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Multiple Choice An apparent solution that does not satisfy the original equation is called a(n) ________ solution. (a) extraneous (b) radical (c) imaginary (d) conditional

Short Answer

Expert verified
The correct answer is (a) extraneous.

Step by step solution

01

Understand the Question

The question asks for the term that describes an apparent solution which does not satisfy the original equation. This is a vocabulary question related to types of solutions in mathematics.
02

Define Each Term

Review the meanings of each given option:- **Extraneous solution**: A solution obtained in the process of solving the equation but does not satisfy the original equation.- **Radical solution**: Usually refers to a solution involving a root, such as the square root.- **Imaginary solution**: A solution involving the imaginary unit, usually represented by 'i', where \(i^2 = -1\).- **Conditional solution**: A solution that holds true only under certain conditions or constraints.
03

Match the Definition

Identify which definition matches the term we are looking for. We need the term that describes a solution which does not satisfy the original equation.
04

Select the Correct Answer

The definition of an extraneous solution matches the description in the question. Therefore, the correct answer is (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Types of Solutions
When solving equations, it is crucial to understand the different types of solutions you might encounter. Here are some key types:

**Extraneous Solutions**: These are solutions that appear during the algebraic manipulation of equations but do not satisfy the original equation. For example, when squaring both sides of an equation, you might introduce solutions that are not valid for the original equation.
**Radical Solutions**: These involve roots, such as square roots or cube roots. For instance, solving \( \sqrt{x} = 3 \) means finding a number which, when squared, equals 3.
**Imaginary Solutions**: These include the imaginary unit 'i', where \(i^2 = -1\). An example is solving \(x^2 + 1 = 0\), which has solutions \(-i\) and \(i\).
**Conditional Solutions**: These are solutions that only hold true under certain conditions or constraints. They depend on specific values for variables or constraints to be accurate.

Knowing these different types can help you better understand the solutions you obtain, and check their validity.
Algebra Vocabulary
Understanding key algebraic vocabulary is essential for mastering the subject. Here are some important terms:

**Equation**: A mathematical statement where two expressions are set equal to each other, such as \(2x + 3 = 7\).
**Variable**: A symbol, usually a letter, that represents an unknown value. Examples include 'x' and 'y'.
**Coefficient**: A number that multiplies a variable in an algebraic expression. For example, in \(3x\), 3 is the coefficient.
**Constant**: A value that does not change. In the equation \(2x + 3 = 7\), 3 and 7 are constants.
**Solution**: A value that satisfies an equation. When you find that \(x = 2\) in \(2x + 3 = 7\), 2 is the solution.

By familiarizing yourself with these terms, you'll be better equipped to understand and solve algebraic problems.
Solving Equations
Solving equations is a fundamental skill in algebra. Here are the general steps to solve an equation:

**Identify the Equation**: Determine what type of equation you are dealing with (linear, quadratic, etc.). This helps determine the strategy you will use.
**Isolate the Variable**: Use inverse operations to get the variable by itself on one side of the equation. For example, in \(2x + 3 = 7\), you would subtract 3 from both sides to get \(2x = 4\), and then divide by 2 to get \(x = 2\).
**Check Your Solution**: Substitute the solution back into the original equation to ensure it satisfies the equation. For \(2x + 3 = 7\) with \(x = 2\), substituting back gives \(2(2) + 3 = 7\), confirming the solution is correct.
**Look for Extraneous Solutions**: Especially in equations involving square roots or other operations that can introduce invalid solutions, always verify that your solutions work in the original equation.

By following these steps, you can systematically approach and solve various algebraic equations effectively.

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Most popular questions from this chapter

Blending Teas The manager of a store that specializes in selling tea decides to experiment with a new blend. She will mix some Earl Grey tea that sells for $6 per pound with some Orange Pekoe tea that sells for 4 per pound to get 100 pounds of the new blend. The selling price of the new blend is to be 5.50 per pound, and there is to be no difference in revenue between selling the new blend and selling the other types. How many pounds of the Earl Grey tea and of the Orange Pekoe tea are required?

Suppose that you want to mix two coffees in order to obtain 100 pounds of a blend. If \(x\) represents the number of pounds of coffee A, which algebraic expression represents the number of pounds of coffee \(\mathrm{B}\) ? (a) \(100-x\) (b) \(x-100\) (c) \(100 x\) (d) \(100+x\)

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