Find the solution(s) to each equation, if there are any. A: 2² = 9 B: VI = 3 C: VI = -3

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
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Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Equations and Solutions:**

Find the solution(s) to each equation, if there are any.

**A:** \( x^2 = 9 \)

**B:** \( \sqrt{x} = 3 \)

**C:** \( \sqrt{x} = -3 \)

**Explanation:**

- **Equation A:** The equation is a quadratic \( x^2 = 9 \). To find the solutions, take the square root of both sides. The solutions are \( x = 3 \) and \( x = -3 \).

- **Equation B:** The equation is \( \sqrt{x} = 3 \). To solve, square both sides to remove the square root, giving \( x = 9 \).

- **Equation C:** The equation is \( \sqrt{x} = -3 \). Since a square root cannot be negative in the set of real numbers, there are no solutions.
Transcribed Image Text:**Equations and Solutions:** Find the solution(s) to each equation, if there are any. **A:** \( x^2 = 9 \) **B:** \( \sqrt{x} = 3 \) **C:** \( \sqrt{x} = -3 \) **Explanation:** - **Equation A:** The equation is a quadratic \( x^2 = 9 \). To find the solutions, take the square root of both sides. The solutions are \( x = 3 \) and \( x = -3 \). - **Equation B:** The equation is \( \sqrt{x} = 3 \). To solve, square both sides to remove the square root, giving \( x = 9 \). - **Equation C:** The equation is \( \sqrt{x} = -3 \). Since a square root cannot be negative in the set of real numbers, there are no solutions.
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