Solving $\sqrt{x+8}-6=x$ A Step-by-Step Guide
Finding solutions to equations involving square roots can be a challenging yet rewarding mathematical exercise. In this article, we will delve into the process of solving the equation $\sqrt{x+8}-6=x$, providing a step-by-step guide to identify all possible solutions. Our focus will be on ensuring clarity and understanding, making this guide accessible to students and math enthusiasts alike. We will explore the algebraic manipulations required, discuss the importance of checking for extraneous solutions, and ultimately determine the correct answers among the provided options.
1. Understanding the Equation and Initial Steps
Our initial focus is on understanding the equation $\sqrtx+8}-6=x$. This equation involves a square root, which means we need to isolate the radical term to effectively solve for x. The presence of the square root also implies that we must be cautious about extraneous solutions, which are solutions that arise from the algebraic manipulation but do not satisfy the original equation. To begin, we will isolate the square root term by adding 6 to both sides of the equation. This gives us = x + 6$. This step is crucial because it sets the stage for eliminating the square root through squaring, a common technique in solving such equations. Before we proceed, it's important to remember that squaring both sides can introduce extraneous solutions, so we must check our final answers in the original equation. The next step involves squaring both sides of the equation to eliminate the square root. This process will transform the equation into a more manageable form, allowing us to solve for x. However, it's not just about the mechanics of squaring; it's about understanding the implications and the potential pitfalls that come with it. By carefully explaining each step and the reasoning behind it, we aim to provide a thorough understanding of the solution process.
2. Squaring Both Sides and Simplifying the Equation
Having isolated the square root, the next step in squaring both sides of the equation $\sqrtx+8} = x + 6$. This is done to eliminate the square root, making the equation easier to solve. When we square both sides, we get\right)^2 = (x + 6)^2$. Simplifying the left side is straightforward: the square root is canceled out, leaving us with x + 8. The right side, however, requires expanding the binomial (x + 6)². Remember that (a + b)² = a² + 2ab + b², so we have: $(x + 6)^2 = x^2 + 2(x)(6) + 6^2 = x^2 + 12x + 36$. Thus, our equation becomes: $x + 8 = x^2 + 12x + 36$. Now, we have a quadratic equation, which we can solve by setting it equal to zero. To do this, we subtract x and 8 from both sides: $0 = x^2 + 12x + 36 - x - 8$. Combining like terms gives us: $0 = x^2 + 11x + 28$. This simplified quadratic equation is now in a standard form that we can solve using factoring, the quadratic formula, or completing the square. Factoring is often the quickest method if the quadratic expression can be easily factored. In this case, we look for two numbers that multiply to 28 and add up to 11. These numbers are 4 and 7. Therefore, we can factor the quadratic as follows: $0 = (x + 4)(x + 7)$. This step is crucial as it breaks down the quadratic equation into simpler linear factors, each of which can be solved independently. By setting each factor equal to zero, we can find the potential solutions for x. However, we must remember that these are just potential solutions, and we will need to verify them in the original equation to ensure they are not extraneous.
3. Solving the Quadratic Equation by Factoring
Now that we have the factored form of the quadratic equation, we will solve the quadratic equation by factoring. We have: $0 = (x + 4)(x + 7)$. To find the solutions for x, we set each factor equal to zero: $x + 4 = 0 \text{ or } x + 7 = 0$. Solving the first equation, $x + 4 = 0$, we subtract 4 from both sides to get: $x = -4$. Solving the second equation, $x + 7 = 0$, we subtract 7 from both sides to get: $x = -7$. So, we have two potential solutions: x = -4 and x = -7. It is critical to remember that these are only potential solutions. Because we squared both sides of the equation earlier, we may have introduced extraneous solutions. These are values that satisfy the transformed equation but not the original equation. Therefore, we must check each of these potential solutions in the original equation to determine whether they are valid. This step is not just a formality; it is a necessary part of the solution process when dealing with radical equations. By verifying each solution, we can ensure that we have accurately identified all the solutions to the original equation and have not included any extraneous values. This careful approach is key to solving equations involving square roots correctly.
4. Checking for Extraneous Solutions
Checking for extraneous solutions is a crucial step in solving equations with square roots. We found two potential solutions: x = -4 and x = -7. Now, we must substitute each of these values back into the original equation, $\sqrtx+8}-6=x$, to see if they hold true. Let's start with x = -4-6=-4$. Simplifying the expression inside the square root: $\sqrt4}-6=-4$. The square root of 4 is 2, so we have-6=-7$. Simplifying the expression inside the square root: $\sqrt{1}-6=-7$. The square root of 1 is 1, so we have: $1-6=-7$. Which simplifies to: $-5=-7$. This statement is false, so x = -7 is an extraneous solution. This means that while x = -7 satisfies the squared equation, it does not satisfy the original equation with the square root. Therefore, we discard x = -7 as a solution. The importance of this step cannot be overstated. It is the final safeguard against incorrect answers and ensures that our solutions are mathematically sound. By carefully checking each potential solution, we can confidently identify the correct answer and avoid the pitfall of extraneous solutions.
5. Identifying the Correct Solution
After identifying the correct solution by checking for extraneous roots, we found that x = -4 is a valid solution, while x = -7 is an extraneous solution. Therefore, the only solution to the equation $\sqrt{x+8}-6=x$ is x = -4. Now, let's revisit the options provided: A. None of the above B. x = -7 C. x = -4 D. x = -4 and x = -7 Based on our analysis, the correct answer is C. x = -4. This entire process underscores the significance of not only solving equations algebraically but also verifying the solutions to ensure they are valid. In equations involving radicals, checking for extraneous solutions is not just a recommendation; it is a necessity. By meticulously following each step—isolating the radical, squaring both sides, solving the resulting equation, and checking for extraneous solutions—we can confidently arrive at the correct answer. In this case, the journey through the equation has led us to the single, valid solution of x = -4, highlighting the elegance and precision of mathematical problem-solving. Understanding these steps is crucial for successfully tackling similar problems in the future.
Conclusion
In conclusion, solving the equation $\sqrt{x+8}-6=x$ involves a series of algebraic steps, with a critical emphasis on checking for extraneous solutions. We started by isolating the square root term, then squared both sides to eliminate the radical. This led us to a quadratic equation, which we solved by factoring. The potential solutions were x = -4 and x = -7. However, only after substituting these values back into the original equation did we discover that x = -7 was an extraneous solution. This left us with x = -4 as the sole valid solution. This process demonstrates the importance of a thorough and careful approach to solving equations, especially those involving radicals. The key takeaways from this exercise include: 1. Isolating the radical term before squaring. 2. Squaring both sides and simplifying the resulting equation. 3. Solving the quadratic equation by factoring or other methods. 4. Checking all potential solutions in the original equation to eliminate extraneous solutions. By mastering these steps, students and math enthusiasts can confidently tackle similar problems and deepen their understanding of algebraic problem-solving. The solution to this equation not only provides a specific answer but also reinforces the broader principles of mathematical rigor and the necessity of verification in the problem-solving process.
