Understanding Extraneous Solutions

Oct 13, 2025 by orovajewels.com 119 views

Introduction

In mathematics, solving equations is a crucial skill that helps us find the value of unknown variables. However, sometimes we may encounter extraneous solutions, which are solutions that do not satisfy the original equation. In this article, we will learn how to solve and check for extraneous solutions using a step-by-step approach.

Understanding Extraneous Solutions

Extraneous solutions occur when we solve an equation and obtain a value that does not satisfy the original equation. This can happen when we square both sides of an equation, which can introduce extraneous solutions. For example, consider the equation $\sqrt{x} = 3$ . If we square both sides, we get $x = 9$ . However, if we plug $x = 9$ back into the original equation, we get $\sqrt{9} = 3$ , which is true. But what if we had obtained $x = -9$ by squaring both sides? If we plug $x = -9$ back into the original equation, we get $\sqrt{-9} = 3$ , which is not true. Therefore, $x = -9$ is an extraneous solution.

Step-by-Step Approach to Solve and Check for Extraneous Solutions

To solve and check for extraneous solutions, we will follow a step-by-step approach:

Step 1: Write Down the Original Equation

The first step is to write down the original equation. In this case, the original equation is $\sqrt{5x - 4} - 6 = 0$ .

Step 2: Add 6 to Both Sides

To isolate the square root term, we add 6 to both sides of the equation. This gives us $\sqrt{5x - 4} = 6$ .

Step 3: Square Both Sides

To eliminate the square root term, we square both sides of the equation. This gives us $5x - 4 = 36$ .

Step 4: Add 4 to Both Sides

To isolate the term with the variable, we add 4 to both sides of the equation. This gives us $5x = 40$ .

Step 5: Divide Both Sides by 5

To solve for the variable, we divide both sides of the equation by 5. This gives us $x = 8$ .

Step 6: Check the Solution

To check if the solution is valid, we plug $x = 8$ back into the original equation. This gives us $\sqrt{5(8) - 4} - 6 = \sqrt{40 - 4} - 6 = \sqrt{36} - 6 = 6 - 6 = 0$ . Since the solution satisfies the original equation, we can conclude that $x = 8$ is a valid solution.

Step 7: Check for Extraneous Solutions

To check for extraneous solutions, we need to plug $x = 8$ back into the original equation and check if it satisfies the equation. If it does, then $x = 8$ is a valid solution. If it does not, then $x = 8$ is an extraneous solution.

Conclusion

In this article, we learned how to solve and check for extraneous solutions using a step-by-step approach. We started with the original equation $\sqrt{5x - 4} - 6 = 0$ and followed the steps to isolate the square root term, eliminate the square root term, and solve for the variable. We then checked the solution by plugging it back into the original equation and verified that it satisfies the equation. Finally, we checked for extraneous solutions by plugging the solution back into the original equation and verifying that it does not satisfy the equation. By following these steps, we can ensure that our solutions are valid and not extraneous.

Example 1: Solve and Check for Extraneous Solutions

Consider the equation $\sqrt{x + 2} = 5$ . To solve and check for extraneous solutions, we follow the steps:

Step 1: Write Down the Original Equation

The original equation is $\sqrt{x + 2} = 5$ .

Step 2: Square Both Sides

To eliminate the square root term, we square both sides of the equation. This gives us $x + 2 = 25$ .

Step 3: Subtract 2 from Both Sides

To isolate the term with the variable, we subtract 2 from both sides of the equation. This gives us $x = 23$ .

Step 4: Check the Solution

To check if the solution is valid, we plug $x = 23$ back into the original equation. This gives us $\sqrt{23 + 2} = \sqrt{25} = 5$ . Since the solution satisfies the original equation, we can conclude that $x = 23$ is a valid solution.

Step 5: Check for Extraneous Solutions

To check for extraneous solutions, we need to plug $x = 23$ back into the original equation and check if it satisfies the equation. If it does, then $x = 23$ is a valid solution. If it does not, then $x = 23$ is an extraneous solution.

Example 2: Solve and Check for Extraneous Solutions

Consider the equation $\sqrt{2x - 3} = 7$ . To solve and check for extraneous solutions, we follow the steps:

Step 1: Write Down the Original Equation

The original equation is $\sqrt{2x - 3} = 7$ .

Step 2: Square Both Sides

To eliminate the square root term, we square both sides of the equation. This gives us $2x - 3 = 49$ .

Step 3: Add 3 to Both Sides

To isolate the term with the variable, we add 3 to both sides of the equation. This gives us $2x = 52$ .

Step 4: Divide Both Sides by 2

To solve for the variable, we divide both sides of the equation by 2. This gives us $x = 26$ .

Step 5: Check the Solution

To check if the solution is valid, we plug $x = 26$ back into the original equation. This gives us $\sqrt{2(26) - 3} = \sqrt{52 - 3} = \sqrt{49} = 7$ . Since the solution satisfies the original equation, we can conclude that $x = 26$ is a valid solution.

Step 6: Check for Extraneous Solutions

To check for extraneous solutions, we need to plug $x = 26$ back into the original equation and check if it satisfies the equation. If it does, then $x = 26$ is a valid solution. If it does not, then $x = 26$ is an extraneous solution.

Conclusion

$In this article, we learned how to solve and check for extraneous solutions using a step-by-step approach. We started with the original equation $\sqrt{5x - 4} - 6 = 0$ and followed the steps to isolate the square root term, eliminate the square root term, and solve for the variable. We then checked the solution by plugging it back into the original equation and verified that it satisfies the equation. Finally, we checked for extraneous solutions by plugging the solution back into the original equation and verifying that it does not satisfy the equation. By following these steps, we can ensure that our solutions are valid and not extraneous.<br/> **Solve and Check for Extraneous Solutions: Q&A**$

Q: What is an extraneous solution?

A: An extraneous solution is a solution that does not satisfy the original equation. It is a solution that is obtained by solving the equation, but it does not meet the conditions of the original equation.

Q: Why do extraneous solutions occur?

A: Extraneous solutions occur when we square both sides of an equation, which can introduce extraneous solutions. This is because squaring both sides of an equation can create new solutions that do not satisfy the original equation.

Q: How do I check for extraneous solutions?

A: To check for extraneous solutions, you need to plug the solution back into the original equation and check if it satisfies the equation. If it does, then the solution is valid. If it does not, then the solution is an extraneous solution.

Q: What are some common mistakes to avoid when solving equations with square roots?

A: Some common mistakes to avoid when solving equations with square roots include:

Squaring both sides of the equation without checking for extraneous solutions
Not checking if the solution satisfies the original equation
Not considering the domain of the square root function
Not simplifying the equation before solving it

Q: How do I simplify an equation with a square root?

A: To simplify an equation with a square root, you can try the following:

Factor the expression inside the square root
Simplify the expression inside the square root
Use the properties of square roots to simplify the equation
Check if the equation can be simplified further

Q: What are some common types of equations that involve square roots?

A: Some common types of equations that involve square roots include:

Equations with a single square root term
Equations with multiple square root terms
Equations with square roots and fractions
Equations with square roots and decimals

Q: How do I solve an equation with a square root and a fraction?

A: To solve an equation with a square root and a fraction, you can try the following:

Simplify the fraction
Use the properties of square roots to simplify the equation
Check if the equation can be simplified further
Solve for the variable

Q: What are some common mistakes to avoid when solving equations with square roots and fractions?

A: Some common mistakes to avoid when solving equations with square roots and fractions include:

Not simplifying the fraction before solving the equation
Not considering the domain of the square root function
Not checking if the solution satisfies the original equation
Not simplifying the equation before solving it

Q: How do I check if a solution is valid?

A: To check if a solution is valid, you need to plug the solution back into the original equation and check if it satisfies the equation. If it does, then the solution is valid. If it does not, then the solution is an extraneous solution.

Q: What are some common types of equations that involve square roots and decimals?

A: Some common types of equations that involve square roots and decimals include:

Equations with a single square root term and a decimal
Equations with multiple square root terms and decimals
Equations with square roots and fractions and decimals
Equations with square roots and decimals and variables

Q: How do I solve an equation with a square root and a decimal?

A: To solve an equation with a square root and a decimal, you can try the following:

Simplify the decimal
Use the properties of square roots to simplify the equation
Check if the equation can be simplified further
Solve for the variable

Q: What are some common mistakes to avoid when solving equations with square roots and decimals?

A: Some common mistakes to avoid when solving equations with square roots and decimals include:

Not simplifying the decimal before solving the equation
Not considering the domain of the square root function
Not checking if the solution satisfies the original equation
Not simplifying the equation before solving it

Conclusion

In this article, we have discussed how to solve and check for extraneous solutions using a step-by-step approach. We have also covered some common types of equations that involve square roots and decimals, and provided tips and tricks for solving these types of equations. By following these steps and avoiding common mistakes, you can ensure that your solutions are valid and not extraneous.