]{\mathrm{-27}}\hfill \\ \begin{array}{c}\text{Rewrite the radicand as a product using}\hfill \\ \text{perfect cube factors.}\hfill \end{array}\hfill & & \sqrt[3]{{\left(\mathrm{-3}\right)}^3}\hfill \\ \text{Take the cube root.}\hfill & & \mathrm{-3}\hfill \end{array}$
   
   
2. ⓑ
   $\begin{array}{ccc}& & \sqrt[4]{\mathrm{-16}}\hfill \\ \text{There is no real number}n\text{where}{n}^4=\mathrm{-16}.\hfill & & \text{Not a real number.}\hfill \end{array}$