Extraneous Solution Of $\sqrt{2p+1} + 2\sqrt{p} = 1$ Explained

$$

In the realm of mathematics, particularly when dealing with radical equations, we often encounter solutions that appear valid but don't satisfy the original equation. These are known as extraneous solutions. This article aims to dissect the equation $\sqrt{2p+1} + 2\sqrt{p} = 1$, identify its extraneous solution, and provide a comprehensive understanding of why these solutions arise. We will explore the step-by-step process of solving the equation, verifying the solutions, and understanding the underlying principles that lead to extraneous results.

What are Extraneous Solutions?

Extraneous solutions are essentially 'false positives' in the solution set of an equation. They emerge when we perform operations that are not reversible, such as squaring both sides of an equation. While these operations help simplify the equation and make it solvable, they can also introduce solutions that do not hold true for the original problem. In the context of radical equations, extraneous solutions are particularly common due to the nature of square roots and other radicals. The squaring operation can turn negative values into positive ones, potentially creating solutions that don't exist in the original equation's domain. It's crucial to verify every solution obtained when solving equations involving radicals or rational expressions to ensure they are genuine and not extraneous. In simpler terms, an extraneous solution is a value that you get when solving an equation that doesn't actually work when you plug it back into the original equation. This often happens when dealing with square roots or other radicals because squaring both sides of an equation can introduce solutions that weren't there initially. It's like adding extra possibilities that don't really fit the original problem. Therefore, always check your answers to make sure they're valid. This concept is a cornerstone of advanced algebra and calculus, where understanding the nature of solutions is critical for accuracy and problem-solving.

Solving the Equation $\sqrt{2p+1} + 2\sqrt{p} = 1$

To find the extraneous solution of the equation $\sqrt{2p+1} + 2\sqrt{p} = 1$, we need to follow a meticulous algebraic process. Our primary goal is to isolate the variable p. This involves several key steps, each designed to simplify the equation while being mindful of potential pitfalls that lead to extraneous solutions. We begin by isolating one of the radical terms, which is a common strategy when dealing with equations involving square roots. This sets the stage for squaring both sides, a pivotal step that eliminates the radical but also introduces the possibility of extraneous solutions. Following the algebraic manipulations, we arrive at potential solutions, which must then be rigorously checked against the original equation. This verification process is not merely a formality; it's the safeguard against accepting extraneous solutions as valid answers. By understanding each step and its implications, we gain a deeper appreciation for the nuances of solving radical equations and the importance of solution verification.

Step 1: Isolate One Radical Term

In the equation $\sqrt{2p+1} + 2\sqrt{p} = 1$, let's isolate the term $\sqrt{2p+1}$ by subtracting $2\sqrt{p}$ from both sides. This gives us:

$\sqrt{2p+1} = 1 - 2\sqrt{p}$

Isolating the radical term is a crucial first step because it sets the stage for eliminating the square root. By having the radical term alone on one side of the equation, we can effectively use the squaring operation to remove the square root symbol. This step is not just about algebraic manipulation; it's about transforming the equation into a form that's easier to work with. However, it's important to recognize that this step, while necessary, also paves the way for the potential introduction of extraneous solutions. The act of squaring both sides can create solutions that satisfy the transformed equation but not the original. Therefore, we must proceed with caution and diligence, keeping in mind the need for verification later on. This careful approach is what distinguishes a thorough problem solver from someone who simply goes through the motions.

Step 2: Square Both Sides

To eliminate the square root, we square both sides of the equation:

$(\sqrt{2p+1})^2 = (1 - 2\sqrt{p})^2$

This simplifies to:

$2p + 1 = 1 - 4\sqrt{p} + 4p$

Squaring both sides is a pivotal step in solving radical equations. It effectively removes the square root, allowing us to work with a more conventional algebraic expression. However, this step is also the primary source of extraneous solutions. The act of squaring can introduce solutions that satisfy the squared equation but not the original radical equation. This is because squaring can mask the sign of a number; for example, both 2 and -2, when squared, result in 4. Thus, the squared equation might accept a negative value that wouldn't be valid under the square root in the original equation. The expansion of $(1 - 2\sqrt{p})^2$ to $1 - 4\sqrt{p} + 4p$ demonstrates the transformation we achieve through squaring. It's a critical maneuver that advances us towards isolating p, but it also underscores the need for meticulous verification later. This step highlights the delicate balance between simplification and the preservation of solution validity.

Step 3: Simplify and Isolate the Remaining Radical

Now, simplify the equation and isolate the remaining radical term:

$2p + 1 = 1 - 4\sqrt{p} + 4p$

Subtract $2p + 1$ from both sides:

$0 = -4\sqrt{p} + 2p$

Rearrange the equation:

$4\sqrt{p} = 2p$

Simplifying and isolating the remaining radical is a strategic move to further disentangle the equation. After the initial squaring, we still have a radical term ($4\sqrt{p}$), which prevents us from directly solving for p. By gathering like terms and rearranging the equation, we set ourselves up for another round of squaring, if necessary. The simplification process involves basic algebraic operations, but its impact on the overall solution cannot be overstated. Each manipulation brings us closer to a form where p can be isolated and its value(s) determined. The rearranged equation, $4\sqrt{p} = 2p$, is a crucial intermediate step. It's simpler than the previous form, but it still retains the essence of the original equation, albeit in a transformed state. This step demonstrates the iterative nature of solving complex equations, where simplification is a continuous process aimed at revealing the underlying structure of the problem.

Step 4: Square Both Sides Again

Square both sides again to eliminate the remaining square root:

$(4\sqrt{p})^2 = (2p)^2$

$16p = 4p^2$

Squaring both sides again is a decisive step in eliminating the remaining radical, but it also doubles down on the risk of introducing extraneous solutions. Each time we square an equation, we potentially broaden the solution set beyond the valid solutions of the original equation. This is because the squaring operation is not one-to-one; it collapses two values (a number and its negative counterpart) into a single result. In this case, squaring $4\sqrt{p}$ and $2p$ results in $16p = 4p^2$, a quadratic equation that is much easier to solve than the original radical equation. However, the simplicity comes at a cost. We must be exceptionally vigilant in verifying the solutions we obtain from this quadratic equation against the original radical equation. The equation $16p = 4p^2$ represents a significant milestone in our solution process, but it also serves as a stark reminder of the importance of the final verification step. It encapsulates the core challenge of solving radical equations: the need to balance simplification with the preservation of solution validity.

Step 5: Solve for p

Divide both sides by 4:

$4p = p^2$

Rearrange into a quadratic equation:

$p^2 - 4p = 0$

Factor out p:

$p(p - 4) = 0$

So, $p = 0$ or $p = 4$.

Solving for p is the culmination of our algebraic manipulations. By transforming the original radical equation into a quadratic equation, we've made it possible to find potential solutions. The factorization of $p^2 - 4p = 0$ into $p(p - 4) = 0$ is a classic technique for solving quadratic equations, revealing two possible values for p: 0 and 4. These values are our candidates for solutions, but they are not guaranteed to be valid. They represent the points where the quadratic equation equals zero, but they might not satisfy the constraints imposed by the original radical equation. The act of factoring and finding the roots is a powerful algebraic tool, but it must be used judiciously in the context of radical equations. Each potential solution must be regarded with skepticism until it has been rigorously verified. This step underscores the importance of not just finding solutions, but also understanding the context in which they are found.

Verifying the Solutions

Verification is the cornerstone of solving radical equations. It is the process of substituting the potential solutions back into the original equation to confirm their validity. This step is not a mere formality; it is a critical safeguard against extraneous solutions, which can arise due to non-reversible operations like squaring. Each potential solution must be scrutinized to ensure it satisfies the original equation's constraints. The verification process highlights the importance of understanding the domain of the equation and the implications of the radical terms. It transforms the act of solving an equation from a purely algebraic exercise into a process of logical deduction, where each solution is tested against the fundamental conditions of the problem. This step underscores the need for precision and attention to detail in mathematical problem-solving.

Check $p = 0$

Substitute $p = 0$ into the original equation:

$\sqrt{2(0) + 1} + 2\sqrt{0} = 1$

$\sqrt{1} + 0 = 1$

$1 = 1$

So, $p = 0$ is a valid solution.

Checking $p = 0$ is a crucial step in the verification process. Substituting this value into the original equation, $\sqrt{2p+1} + 2\sqrt{p} = 1$, allows us to determine whether it satisfies the equation's constraints. The substitution yields $\sqrt{2(0) + 1} + 2\sqrt{0} = 1$, which simplifies to $\sqrt{1} + 0 = 1$, and further to $1 = 1$. This equality confirms that $p = 0$ is indeed a valid solution. The process of verifying a solution is not just about plugging in a value; it's about ensuring that the entire equation holds true. Each term must be evaluated correctly, and the resulting equality must be valid. This step reinforces the importance of careful arithmetic and algebraic manipulation. It also highlights the fundamental principle that a solution must satisfy the original problem statement. The successful verification of $p = 0$ marks a significant milestone in our solution process, but it does not negate the need to verify other potential solutions.

Check $p = 4$

Substitute $p = 4$ into the original equation:

$\sqrt{2(4) + 1} + 2\sqrt{4} = 1$

$\sqrt{9} + 2(2) = 1$

$3 + 4 = 1$

$7 = 1$

This is not true, so $p = 4$ is an extraneous solution.

Checking $p = 4$ is the final step in our verification process, and it's here that we identify the extraneous solution. Substituting $p = 4$ into the original equation, $\sqrt{2p+1} + 2\sqrt{p} = 1$, yields $\sqrt{2(4) + 1} + 2\sqrt{4} = 1$. This simplifies to $\sqrt{9} + 2(2) = 1$, and further to $3 + 4 = 1$, which leads to the false statement $7 = 1$. This discrepancy clearly indicates that $p = 4$ does not satisfy the original equation and is, therefore, an extraneous solution. This process underscores the critical importance of verification in solving radical equations. It demonstrates that not all solutions obtained through algebraic manipulation are valid. The extraneous solution arises because the squaring operation, while necessary for solving the equation, introduces potential solutions that do not adhere to the original equation's constraints. The identification of $p = 4$ as an extraneous solution reinforces the need for meticulousness and attention to detail in mathematical problem-solving. It also highlights the nuanced nature of solutions in the context of radical equations.

Identifying the Extraneous Solution

After verifying both potential solutions, we find that $p = 0$ is a valid solution, but $p = 4$ does not satisfy the original equation. Therefore, $p = 4$ is the extraneous solution. This outcome underscores the importance of the verification step in solving radical equations. Extraneous solutions can arise due to the nature of the operations performed, such as squaring both sides, which can introduce solutions that do not exist in the original equation's domain. The ability to identify and discard these extraneous solutions is a critical skill in algebra and beyond. It reflects a deeper understanding of the problem-solving process, where solutions are not merely computed but also validated against the initial conditions. The extraneous solution $p = 4$ serves as a tangible reminder of the potential pitfalls in algebraic manipulations and the necessity of rigorous verification.

Why do Extraneous Solutions Occur?

Extraneous solutions occur primarily because of the non-reversible nature of certain algebraic operations, particularly squaring. When we square both sides of an equation, we are essentially saying that if A = B, then A² = B². While this is true, the converse is not necessarily true. That is, if A² = B², it does not automatically follow that A = B; it could also be the case that A = -B. This is because the squaring operation eliminates the sign of a number, making both positive and negative values result in the same square. In the context of radical equations, this can lead to extraneous solutions. For example, consider the simple equation \sqrt{x} = -2. Squaring both sides gives x = 4, but if we substitute x = 4 back into the original equation, we get \sqrt{4} = -2, which simplifies to 2 = -2, a clear contradiction. The extraneous solution x = 4 arose because the squaring operation masked the fact that the square root of a number is always non-negative. Understanding the origins of extraneous solutions is crucial for developing a robust problem-solving approach. It highlights the importance of considering the domain of the equation and the potential impact of non-reversible operations. It also reinforces the need for verification as a standard practice in solving equations.

Conclusion

In summary, the extraneous solution to the equation $\sqrt{2p+1} + 2\sqrt{p} = 1$ is $p = 4$. This solution arises due to the squaring operations performed to eliminate the square roots, which introduce potential solutions that do not satisfy the original equation. This exercise underscores the critical importance of verifying all solutions when dealing with radical equations to ensure their validity. The process of solving and verifying radical equations is a cornerstone of algebraic problem-solving. It requires a combination of algebraic manipulation skills, logical reasoning, and a deep understanding of the potential pitfalls that can lead to extraneous solutions. The extraneous solution $p = 4$ serves as a valuable lesson in the importance of rigor and precision in mathematical problem-solving. It also highlights the nuanced nature of solutions and the need to go beyond mere computation to ensure their validity. By mastering the techniques for solving radical equations and understanding the concept of extraneous solutions, students can develop a more robust and confident approach to algebraic problem-solving.