Solving √(r+3) = R-33 A Step-by-Step Mathematical Guide
Introduction
In the realm of mathematics, solving equations involving square roots can sometimes appear daunting. However, with a systematic approach and a clear understanding of the underlying principles, such equations can be tackled effectively. This article aims to provide a comprehensive guide on how to solve the equation √(r+3) = r-33, walking you through each step of the process with detailed explanations and insights. Understanding how to manipulate and solve equations like this is a fundamental skill in algebra and serves as a building block for more advanced mathematical concepts. This article is designed for students, educators, and anyone with an interest in mathematics who wants to deepen their understanding of algebraic equations and problem-solving techniques. By the end of this guide, you will not only be able to solve this specific equation but also have a solid foundation for approaching similar problems with confidence.
Understanding the Equation
The equation we are dealing with is √(r+3) = r-33. At its core, this equation is an algebraic statement that equates the square root of the expression r+3 to the expression r-33. To solve for r, we need to isolate the variable r by performing valid mathematical operations on both sides of the equation. The presence of the square root introduces a unique challenge, as we need to eliminate it to simplify the equation. This typically involves squaring both sides, but it's crucial to remember that this operation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. Therefore, verification of the solutions is a crucial step in the process. Furthermore, understanding the domain of the equation is important. Since we are dealing with a square root, the expression inside the square root must be non-negative, meaning r+3 ≥ 0. This implies that r ≥ -3. This constraint will help us filter out any extraneous solutions later on. By grasping these fundamental aspects of the equation, we set the stage for a methodical and accurate solution process. This initial understanding ensures that we proceed with a clear roadmap, reducing the likelihood of errors and enhancing our problem-solving efficiency.
Step-by-Step Solution
1. Isolating the Square Root:
The first step in solving the equation √(r+3) = r-33 is to ensure that the square root term is isolated on one side of the equation. In this case, the square root term, √(r+3), is already isolated on the left side. This isolation is crucial because it allows us to directly address the square root by squaring both sides of the equation in the next step. If the square root term were not isolated, we would need to perform algebraic manipulations to achieve this before proceeding. This might involve adding or subtracting terms from both sides of the equation to ensure that the square root is the only term on one side. However, since our equation already has the square root isolated, we can move directly to the next phase of the solution. This streamlined start helps to simplify the process and focus our efforts on eliminating the square root and solving for the variable r.
2. Squaring Both Sides:
To eliminate the square root in the equation √(r+3) = r-33, we square both sides of the equation. This operation is based on the principle that if two quantities are equal, then their squares are also equal. Squaring the left side, (√(r+3))^2, effectively cancels out the square root, leaving us with r+3. Squaring the right side, (r-33)^2, requires expanding the binomial. This expansion can be done using the formula (a-b)^2 = a^2 - 2ab + b^2. Applying this formula, we get (r-33)^2 = r^2 - 2(r)(33) + 33^2 = r^2 - 66r + 1089. So, after squaring both sides, the equation becomes r+3 = r^2 - 66r + 1089. This transformation is a significant step because it converts the original equation with a square root into a quadratic equation, which we can solve using standard algebraic techniques. However, it's essential to remember that squaring both sides can introduce extraneous solutions, so we must verify our solutions later.
3. Rearranging into a Quadratic Equation:
Now that we have the equation r+3 = r^2 - 66r + 1089, the next step is to rearrange it into the standard form of a quadratic equation, which is ax^2 + bx + c = 0. To do this, we want to set one side of the equation to zero. We can achieve this by subtracting r and subtracting 3 from both sides of the equation. This gives us 0 = r^2 - 66r + 1089 - r - 3, which simplifies to 0 = r^2 - 67r + 1086. This is now a quadratic equation in the standard form, where a = 1, b = -67, and c = 1086. Having the equation in 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. The standard form provides a structured framework for identifying the coefficients, which are necessary for applying these solution methods. This rearrangement is a key step in our journey to find the values of r that satisfy the original equation.
4. Solving the Quadratic Equation:
We now have the quadratic equation r^2 - 67r + 1086 = 0. There are several methods to solve this, but in this case, factoring is a viable option. We need to find two numbers that multiply to 1086 and add up to -67. These numbers are -42 and -25. Thus, we can factor the quadratic equation as (r - 42)(r - 25) = 0. Setting each factor equal to zero gives us two potential solutions for r: r - 42 = 0 which implies r = 42, and r - 25 = 0 which implies r = 25. These values, r = 42 and r = 25, are the roots of the quadratic equation. However, it is crucial to remember that since we squared both sides of the original equation, we need to verify these solutions to ensure they are not extraneous. This verification process is the next essential step in our solution.
5. Verifying the Solutions:
After solving the quadratic equation, we obtained two potential solutions: r = 42 and r = 25. However, squaring both sides of an equation can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, it is essential to verify each potential solution in the original equation, √(r+3) = r-33.
Let's start with r = 42. Substituting this value into the original equation gives us √(42+3) = 42-33, which simplifies to √45 = 9. Since √45 is approximately 6.708, this solution does not satisfy the original equation. Thus, r = 42 is an extraneous solution and must be discarded.
Now, let's verify r = 25. Substituting this value into the original equation gives us √(25+3) = 25-33, which simplifies to √28 = -8. Since √28 is approximately 5.292, this solution also does not satisfy the original equation. Thus, r = 25 is also an extraneous solution and must be discarded.
6. Considering the Domain:
Before declaring that there are no solutions, it's crucial to revisit the original equation and the domain of the variable r. Recall that we have √(r+3) = r-33. Since the expression inside the square root must be non-negative, we have the condition r+3 ≥ 0, which implies r ≥ -3. Additionally, the square root of a number is always non-negative, so √(r+3) ≥ 0. This means that r-33 must also be non-negative, leading to the condition r-33 ≥ 0, which implies r ≥ 33.
We found two potential solutions, r = 42 and r = 25. While r = 42 satisfies the condition r ≥ -3, and r = 25 also satisfies r ≥ -3, neither of them satisfy the condition r ≥ 33. Furthermore, our verification step showed that neither of these values satisfy the original equation. Therefore, we need to carefully reconsider our analysis.
We made an error in our verification step. Let's re-evaluate the solutions. For r = 42, the original equation becomes √(42+3) = 42-33, which simplifies to √45 = 9. This is incorrect, as √45 ≈ 6.708, so r = 42 is indeed extraneous. For r = 25, the original equation becomes √(25+3) = 25-33, which simplifies to √28 = -8. This is also incorrect, as the square root of a number cannot be negative. Thus, r = 25 is also extraneous.
Considering the domain r ≥ 33, we realize that our potential solutions did not meet this criterion, and our verification confirmed they are extraneous. Therefore, based on our analysis, there are no solutions to the equation √(r+3) = r-33.
Conclusion
In conclusion, solving equations involving square roots requires a meticulous approach. For the equation √(r+3) = r-33, we followed a step-by-step process: isolating the square root, squaring both sides, rearranging into a quadratic equation, solving the quadratic equation, and crucially, verifying the solutions. Our initial solutions, r = 42 and r = 25, were found to be extraneous upon verification. Additionally, considering the domain of the equation, we established that r must be greater than or equal to 33. Since neither of our potential solutions met this criterion and failed the verification step, we conclude that there are no real solutions to the equation √(r+3) = r-33. This exercise highlights the importance of not only applying algebraic techniques correctly but also understanding the implications of each step and the need for thorough verification.
