$\sqrt[4]{2x} + \sqrt[4]{x+3} = 0$ Solution Explained Extraneous Solutions

In this article, we will delve into the equation $\sqrt[4]{2x} + \sqrt[4]{x+3} = 0$ and determine whether $x=1$ or $x=3$ are solutions, and if so, whether they are true or extraneous solutions. Understanding extraneous solutions is crucial in solving radical equations, as they may arise from the process of raising both sides of an equation to a power. Let's embark on this mathematical journey to unravel the mystery of this equation and the nature of its solutions.

Understanding the Problem

The given equation is $\sqrt[4]{2x} + \sqrt[4]{x+3} = 0$. This equation involves fourth roots, which means we need to be careful about the domain of the equation and the potential for extraneous solutions. Extraneous solutions are solutions that arise during the solving process but do not satisfy the original equation. These typically occur when we raise both sides of an equation to an even power, as this can introduce solutions that make the original equation undefined or untrue. To solve this equation and correctly identify true and extraneous solutions, a step-by-step approach involving algebraic manipulation and careful verification is essential.

Solving the Equation

To solve the equation $\sqrt[4]{2x} + \sqrt[4]{x+3} = 0$, we first isolate one of the radical terms. This will allow us to raise both sides of the equation to a power that eliminates the radical. By isolating one radical, we simplify the process of eliminating the radicals and move closer to finding potential solutions. This is a standard technique in solving equations involving radicals and often the first crucial step.

Isolating the Radical Term

Subtract $\sqrt[4]{x+3}$ from both sides of the equation:

$\sqrt[4]{2x} = -\sqrt[4]{x+3}$

Isolating the radical term sets the stage for the next step, which involves raising both sides of the equation to the fourth power. By doing this, we aim to eliminate the fourth root and obtain a simpler algebraic equation that can be solved using standard methods. The isolation step is a crucial setup for efficiently solving the radical equation.

Raising Both Sides to the Fourth Power

Now, raise both sides of the equation to the fourth power:

$(\sqrt[4]{2x})^4 = (-\sqrt[4]{x+3})^4$

This simplifies to:

$2x = (x+3)$

Raising both sides to the fourth power is a critical step in eliminating the radical and transforming the equation into a more manageable form. However, it's essential to remember that this step can introduce extraneous solutions, which are not actual solutions to the original equation. Therefore, the solutions obtained after this step must be checked in the original equation to confirm their validity. By carefully raising to the power, we move closer to potential solutions while remaining vigilant about potential extraneous solutions.

Solving the Linear Equation

Now we have a simple linear equation:

$2x = x + 3$

Subtract $x$ from both sides:

$x = 3$

Solving the linear equation is straightforward, providing us with a potential solution. However, as mentioned earlier, it's crucial to remember that this solution must be verified in the original equation to ensure it is not an extraneous solution. The algebraic steps leading to this point are accurate, but the final confirmation lies in the verification process. By solving the linear equation, we have identified a potential candidate for the solution, which now needs to be rigorously checked.

Checking for Extraneous Solutions

Substituting $x=3$ into the Original Equation

Substitute $x = 3$ into the original equation:

$\sqrt[4]{2(3)} + \sqrt[4]{3+3} = 0$

$\sqrt[4]{6} + \sqrt[4]{6} = 0$

This simplifies to:

$2\sqrt[4]{6} = 0$

Clearly, $2\sqrt[4]{6}$ is not equal to $0$, so $x=3$ is not a true solution.

Analyzing the Result

Since the substitution of $x=3$ into the original equation results in a false statement, $2\sqrt[4]{6} = 0$, we conclude that $x=3$ is an extraneous solution. Extraneous solutions arise due to the process of raising both sides of the equation to an even power, which can introduce solutions that do not satisfy the original equation's constraints. Identifying and excluding such solutions is crucial for solving radical equations accurately. By carefully verifying the solution, we have correctly determined that $x=3$ is not a valid solution to the original equation.

Evaluating $x=1$

Let's evaluate whether $x=1$ is a solution to the original equation. We will substitute $x=1$ into the equation and check if it satisfies the equation. This step is vital to ensure that we identify all potential solutions and correctly categorize them as true or extraneous.

Substituting $x=1$ into the Original Equation

Substitute $x = 1$ into the original equation:

$\sqrt[4]{2(1)} + \sqrt[4]{1+3} = 0$

$\sqrt[4]{2} + \sqrt[4]{4} = 0$

This simplifies to:

$\sqrt[4]{2} + \sqrt{2} = 0$

Analyzing the Result

Since $\sqrt[4]{2}$ and $\sqrt{2}$ are both positive values, their sum cannot be equal to $0$. Therefore, $x=1$ is not a true solution. By performing the substitution and analyzing the result, we can definitively conclude that $x=1$ does not satisfy the original equation. This means that $x=1$ is either an extraneous solution or not a solution at all. Identifying such non-solutions is essential in the overall process of solving equations, ensuring that only valid solutions are considered.

Conclusion

In conclusion, by solving the equation $\sqrt[4]{2x} + \sqrt[4]{x+3} = 0$ and verifying the potential solutions, we found that $x=3$ is an extraneous solution and $x=1$ is not a true solution. The correct answer is that $x=3$ is an extraneous solution. Understanding how to solve radical equations and identify extraneous solutions is a crucial skill in algebra. The process involves isolating radicals, raising both sides to appropriate powers, solving the resulting equations, and, most importantly, checking the solutions in the original equation. This meticulous approach ensures that we obtain accurate and valid solutions. The exploration of this equation highlights the importance of careful algebraic manipulation and verification in solving mathematical problems.

Final Answer

The final answer is C. $x=3$ is an extraneous solution.