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Chapter 8: Problem 63

What is an extraneous solution of a radical equation?

Short Answer

Expert verified
An extraneous solution is a solution which does not satisfy the original equation but is developed during the solving process, primarily when working with radical equations. To identify an extraneous solution, it's critical to plug the potential solutions back into the original equation to verify if they hold true.

Step by step solution

01

Define Extraneous Solution

An extraneous solution is a solution that is derived from the process of solving the equation, but does not actually satisfy the original equation.
02

Describe How Extraneous Solutions Occur

Extraneous solutions occur when the process of solving the equation creates additional solutions that do not satisfy the original equation. This usually happens during the process of solving radical equations, where squaring both sides is a common step.
03

Explain Checking for Extraneous Solutions

When solving a radical equation, it is important to check all potential solutions in the original equation to ensure they are not extraneous. This involves substituting the solutions back into the original equation and seeing if they hold true.

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Key Concepts

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

Radical Equation
A radical equation is an equation in which the variable is located under a radical sign, like a square root or cube root. These equations can be a bit tricky because they often involve additional steps to isolate the variable.
  • Radicals can make equations more complex, as they hide the variable within roots.
  • Consider a simple radical equation like \( \sqrt{x} = 5 \). To remove the radical, you square both sides, resulting in \( x = 25 \).
  • However, squaring both sides can often produce solutions that aren't valid for the original equation. These are called extraneous solutions, which we'll talk more about in later sections.
Understanding radical equations is crucial because they involve finding solutions that genuinely fit the equation鈥攏ot just mathematically manipulated results.
Checking Solutions
Checking solutions is a vital step when dealing with radical equations. It ensures the solutions we've found actually satisfy the original equation without introducing any errors.
  • Once you think you have found a solution, substitute it back into the original equation.
  • If the left and right sides of the equation equal, then the solution is valid.
  • If not, you have encountered an extraneous solution鈥攁 solution that doesn鈥檛 satisfy the original equation.
Always remember that checking your solutions is like a safety net. It helps you catch those trickier cases where a solution might seem right but doesn鈥檛 actually work when put back into the original context.
Solving Equations
Solving equations, especially radical ones, requires a keen understanding of algebraic principles to ensure valid results.
  • Begin by isolating the radical on one side of the equation.
  • Square both sides to eliminate the radical. Be cautious, as this step is where extraneous solutions might originate.
  • Simplify the equation and solve for the variable.
  • Finally, check your solutions to weed out any that don't satisfy the initial equation.
Taking a careful, step-by-step approach minimizes errors and helps you deal effectively with potential pitfalls. Always keep the concept of an extraneous solution in mind, as these false results often appear in equations involving radicals.

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Most popular questions from this chapter

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. By adding the exponents, I simplified \(5^{\frac{1}{2}} \cdot 5^{\frac{1}{2}}\) and obtained 25

Without using a calculator, simplify the expressions completely. $$\frac{3^{-1} \cdot 3^{\frac{1}{2}}}{3^{-\frac{3}{2}}}$$

Simplify by first writing the expression in radical form. If applicable, use a calculator to verify your answer. $$81^{-\frac{5}{4}}$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$2^{\frac{1}{2}} \cdot 2^{\frac{3}{2}}=\left(\frac{1}{4}\right)^{-1}$$

In Exercises \(53-74\), rationalize each denominator. Simplify, if possible. $$\frac{2}{\sqrt{5}-\sqrt{3}}$$
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