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Expert-verified Found in: Page 267 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Rationalize Put each fractional expression into standard form by rationalizing the denominator.a. $\frac{\mathbf{1}}{\sqrt{\mathbf{5}\mathbf{x}}}$b. $\sqrt{\frac{\mathbf{x}}{\mathbf{5}}}$c. $\sqrt[\mathbf{5}]{\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}}}$

1. The simplified form of $\frac{1}{\sqrt{5x}}$ is $\frac{\sqrt{5x}}{5x}$.
2. The simplified form of $\sqrt{\frac{x}{5}}$ is $\frac{\sqrt{5x}}{5}$.
3. The simplified form of $\sqrt{\frac{1}{{x}^{3}}}$ is $\frac{\sqrt{{x}^{2}}}{x}$.
See the step by step solution

## Part a. Step 1. Given information.

The given expression is $\frac{1}{\sqrt{5x}}$.

## part a. Step 2. Write the concept.

Rationalizing the denominator is a procedure to eliminate the radical in the denominator by multiplying both numerator and denominator by an appropriate expression.

If the denominator is of the form $\sqrt{a}$, then multiply numerator and denominator by $\sqrt{a}$.

If the denominator is of the form $\sqrt[n]{{a}^{m}},m, then multiply numerator and denominator by $\sqrt[n]{{a}^{n-m}}$.

## Part a. Step 3. Determine the simplified form of the expression.

The given expression can be written as:

$\begin{array}{c}\frac{1}{\sqrt{5x}}=\frac{1}{\sqrt{5x}}×\frac{\sqrt{5x}}{\sqrt{5x}}\\ =\frac{\sqrt{5x}}{5x}\end{array}$

## Part b. Step 1. Given information.

The given expression is $\sqrt{\frac{x}{5}}$.

## Part b. Step 2. Write the concept.

Rationalizing the denominator is a procedure to eliminate the radical in the denominator by multiplying both numerator and denominator by an appropriate expression.

If the denominator is of the form $\sqrt{a}$, then multiply numerator and denominator by $\sqrt{a}$.

If the denominator is of the form $\sqrt[n]{{a}^{m}},m, then multiply numerator and denominator by $\sqrt[n]{{a}^{n-m}}$.

## Part c. Step 3. Determine the simplified form of the expression.

The given expression can be written as:

$\begin{array}{c}\sqrt{\frac{x}{5}}=\frac{\sqrt{x}}{\sqrt{5}}\\ =\frac{\sqrt{x}}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\\ =\frac{\sqrt{x×5}}{5}\\ =\frac{\sqrt{5x}}{5}\end{array}$

## Part c. Step 1. Given information.

The given expression is $\sqrt{\frac{1}{{x}^{3}}}$.

## Part c.  Step 2. Write the concept.

Rationalizing the denominator is a procedure to eliminate the radical in the denominator by multiplying both numerator and denominator by an appropriate expression.

If the denominator is of the form $\sqrt{a}$, then multiply numerator and denominator by $\sqrt{a}$.

If the denominator is of the form $\sqrt[n]{{a}^{m}},m, then multiply numerator and denominator by $\sqrt[n]{{a}^{n-m}}$.

## Part c. Step 3. Determine the simplified form of the expression.

The given expression can be written as:

$\begin{array}{c}\sqrt{\frac{1}{{x}^{3}}}=\frac{1}{\sqrt{{x}^{3}}}\\ =\frac{1}{\sqrt{{x}^{3}}}×\frac{\sqrt{{x}^{5-3}}}{\sqrt{{x}^{5-3}}}\\ =\frac{1}{\sqrt{{x}^{3}}}×\frac{\sqrt{{x}^{2}}}{\sqrt{{x}^{2}}}\\ =\frac{\sqrt{{x}^{2}}}{\sqrt{{x}^{3+2}}}\\ =\frac{\sqrt{{x}^{2}}}{\sqrt{{x}^{5}}}\\ =\frac{\sqrt{{x}^{2}}}{x}\end{array}$ ### Want to see more solutions like these? 