Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Problem:**
Rationalize the denominator and simplify:
\[
\frac{4 + 3\sqrt{x}}{8 + \sqrt{x}}
\]
**Solution Explanation:**
To rationalize the denominator of the fraction, multiply both the numerator and the denominator by the conjugate of the denominator, which is \(8 - \sqrt{x}\).
**Step 1:** Multiply the numerator and denominator by the conjugate.
\[
\frac{(4 + 3\sqrt{x})(8 - \sqrt{x})}{(8 + \sqrt{x})(8 - \sqrt{x})}
\]
**Step 2:** Apply the difference of squares formula to the denominator.
The formula for the difference of squares is:
\[
(a + b)(a - b) = a^2 - b^2
\]
Here, \(a = 8\) and \(b = \sqrt{x}\),
\[
(8)^2 - (\sqrt{x})^2 = 64 - x
\]
**Step 3:** Expand the numerator.
Use the distributive property to expand \( (4 + 3\sqrt{x})(8 - \sqrt{x}) \).
\[
= 4 \cdot 8 - 4 \cdot \sqrt{x} + 3\sqrt{x} \cdot 8 - 3\sqrt{x} \cdot \sqrt{x}
\]
Simplify each term:
\[
= 32 - 4\sqrt{x} + 24\sqrt{x} - 3x
\]
Combine like terms:
\[
= 32 + 20\sqrt{x} - 3x
\]
**Final Result:**
After rationalizing and simplifying, the expression becomes:
\[
\frac{32 + 20\sqrt{x} - 3x}{64 - x}
\]
This simplified form can be further evaluated or used as needed in calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd044583c-5a7a-479c-b47f-e9f2bc60a85c%2F0f8e6480-ea31-424a-bc71-92d3c0c0c997%2Frjd2bkb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Rationalize the denominator and simplify:
\[
\frac{4 + 3\sqrt{x}}{8 + \sqrt{x}}
\]
**Solution Explanation:**
To rationalize the denominator of the fraction, multiply both the numerator and the denominator by the conjugate of the denominator, which is \(8 - \sqrt{x}\).
**Step 1:** Multiply the numerator and denominator by the conjugate.
\[
\frac{(4 + 3\sqrt{x})(8 - \sqrt{x})}{(8 + \sqrt{x})(8 - \sqrt{x})}
\]
**Step 2:** Apply the difference of squares formula to the denominator.
The formula for the difference of squares is:
\[
(a + b)(a - b) = a^2 - b^2
\]
Here, \(a = 8\) and \(b = \sqrt{x}\),
\[
(8)^2 - (\sqrt{x})^2 = 64 - x
\]
**Step 3:** Expand the numerator.
Use the distributive property to expand \( (4 + 3\sqrt{x})(8 - \sqrt{x}) \).
\[
= 4 \cdot 8 - 4 \cdot \sqrt{x} + 3\sqrt{x} \cdot 8 - 3\sqrt{x} \cdot \sqrt{x}
\]
Simplify each term:
\[
= 32 - 4\sqrt{x} + 24\sqrt{x} - 3x
\]
Combine like terms:
\[
= 32 + 20\sqrt{x} - 3x
\]
**Final Result:**
After rationalizing and simplifying, the expression becomes:
\[
\frac{32 + 20\sqrt{x} - 3x}{64 - x}
\]
This simplified form can be further evaluated or used as needed in calculations.
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