Solving The Radical Equation Sqrt(2x-5) = X-4

In this article, we delve into solving the algebraic equation $\sqrt{2x-5} = x-4$. This type of equation involves a square root, making it a radical equation. Solving radical equations requires a specific approach to eliminate the radical and find the possible solutions for x. However, it's crucial to remember that not all solutions obtained algebraically are valid. We must check our solutions to ensure they do not lead to extraneous results, which arise from the process of squaring both sides of the equation. This article aims to provide a step-by-step guide on how to solve the equation, verify the solutions, and understand the underlying concepts. By the end of this discussion, you should have a clear understanding of how to approach similar radical equations and ensure the accuracy of your solutions. The methodology we will employ involves isolating the square root term, squaring both sides to eliminate the radical, solving the resulting quadratic equation, and finally, verifying the solutions to discard any extraneous roots. This comprehensive approach is essential for mastering the technique of solving radical equations. Understanding each step and its significance is key to successfully tackling these types of mathematical problems. Mastering these techniques is fundamental for further studies in algebra and calculus, where radical equations frequently appear. This guide will not only provide the solution but also emphasize the importance of verifying solutions in the context of radical equations.

1. Isolate the Square Root

The given equation is $\sqrt{2x-5} = x-4$. In this case, the square root term is already isolated on the left side of the equation, which simplifies our initial steps. Isolating the radical is a crucial first step in solving radical equations. This ensures that when we square both sides, we effectively eliminate the square root. If the equation had additional terms outside the square root, we would first need to move those terms to the other side. However, in this specific equation, we can directly proceed to the next step because the square root term, $\sqrt{2x-5}$, is the sole term on the left side. This isolation is essential because it sets the stage for squaring both sides, which will eliminate the radical. The absence of additional terms simplifies the algebraic manipulation and reduces the chances of introducing extraneous solutions during the squaring process. Therefore, ensuring the square root is isolated is a fundamental aspect of solving radical equations. This initial isolation sets the stage for the subsequent steps and ensures the accuracy of the solution process. By having the radical term by itself, we are better positioned to apply the inverse operation (squaring) to both sides and simplify the equation. This strategic approach is a hallmark of solving radical equations and highlights the importance of algebraic manipulation in simplifying complex expressions. The emphasis on isolation as the first step underscores the methodical approach required for solving such problems.

2. Square Both Sides

To eliminate the square root, we square both sides of the equation. This gives us $(\sqrt{2x-5})^2 = (x-4)^2$. Squaring the left side removes the square root, resulting in $2x-5$. Squaring the right side involves expanding $(x-4)^2$, which can be done using the formula $(a-b)^2 = a^2 - 2ab + b^2$. Thus, $(x-4)^2 = x^2 - 8x + 16$. So, our equation becomes $2x-5 = x^2 - 8x + 16$. Squaring both sides is a critical step in solving radical equations, as it transforms the equation into a more manageable form, often a polynomial equation. This step allows us to apply standard algebraic techniques to solve for the variable x. However, it is also the step that can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original radical equation. Therefore, it is crucial to verify any solutions obtained after squaring both sides. The process of squaring eliminates the radical, but it also eliminates the information about the sign of the expression under the radical. This is why extraneous solutions can arise. The transformation from a radical equation to a polynomial equation simplifies the solving process, but it also adds the responsibility of checking the validity of the solutions. The act of squaring both sides is a double-edged sword – it simplifies the equation but necessitates careful verification of the solutions. This step is fundamental to the method of solving radical equations and highlights the importance of understanding the potential consequences of algebraic manipulations.

3. Simplify and Rearrange

Now, we rearrange the equation to form a quadratic equation. Starting with $2x-5 = x^2 - 8x + 16$, we want to set one side to zero. Subtracting $2x$ from both sides gives $-5 = x^2 - 10x + 16$. Then, adding 5 to both sides results in $0 = x^2 - 10x + 21$. This quadratic equation is now in the standard form $ax^2 + bx + c = 0$, where $a=1$, $b=-10$, and $c=21$. Rearranging the equation into the standard quadratic form is essential for applying methods such as factoring or the quadratic formula. This step allows us to leverage the well-established techniques for solving quadratic equations. The goal is to bring all terms to one side, leaving zero on the other side, which provides the structure needed to find the roots of the equation. The process involves algebraic manipulations such as addition and subtraction to maintain the equality while transforming the equation into its standard form. This rearrangement is not only a procedural step but also a strategic one, as it sets the stage for the subsequent solution process. The act of rearranging terms underscores the importance of algebraic fluency in solving equations. By manipulating the equation into a familiar form, we can apply standard methods and increase the likelihood of finding the correct solutions. The significance of this step lies in its role as a bridge between the radical equation and the quadratic equation, highlighting the interconnectedness of different algebraic concepts.

4. Solve the Quadratic Equation

We have the quadratic equation $x^2 - 10x + 21 = 0$. To solve this, we can try factoring. We look for two numbers that multiply to 21 and add to -10. These numbers are -3 and -7. Therefore, we can factor the equation as $(x-3)(x-7) = 0$. Setting each factor equal to zero gives us two possible solutions: $x-3 = 0$ which implies $x=3$, and $x-7 = 0$ which implies $x=7$. Thus, we have two potential solutions, $x=3$ and $x=7$. Factoring is a common and efficient method for solving quadratic equations, especially when the coefficients are integers and the roots are rational. This technique relies on decomposing the quadratic expression into the product of two linear factors. The ability to quickly and accurately factor a quadratic equation is a valuable skill in algebra. When factoring is not straightforward, we can also use the quadratic formula to find the solutions. However, in this case, factoring provides a direct and concise approach. The solutions obtained from factoring represent the values of x that make the quadratic expression equal to zero. These potential solutions are critical points that must be further evaluated in the context of the original radical equation. The act of factoring not only provides the roots but also enhances our understanding of the structure of the quadratic equation. It highlights the relationship between the coefficients and the solutions, which is a fundamental concept in algebra.

5. Check for Extraneous Solutions

It is essential to check our solutions in the original equation to ensure they are valid. This is because squaring both sides can introduce extraneous solutions. Let's check $x=3$: $\sqrt{2(3)-5} = 3-4$ which simplifies to $\sqrt{6-5} = -1$, so $\sqrt{1} = -1$, which gives $1 = -1$. This is false, so $x=3$ is an extraneous solution. Now let's check $x=7$: $\sqrt{2(7)-5} = 7-4$ which simplifies to $\sqrt{14-5} = 3$, so $\sqrt{9} = 3$, which gives $3 = 3$. This is true, so $x=7$ is a valid solution. Checking for extraneous solutions is a critical step in solving radical equations. This step ensures that the solutions we obtained after squaring both sides actually satisfy the original equation. Extraneous solutions arise because squaring both sides can introduce new solutions that were not present in the original equation. The process of checking involves substituting each potential solution back into the original equation and verifying whether the equation holds true. If a solution does not satisfy the original equation, it is deemed extraneous and must be discarded. This verification process is not merely a procedural formality but a fundamental aspect of solving radical equations. It highlights the importance of understanding the potential impact of algebraic manipulations on the solution set. The act of checking underscores the need for mathematical rigor and attention to detail in solving equations. It ensures that we only accept solutions that are consistent with the original problem.

Therefore, the only valid solution to the equation $\sqrt{2x-5} = x-4$ is $x=7$. We found $x=3$ as a potential solution, but it turned out to be extraneous. This example illustrates the importance of verifying solutions when dealing with radical equations. The final solution is obtained by systematically following the steps of isolating the radical, squaring both sides, solving the resulting equation, and crucially, checking for extraneous solutions. This process ensures the accuracy and validity of the solution. Solving radical equations requires a thorough understanding of algebraic manipulations and a keen awareness of the potential for extraneous solutions. The emphasis on verification is a hallmark of mathematical problem-solving, highlighting the need for rigor and attention to detail. The ability to solve radical equations is a fundamental skill in algebra and is essential for further studies in mathematics. This example provides a comprehensive illustration of the techniques and considerations involved in solving such equations. The methodical approach ensures that the correct solution is identified, and any extraneous solutions are discarded. The importance of each step, from isolating the radical to checking the solutions, is underscored throughout the process, reinforcing the need for a comprehensive understanding of the underlying concepts.