# Which of the Following Is an Extraneous Solution of mc015-1.jpg? X = –12 X = –3 X = 3 X = 12

Which of the Following Is an Extraneous Solution of mc015-1.jpg? X = –12 X = –3 X = 3 X = 12

In mathematics, solving equations is a fundamental concept that students encounter throughout their academic journey. While solving equations, it is crucial to identify the extraneous solutions that may arise during the process. These extraneous solutions are values that satisfy the equation but do not satisfy the original problem. In this article, we will explore the concept of extraneous solutions and determine which of the given solutions is extraneous.

Understanding Extraneous Solutions:

Before delving into the given problem, let’s first gain a clear understanding of what extraneous solutions are. Extraneous solutions occur when solving an equation leads to a value that satisfies the equation but does not satisfy the original problem. They typically arise when there are steps in the equation-solving process that involve squaring both sides or taking square roots.

To identify extraneous solutions, it is essential to verify each solution by substituting it back into the original equation. If the substituted value does not satisfy the original equation, it is considered an extraneous solution.

Determining the Extraneous Solution:

Now, let’s apply this knowledge to the given problem and determine which of the following solutions is extraneous: X = –12, X = –3, X = 3, or X = 12.

Since we do not have the original equation provided, we are unable to provide a definitive answer. However, we can walk through the process of identifying extraneous solutions using a hypothetical equation.

Let’s consider the equation: (x + 5)² = 25

To solve this equation, we take the square root of both sides:

√[(x + 5)²] = √25

Simplifying further:

(x + 5) = ±5

Now, we subtract 5 from both sides:

x = -5 ± 5

This gives us two possible solutions:

x = 0 or x = -10

To determine if either of these solutions is extraneous, we substitute them back into the original equation:

For x = 0: (0 + 5)² = 25
25 = 25

For x = -10: (-10 + 5)² = 25
25 = 25

Both solutions satisfy the original equation, indicating that neither of them is extraneous.

FAQs:

Q: How do extraneous solutions occur?
Extraneous solutions occur when solving an equation involves steps that introduce additional solutions that do not satisfy the original problem.

Q: Are extraneous solutions common?
Extraneous solutions are not uncommon. They frequently arise when taking square roots or squaring both sides of an equation.

Q: Can extraneous solutions be avoided?
While extraneous solutions cannot always be avoided, being aware of their possibility and verifying solutions by substituting them back into the original equation can help identify and eliminate them.

Q: Why are extraneous solutions important to consider?
Identifying extraneous solutions is crucial as they can lead to incorrect conclusions or solutions that do not apply to the original problem. It is essential to ensure that the solutions obtained are valid in the context of the problem being solved.

In conclusion, identifying extraneous solutions is an important aspect of solving equations. It ensures that the solutions obtained are valid and applicable to the original problem. While we were unable to determine the specific extraneous solution for the given problem of X = –12, X = –3, X = 3, and X = 12 due to the absence of the original equation, we provided a comprehensive understanding of extraneous solutions and the process involved in identifying them.