Solving Sqrt(x-5) = -3 And Checking For Extraneous Solutions

Introduction

In the realm of mathematics, particularly in algebra, solving equations involving radicals is a fundamental skill. However, equations are not always straightforward, and they may present us with solutions that, while appearing correct algebraically, do not satisfy the original equation. These are known as extraneous solutions. In this comprehensive guide, we will dissect the equation $\sqrt{x-5} = -3$, systematically solve it, and rigorously check whether the solution obtained is valid or an extraneous solution. Our focus is not only on finding a numerical answer but also on understanding the underlying concepts and the importance of verifying solutions in the context of radical equations. This article serves as a valuable resource for students, educators, and anyone keen on deepening their understanding of algebraic problem-solving techniques and the subtleties involved in dealing with square roots and other radicals.

Understanding Radical Equations

Before diving into the specifics of our equation, let's first establish a solid foundation by understanding what radical equations are and the potential pitfalls they present. A radical equation is an equation that contains a radical expression, most commonly a square root. The presence of radicals introduces a critical consideration: the domain of the radical. For real numbers, the expression inside a square root (the radicand) must be non-negative. This restriction arises because the square root of a negative number is not a real number. When solving radical equations, we often perform operations that can introduce solutions that don't adhere to this domain restriction, leading to extraneous solutions. Therefore, checking our solutions against the original equation is not merely a formality but an essential step in the problem-solving process. In the following sections, we will methodically address the equation $\sqrt{x-5} = -3$, applying these principles to determine the true solution and identify any extraneous ones. We will explore the algebraic steps required to isolate the variable x, and more importantly, we will emphasize the critical process of validating the solution within the context of the original equation and the domain restrictions imposed by the square root.

Step-by-Step Solution of $\sqrt{x-5} = -3$

Now, let's embark on a step-by-step journey to solve the equation $\sqrtx-5} = -3$. This process will not only yield a potential solution but also illuminate the intricacies of dealing with radical equations. Our primary goal is to isolate the variable x, and to achieve this, we must first eliminate the square root. The standard method for doing this is to square both sides of the equation. By squaring both sides, we effectively undo the square root operation, allowing us to access the expression inside. However, this is also the step where extraneous solutions can creep in, underscoring the need for careful verification later. So, let's begin Squaring both sides of $\sqrt{x-5 = -3$ gives us $(\sqrt{x-5})^2 = (-3)^2$, which simplifies to $x - 5 = 9$. Now, we have a simple linear equation that we can solve for x. To isolate x, we add 5 to both sides of the equation: $x - 5 + 5 = 9 + 5$. This simplifies to $x = 14$. At this juncture, we have found a potential solution: x = 14. However, our work is not yet complete. We must now meticulously verify this solution to ensure it satisfies the original equation and does not violate any domain restrictions. The next section will delve into the crucial process of checking this solution and determining whether it is valid or an extraneous solution.

Checking the Solution: Validity and Extraneous Roots

With a potential solution in hand, x = 14, the critical next step is to check its validity within the original equation: $\sqrt{x-5} = -3$. This verification process is not merely a formality; it is a crucial safeguard against accepting extraneous solutions, which can arise when dealing with radical equations. To check the solution, we substitute x = 14 back into the original equation and evaluate both sides. Substituting x = 14 into the equation gives us $\sqrt{14-5} = -3$. Simplifying the expression under the square root, we get $\sqrt{9} = -3$. The square root of 9 is 3, so the equation becomes $3 = -3$. This statement is clearly false. The left side of the equation equals 3, while the right side is -3. These two values are not equal, indicating that x = 14 does not satisfy the original equation. This result leads us to a significant conclusion: x = 14 is an extraneous solution. It is a solution that arises from the algebraic steps we took to solve the equation, but it does not actually make the original equation true. The presence of this extraneous solution highlights the importance of the verification step when solving radical equations. In the next section, we will explore why this extraneous solution occurred and discuss the implications for the overall solution of the equation.

Why Extraneous Solutions Occur

Having identified x = 14 as an extraneous solution, it's crucial to understand why such solutions arise in the context of radical equations. The primary reason lies in the process of squaring both sides of the equation, which we performed to eliminate the square root. Squaring both sides can introduce extraneous solutions because it effectively ignores the sign of the expressions. To illustrate this, consider two numbers, a and b, such that a = b. Squaring both sides gives us $a^2 = b^2$. However, the converse is not necessarily true. If $a^2 = b^2$, it does not automatically imply that a = b. It could also be the case that a = -b. This is precisely what happens in our equation. When we squared both sides of $\sqrt{x-5} = -3$, we created a situation where the information about the sign was lost. The square root function, by definition, yields a non-negative result. In other words, $\sqrt{x-5}$ can never be equal to a negative number like -3. However, by squaring both sides, we transformed the equation into one that allowed for both positive and negative solutions, leading to the extraneous solution x = 14. This understanding underscores the importance of not just solving radical equations algebraically but also interpreting the solutions within the context of the original equation and the properties of the radical function. In the concluding section, we will summarize our findings and provide a definitive answer to the equation.

Conclusion: The Final Answer

In this comprehensive exploration, we embarked on a journey to solve the equation $\sqrtx-5} = -3$. Through a methodical, step-by-step approach, we arrived at a potential solution, x = 14. However, we did not stop there. We recognized the critical importance of verifying solutions in the context of radical equations, where extraneous solutions can often arise. Our verification process revealed that x = 14 is indeed an extraneous solution. When we substituted x = 14 back into the original equation, we found that it did not hold true, indicating that it is not a valid solution. Furthermore, we delved into the reasons behind the occurrence of extraneous solutions, emphasizing the role of squaring both sides of the equation in potentially introducing solutions that do not satisfy the original equation's constraints. Considering these findings, we arrive at a conclusive answer The equation $\sqrt{x-5 = -3$ has no real solutions. The apparent solution, x = 14, is an artifact of the algebraic manipulation and does not reflect a true solution to the original equation. This exploration underscores the importance of a thorough understanding of radical equations, the process of solving them, and the critical step of verifying solutions. By mastering these concepts, we can confidently navigate the world of algebra and avoid the pitfalls of extraneous solutions.