Solving The Equation √(w²+7w-39) = √(10w+1) A Step-by-Step Guide

In the realm of mathematics, solving equations is a fundamental skill. Today, we embark on a journey to dissect and conquer the equation √(w²+7w-39) = √(10w+1). This equation, a captivating blend of algebraic expressions and square roots, presents a unique challenge that demands a methodical approach and a keen eye for detail. Let's delve into the intricacies of this equation, exploring the steps required to arrive at the correct solution and understand the underlying mathematical principles involved.

The Quest for 'w': A Step-by-Step Solution

Our primary objective is to isolate the variable 'w' and determine its value(s) that satisfy the given equation. To achieve this, we'll employ a series of algebraic manipulations, each step carefully designed to bring us closer to the solution. Here's a breakdown of the process:

1. Squaring Both Sides: A Necessary First Step

   The presence of square roots can often complicate equations. To eliminate these radicals, our initial move involves squaring both sides of the equation. This seemingly simple step is crucial as it transforms the equation into a more manageable form, paving the way for further simplification. By squaring both sides, we effectively undo the square root operation, allowing us to work with the expressions within the radicals. This transformation yields a new equation, one that is devoid of square roots and more amenable to algebraic manipulation. This process is guided by the principle that performing the same operation on both sides of an equation maintains its balance, ensuring that the solutions remain unchanged. Let's proceed with squaring both sides of the equation √(w²+7w-39) = √(10w+1). Squaring both sides of the equation is a critical step in solving radical equations. It eliminates the square roots, allowing us to work with a more familiar algebraic expression. However, it's crucial to remember that squaring both sides can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, it's essential to check our solutions at the end of the process to ensure they are valid. Squaring both sides gives us: (√(w²+7w-39))² = (√(10w+1))², which simplifies to w²+7w-39 = 10w+1. This new equation is a quadratic equation, and we can solve it using various techniques, such as factoring, completing the square, or using the quadratic formula. In this case, we will proceed by rearranging the equation to bring all terms to one side, setting the equation equal to zero.

2. Simplifying and Rearranging: Setting the Stage for Factoring

   After squaring both sides, we obtain the equation w²+7w-39 = 10w+1. To solve this equation, we need to rearrange it into a standard form that facilitates the application of solution techniques. The standard form for a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants. To achieve this form, we'll subtract 10w and 1 from both sides of the equation. This process involves combining like terms, specifically the 'w' terms and the constant terms. By subtracting 10w from both sides, we eliminate the '10w' term on the right side and introduce a '-10w' term on the left side. Similarly, subtracting 1 from both sides eliminates the constant term '1' on the right side and introduces a '-1' term on the left side. This rearrangement is a fundamental algebraic manipulation that allows us to consolidate the terms and express the equation in a form suitable for solving. The resulting equation, after simplification, will be a quadratic equation in the standard form, making it easier to identify the coefficients and apply appropriate solution methods. This step is crucial for progressing towards the solution of the equation. Subtracting 10w and 1 from both sides of the equation w²+7w-39 = 10w+1, we get: w²+7w-39 - 10w - 1 = 10w+1 - 10w - 1, which simplifies to w² - 3w - 40 = 0. This is a quadratic equation in the standard form, ax² + bx + c = 0, where a = 1, b = -3, and c = -40. Now, we can proceed to solve this quadratic equation.

3. Factoring the Quadratic: Unveiling the Roots

   Now we have a quadratic equation: w² - 3w - 40 = 0. Factoring is a powerful technique for solving quadratic equations, especially when the equation can be expressed as a product of two binomials. The goal of factoring is to find two numbers that, when multiplied, give the constant term (-40 in this case) and, when added, give the coefficient of the linear term (-3 in this case). These two numbers will help us rewrite the quadratic expression as a product of two binomials. In this particular equation, we need to find two numbers that multiply to -40 and add up to -3. After careful consideration, we can identify these numbers as -8 and 5. The product of -8 and 5 is -40, and their sum is -3, satisfying the required conditions. Once we have identified these numbers, we can rewrite the quadratic expression as a product of two binomials: (w - 8)(w + 5). This factorization transforms the equation into a form where we can easily identify the solutions. The factored form allows us to 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 property is crucial for finding the values of 'w' that satisfy the equation. By setting each factor equal to zero, we obtain two simple linear equations, which can be easily solved to find the values of 'w'. Factoring the quadratic equation is a key step in solving for 'w'.

4. The Zero-Product Property: Isolating Potential Solutions

   The factored form of our equation is (w - 8)(w + 5) = 0. This is where the zero-product property comes into play. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, for the equation (w - 8)(w + 5) = 0 to hold true, either (w - 8) must be equal to zero, or (w + 5) must be equal to zero, or both. This property allows us to break down the quadratic equation into two simpler linear equations. By setting each factor equal to zero, we create two independent equations that can be solved individually. These linear equations are much easier to solve than the original quadratic equation. Solving each linear equation will give us a potential value for 'w' that satisfies the original equation. However, it's important to remember that these are just potential solutions. We still need to verify whether these solutions are valid by substituting them back into the original equation. This is because squaring both sides of the equation in the initial steps can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. The zero-product property is a fundamental tool in solving factored equations. Applying the zero-product property, we set each factor equal to zero: w - 8 = 0 or w + 5 = 0.

5. Solving for 'w': Unveiling the Candidate Solutions

   From the previous step, we have two linear equations: w - 8 = 0 and w + 5 = 0. Solving these equations is straightforward. To solve w - 8 = 0, we simply add 8 to both sides of the equation. This isolates 'w' on the left side and gives us w = 8. Similarly, to solve w + 5 = 0, we subtract 5 from both sides of the equation. This isolates 'w' on the left side and gives us w = -5. These two values, w = 8 and w = -5, are our candidate solutions. They are the potential values of 'w' that could satisfy the original equation. However, as mentioned earlier, we need to verify these solutions to ensure they are valid. Squaring both sides of an equation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, it's crucial to substitute each candidate solution back into the original equation and check if it holds true. If a candidate solution does not satisfy the original equation, it is considered an extraneous solution and must be discarded. The process of verifying solutions is an essential part of solving equations, especially those involving radicals or rational expressions. Solving the linear equations, we get two potential solutions: w = 8 and w = -5. These are the values we need to check in the original equation.

6. The Crucial Check: Verifying the Solutions

   Now comes the critical step of verifying our potential solutions. We have two candidate solutions: w = 8 and w = -5. To determine if these solutions are valid, we must substitute each one back into the original equation, √(w²+7w-39) = √(10w+1), and check if the equation holds true. This step is crucial because squaring both sides of an equation, as we did in the beginning, can sometimes introduce extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original equation. Therefore, it's essential to verify each candidate solution to ensure it is a genuine solution to the original equation. Let's start by checking w = 8. Substituting w = 8 into the original equation, we get: √(8²+7(8)-39) = √(10(8)+1). Simplifying the expressions inside the square roots, we have: √(64+56-39) = √(80+1), which further simplifies to √81 = √81. This is a true statement, as the square root of 81 is indeed 9. Therefore, w = 8 is a valid solution. Now, let's check w = -5. Substituting w = -5 into the original equation, we get: √((-5)²+7(-5)-39) = √(10(-5)+1). Simplifying the expressions inside the square roots, we have: √(25-35-39) = √(-50+1), which further simplifies to √(-49) = √(-49). This statement involves taking the square root of a negative number, which results in an imaginary number. However, the square root of a negative number is not a real number, and the original equation is defined in the realm of real numbers. Therefore, w = -5 is not a valid solution. This demonstrates the importance of checking solutions, as w = -5 is an extraneous solution that arises from the squaring process. The verification step confirms that only w = 8 is a valid solution to the original equation. This crucial step ensures that we have identified the correct solution and avoided any extraneous solutions. Checking our solutions is vital to ensure accuracy. For w = 8: √(8²+7(8)-39) = √(64+56-39) = √81 = 9 and √(10(8)+1) = √(80+1) = √81 = 9. So, w = 8 is a valid solution. For w = -5: √((-5)²+7(-5)-39) = √(25-35-39) = √(-49), which is not a real number, and √(10(-5)+1) = √(-50+1) = √(-49), which is also not a real number. Since we are dealing with real numbers, w = -5 is an extraneous solution.

7. The Solution: A Triumphant Conclusion

   After meticulously working through the steps, we've arrived at the solution. We started by squaring both sides of the equation to eliminate the square roots, then simplified and rearranged the equation into a standard quadratic form. Factoring the quadratic expression allowed us to apply the zero-product property and identify potential solutions. However, the journey didn't end there. We recognized the importance of verifying our solutions to avoid extraneous results. By substituting each candidate solution back into the original equation, we determined that only one value, w = 8, satisfied the equation. The other candidate, w = -5, turned out to be an extraneous solution, a consequence of the squaring process. This final step of verification underscores the importance of rigor and attention to detail in mathematical problem-solving. It ensures that we arrive at the correct solution and avoid any potential pitfalls along the way. The solution to the equation √(w²+7w-39) = √(10w+1) is w = 8. This is the only value of 'w' that satisfies the original equation. We have successfully navigated the complexities of this equation, employing algebraic techniques and careful verification to arrive at the final answer. The process highlights the importance of each step in solving equations, from the initial simplification to the final check. The solution to the equation √(w²+7w-39) = √(10w+1) is w = 8. This is the correct answer, as it satisfies the original equation after verification.

The Answer: D. w=8

Therefore, the solution to the equation √(w²+7w-39) = √(10w+1) is D. w=8. This meticulous step-by-step solution demonstrates the process of solving radical equations, highlighting the importance of verification to avoid extraneous solutions. By squaring both sides, simplifying, factoring, applying the zero-product property, and ultimately verifying our solutions, we have confidently arrived at the correct answer.