Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
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Kindly elaborate on the steps to rationalizing the denominator as well
![**Rationalize the Denominator**
**Problem:**
Rationalize the denominator.
\[
4) \frac{\sqrt{20}}{\sqrt{75}}
\]
**Options:**
A) \(\frac{2\sqrt{5}}{3}\)
B) \(\frac{2\sqrt{15}}{3}\)
C) \(\frac{2\sqrt{3}}{15}\)
D) \(\frac{2\sqrt{15}}{15}\)
**Explanation:**
When rationalizing the denominator, the goal is to eliminate any square roots present in the denominator. This can be achieved by multiplying both the numerator and the denominator by an appropriate value.
1. Consider the original expression: \(\frac{\sqrt{20}}{\sqrt{75}}\)
- Break down the square roots:
\[
\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}
\]
\[
\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}
\]
- Substitute these values back into the original expression:
\[
\frac{2\sqrt{5}}{5\sqrt{3}}
\]
2. To rationalize the denominator, multiply both the numerator and the denominator by \(\sqrt{3}\):
\[
\frac{2\sqrt{5}}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{5} \cdot \sqrt{3}}{5\sqrt{3} \cdot \sqrt{3}} = \frac{2\sqrt{15}}{15}
\]
So the rationalized form of the given expression is \(\frac{2\sqrt{15}}{15}\).
**Answer:**
D) \(\frac{2\sqrt{15}}{15}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8311114-a7cb-49e3-b44e-1e05de4bc29a%2Fb36a8354-e402-4c80-a1a6-3530730f510e%2Fvo1543.png&w=3840&q=75)
Transcribed Image Text:**Rationalize the Denominator**
**Problem:**
Rationalize the denominator.
\[
4) \frac{\sqrt{20}}{\sqrt{75}}
\]
**Options:**
A) \(\frac{2\sqrt{5}}{3}\)
B) \(\frac{2\sqrt{15}}{3}\)
C) \(\frac{2\sqrt{3}}{15}\)
D) \(\frac{2\sqrt{15}}{15}\)
**Explanation:**
When rationalizing the denominator, the goal is to eliminate any square roots present in the denominator. This can be achieved by multiplying both the numerator and the denominator by an appropriate value.
1. Consider the original expression: \(\frac{\sqrt{20}}{\sqrt{75}}\)
- Break down the square roots:
\[
\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}
\]
\[
\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}
\]
- Substitute these values back into the original expression:
\[
\frac{2\sqrt{5}}{5\sqrt{3}}
\]
2. To rationalize the denominator, multiply both the numerator and the denominator by \(\sqrt{3}\):
\[
\frac{2\sqrt{5}}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{5} \cdot \sqrt{3}}{5\sqrt{3} \cdot \sqrt{3}} = \frac{2\sqrt{15}}{15}
\]
So the rationalized form of the given expression is \(\frac{2\sqrt{15}}{15}\).
**Answer:**
D) \(\frac{2\sqrt{15}}{15}\)
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