Q. 111 - Intermediate Algebra

Question

Use the Quotient Property to simplify square roots.

(a) $\frac{\sqrt{45p^9}}{\sqrt{5q^2}}$

(b) $\frac{\sqrt[4]{64}}{\sqrt[4]{2}}$

(c) $\frac{\sqrt[5]{128x^8}}{\sqrt[5]{2x^2}}$

Step-by-Step Solution

Answer

(a) $\frac{3p^4\sqrt{p}}{\left|q\right|}$

(b) $2\sqrt[4]{2}$

(c) $2x\sqrt[5]{2x}$

Part (a) Step 1. Given information.

Given expression:

$\frac{\sqrt{45p^9}}{\sqrt{5q^2}}$.

Part (a) Step 2. Write the concept.

For simplifying the radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a} and \sqrt[n]{b}$ are the real number, $b\ne 0$for any integer $n\ge 2$.

Part (a) Step 3. Calculation

$$\begin{gathered} \frac{\sqrt{45p^9}}{\sqrt{5q^2}}=\sqrt{\frac{45p^9}{5q^2}} \\ =\sqrt{\frac{9.5.p^8.p}{5.q^2}} \\ =\sqrt{\frac{{\left(3\right)}^2.{\left(p^4\right)}^2.p}{q^2}} \\ =\frac{3p^4\sqrt{p}}{\left|q\right|} \end{gathered}$$

Part (b) Step 1. Given information.

Given radical expression:

$\frac{\sqrt[4]{64}}{\sqrt[4]{2}}$.

Part (b) Step 2. Write the concept.

For simplifying the radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a} and \sqrt[n]{b}$ are the real number, $b\ne 0$ for any integer $n\ge 2$.

Part (b) Step 3. Calculation

$$\begin{gathered} \frac{\sqrt[4]{64}}{\sqrt[4]{2}}=\sqrt[4]{\frac{64}{2}} \\ =\sqrt[4]{\frac{2.32}{2}} \\ =\sqrt[4]{{\left(2\right)}^4.2} \\ =2\sqrt[4]{2} \end{gathered}$$

Part (c) Step 1. Given information.

Given expression:

$\frac{\sqrt[5]{128x^8}}{\sqrt[5]{2x^2}}$.

Part (c) Step 2. Write the concept.

radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}} and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if  $\sqrt[n]{a} and \sqrt[n]{b}$are the real number, $b\ne 0$ for any integer $n\ge 2$.

Part (c) Step 3. Calculation

$$\begin{gathered} \frac{\sqrt[5]{128x^8}}{\sqrt[5]{2x^2}}=\sqrt[5]{\frac{128x^8}{2x^2}} \\ =\sqrt[5]{\frac{2.64.x^8}{2.x^2}} \\ =\sqrt[5]{{\left(2x\right)}^5.2x} \\ =2x\sqrt[5]{2x} \end{gathered}$$