Solving The Equation √(5x + 11) = X + 1 A Step-by-Step Guide

When dealing with radical equations like √(5x + 11) = x + 1, the primary goal is to isolate the variable x. Radical equations often require a methodical approach to ensure accurate solutions. This involves understanding the properties of radicals and applying algebraic techniques effectively. In this comprehensive guide, we will walk through the step-by-step process of solving this particular equation, highlighting key concepts and potential pitfalls along the way. Understanding these steps is essential for tackling more complex radical equations. We’ll cover squaring both sides, simplifying the resulting quadratic equation, and crucially, checking for extraneous solutions. This last step is a vital part of solving radical equations, as the squaring process can sometimes introduce solutions that don't actually satisfy the original equation. By mastering this process, you'll be well-equipped to handle a variety of similar problems and gain a deeper understanding of algebraic manipulations. The importance of checking solutions cannot be overstated, as it ensures the validity of your answers and reinforces your grasp of the underlying mathematical principles. So, let's delve into the specifics of solving √(5x + 11) = x + 1 and develop a robust strategy for handling radical equations.

Our journey to solve the equation √(5x + 11) = x + 1 begins with isolating the radical term. Currently, the square root term, √(5x + 11), is already isolated on the left side of the equation, which simplifies our initial step. The next crucial step involves eliminating the square root. To do this, we square both sides of the equation. Squaring both sides is a fundamental algebraic operation, but it's essential to remember that this can sometimes introduce extraneous solutions. This is why checking our solutions later is a critical part of the process. When we square the left side, (√(5x + 11))², the square root is effectively cancelled out, leaving us with 5x + 11. On the right side, (x + 1)², we need to expand the binomial. Expanding (x + 1)² gives us (x + 1)(x + 1), which equals x² + 2x + 1. Now our equation looks like this: 5x + 11 = x² + 2x + 1. This transformation is a key step in converting the radical equation into a more manageable form, specifically a quadratic equation. Understanding how to move from a radical equation to a polynomial equation is crucial for solving a wide range of algebraic problems. This step sets the stage for the subsequent algebraic manipulations that will lead us to the solution(s) for x. We’ll now proceed to rearrange the equation into the standard quadratic form to facilitate solving for x.

Rearranging the Equation

Having squared both sides of the original equation, we now have 5x + 11 = x² + 2x + 1. To solve for x, the next step is to rearrange this equation into the standard form of a quadratic equation, which is ax² + bx + c = 0. This form is crucial because it allows us to apply various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. To get our equation into this form, we need to move all terms to one side, leaving zero on the other side. We can achieve this by subtracting 5x and 11 from both sides of the equation. Subtracting 5x from both sides gives us 11 = x² - 3x + 1. Next, subtracting 11 from both sides results in 0 = x² - 3x - 10. Now our equation is in the standard quadratic form, where a = 1, b = -3, and c = -10. This rearrangement is a fundamental algebraic manipulation, and mastering it is essential for solving quadratic equations. By bringing all terms to one side and setting the equation equal to zero, we create a clear pathway for applying techniques such as factoring or the quadratic formula. This step transforms a complex equation into a more solvable form, paving the way for finding the values of x that satisfy the equation. With the equation now in standard form, we can proceed to solve it using appropriate methods.

Solving the Quadratic Equation

With the equation rearranged into the standard quadratic form, x² - 3x - 10 = 0, we can now solve for x. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, the equation is easily factorable, which makes factoring the most efficient method. Factoring involves breaking down the quadratic expression into two binomial factors. We are looking for two numbers that multiply to -10 (the constant term) and add up to -3 (the coefficient of the x term). The numbers -5 and 2 satisfy these conditions since (-5) * (2) = -10 and (-5) + 2 = -3. Therefore, we can factor the quadratic expression as (x - 5)(x + 2) = 0. Now, we apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This means either x - 5 = 0 or x + 2 = 0. Solving these two linear equations gives us two potential solutions for x. For x - 5 = 0, we add 5 to both sides, yielding x = 5. For x + 2 = 0, we subtract 2 from both sides, yielding x = -2. These values, x = 5 and x = -2, are the potential solutions to the quadratic equation. However, because we squared the original equation, it's crucial to check these solutions in the original radical equation to ensure they are valid and not extraneous solutions. The process of solving quadratic equations by factoring is a fundamental skill in algebra, and this step demonstrates its practical application.

Checking for Extraneous Solutions

After solving the quadratic equation, we obtained two potential solutions: x = 5 and x = -2. However, when dealing with radical equations, it's essential to check these solutions in the original equation to ensure they are valid. This is because squaring both sides of an equation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. To check x = 5, we substitute it into the original equation: √(5x + 11) = x + 1. Plugging in x = 5 gives us √(5(5) + 11) = 5 + 1, which simplifies to √(25 + 11) = 6, and further to √36 = 6. Since √36 is indeed 6, the equation holds true for x = 5, making it a valid solution. Next, we check x = -2. Substituting x = -2 into the original equation gives us √(5(-2) + 11) = -2 + 1, which simplifies to √(-10 + 11) = -1, and further to √1 = -1. However, √1 is 1, not -1. Therefore, the equation does not hold true for x = -2, making it an extraneous solution. This step highlights the critical importance of checking solutions in the original equation when dealing with radical equations. The process of verifying solutions ensures that we only accept valid answers and discard extraneous ones, reinforcing our understanding of the solution set. In this case, only x = 5 is a valid solution, and x = -2 is an extraneous solution that must be discarded. This thorough check provides confidence in our final answer.

Final Answer

After carefully solving the equation √(5x + 11) = x + 1 and diligently checking for extraneous solutions, we have arrived at the final answer. We initially squared both sides of the equation to eliminate the square root, which led us to the quadratic equation x² - 3x - 10 = 0. Solving this quadratic equation by factoring gave us two potential solutions: x = 5 and x = -2. However, it was crucial to verify these solutions in the original radical equation to ensure their validity. Upon checking, we found that x = 5 satisfies the original equation, while x = -2 does not. The solution x = -2 was identified as an extraneous solution, resulting from the squaring process, and therefore must be discarded. Thus, the only valid solution to the equation √(5x + 11) = x + 1 is x = 5. This comprehensive process highlights the importance of not only solving equations accurately but also verifying solutions to avoid extraneous results. The final answer, x = 5, represents the value of x that truly satisfies the given equation, demonstrating a clear understanding of radical equations and the steps required to solve them correctly. This methodical approach ensures that we arrive at the correct solution with confidence.