Squaring both sides of an equation

Oct 13, 2025 by miodeya.com 128 views

**Which of the following is an extraneous solution of $$\sqrt{4x+41}=x+5$$?**

Understanding Extraneous Solutions

In mathematics, an extraneous solution is a solution that appears to be valid but is actually incorrect. This can occur when solving equations involving square roots, as the square root function can introduce extraneous solutions. In this article, we will explore the concept of extraneous solutions and determine which of the given options is an extraneous solution of the equation $$\sqrt{4x+41}=x+5$$.

What are Extraneous Solutions?

Extraneous solutions are solutions that are not valid or are not true solutions to an equation. They can occur when solving equations involving square roots, as the square root function can introduce extraneous solutions. Extraneous solutions can be caused by various factors, including:

Squaring both sides of an equation: When we square both sides of an equation, we can introduce extraneous solutions. This is because the square root function is not a one-to-one function, meaning that different inputs can produce the same output.
Not checking for domain restrictions: When solving equations involving square roots, we must check for domain restrictions. If the expression inside the square root is negative, the equation has no real solutions.
Not checking for extraneous solutions: After solving an equation, we must check for extraneous solutions. This involves substituting the solution back into the original equation to verify that it is true.

Solving the Equation

To solve the equation $$\sqrt{4x+41}=x+5$$, we can start by squaring both sides of the equation:

\left(\sqrt{4x+41}\right)^2 = (x+5)^2

This simplifies to:

4x+41 = x^2 + 10x + 25

Next, we can rearrange the equation to form a quadratic equation:

x^2 + 6x - 16 = 0

We can factor the quadratic equation:

(x+8)(x-2) = 0

This gives us two possible solutions:

x+8 = 0 \quad \text{or} \quad x-2 = 0

Solving for x, we get:

x = -8 \quad \text{or} \quad x = 2

Checking for Extraneous Solutions

Now that we have found two possible solutions, we must check for extraneous solutions. We can do this by substituting the solutions back into the original equation:

x = -8: $$\sqrt{4(-8)+41} = -8+5$$ $$\sqrt{-16} = -3$$ This is not true, so x = -8 is an extraneous solution.
x = 2: $$\sqrt{4(2)+41} = 2+5$$ $$\sqrt{49} = 7$$ This is true, so x = 2 is a valid solution.

Conclusion

In conclusion, the extraneous solution of the equation $$\sqrt{4x+41}=x+5$$ is x = -8. This is because x = -8 is not a valid solution to the equation, as it does not satisfy the original equation.

Frequently Asked Questions

Q: What is an extraneous solution? A: An extraneous solution is a solution that appears to be valid but is actually incorrect.
Q: How can I determine if a solution is extraneous? A: You can determine if a solution is extraneous by substituting the solution back into the original equation to verify that it is true.
Q: Why do extraneous solutions occur? A: Extraneous solutions can occur when solving equations involving square roots, as the square root function can introduce extraneous solutions.
Q: How can I avoid extraneous solutions? A: You can avoid extraneous solutions by checking for domain restrictions and not checking for extraneous solutions after solving an equation.

Additional Resources

Mathematics textbooks: For more information on solving equations involving square roots and extraneous solutions, consult a mathematics textbook.
Online resources: For additional resources and practice problems, visit online resources such as Khan Academy or Mathway.
Mathematics communities: Join online mathematics communities, such as Reddit's r/learnmath, to ask questions and get help from other mathematicians.