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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![### Solving Equations
#### Problem Statement:
Solve the following equation:
\[ x = \sqrt{91 - x + 1} \]
In this problem, we need to find the value of \( x \) that satisfies the given equation. Let's break down the steps to solve this equation.
1. Start with the given equation:
\[ x = \sqrt{91 - x + 1} \]
2. Simplify the expression inside the square root:
\[ x = \sqrt{92 - x} \]
3. Square both sides of the equation to eliminate the square root:
\[ x^2 = 92 - x \]
4. Move all terms to one side to set the equation to zero:
\[ x^2 + x - 92 = 0 \]
5. Now, solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -92 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -92}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 368}}{2} \]
\[ x = \frac{-1 \pm \sqrt{369}}{2} \]
6. Calculate the solutions:
\[ x = \frac{-1 + \sqrt{369}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{369}}{2} \]
Since \( \sqrt{369} \) is approximately 19.2:
\[ x = \frac{-1 + 19.2}{2} \approx 9.1 \]
\[ x = \frac{-1 - 19.2}{2} \approx -10.1 \]
7. Check the solutions in the original equation:
For \( x \approx 9.1 \):
\[ 9.1 = \sqrt{91 - 9.1 + 1} \]
\[ 9.1 = \sqrt{82.9} \]
Since \( \sqrt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d195721-de89-428b-98bb-2769b26cd5c0%2F5287e46e-548a-41ec-817d-ca42448c2458%2Feoakszb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Equations
#### Problem Statement:
Solve the following equation:
\[ x = \sqrt{91 - x + 1} \]
In this problem, we need to find the value of \( x \) that satisfies the given equation. Let's break down the steps to solve this equation.
1. Start with the given equation:
\[ x = \sqrt{91 - x + 1} \]
2. Simplify the expression inside the square root:
\[ x = \sqrt{92 - x} \]
3. Square both sides of the equation to eliminate the square root:
\[ x^2 = 92 - x \]
4. Move all terms to one side to set the equation to zero:
\[ x^2 + x - 92 = 0 \]
5. Now, solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -92 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -92}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 368}}{2} \]
\[ x = \frac{-1 \pm \sqrt{369}}{2} \]
6. Calculate the solutions:
\[ x = \frac{-1 + \sqrt{369}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{369}}{2} \]
Since \( \sqrt{369} \) is approximately 19.2:
\[ x = \frac{-1 + 19.2}{2} \approx 9.1 \]
\[ x = \frac{-1 - 19.2}{2} \approx -10.1 \]
7. Check the solutions in the original equation:
For \( x \approx 9.1 \):
\[ 9.1 = \sqrt{91 - 9.1 + 1} \]
\[ 9.1 = \sqrt{82.9} \]
Since \( \sqrt
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