Solving √[5x-5] + √[x+3] = 2 A Step-by-Step Guide
Radical equations, equations containing variables within radicals (like square roots, cube roots, etc.), often appear intimidating, but they can be systematically solved using algebraic techniques. In this comprehensive guide, we will delve into the process of solving the radical equation √[5x-5] + √[x+3] = 2, providing a detailed step-by-step solution along with explanations to ensure clarity and understanding. Understanding how to solve radical equations is a crucial skill in algebra, and mastering these techniques will empower you to tackle more complex mathematical problems with confidence. Before diving into the specifics of this equation, let's establish a general strategy for solving equations involving radicals.
When solving equations with radicals, the primary goal is to isolate the radical terms and then eliminate them by raising both sides of the equation to the appropriate power. For example, if we have a square root, we would square both sides; for a cube root, we would cube both sides, and so on. This process aims to transform the radical equation into a more manageable algebraic form, typically a polynomial equation, which we can then solve using standard methods. However, it is crucially important to check our solutions in the original equation, as this process of raising both sides to a power can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original radical equation. Therefore, verification is an indispensable step in the process of solving radical equations.
As we proceed, we will emphasize the importance of each step, including isolating radicals, squaring both sides, simplifying the resulting equations, and, most importantly, verifying the solutions. This detailed approach will not only help in solving the given equation but also build a strong foundation for solving other radical equations. We will also highlight common pitfalls and strategies to avoid them, ensuring a robust understanding of the solution process. The ability to solve equations like this is fundamental in many areas of mathematics and its applications, making this topic an essential part of mathematical education.
Step 1: Isolate One Radical
Our initial equation is √[5x-5] + √[x+3] = 2. To begin, we need to isolate one of the radicals. Let's isolate the first radical, √[5x-5], by subtracting √[x+3] from both sides of the equation. This gives us:
√[5x-5] = 2 - √[x+3]
Isolating a radical is a pivotal first step because it allows us to eliminate the radical by raising both sides of the equation to the appropriate power in the subsequent step. This step is particularly crucial when dealing with multiple radicals, as isolating one at a time simplifies the process. Without this step, squaring both sides immediately would result in a more complex equation with cross-terms involving radicals, making it significantly harder to solve. This careful isolation of the radical term is a testament to the importance of strategic manipulation in algebra, where the right sequence of steps can drastically simplify a problem.
The importance of this initial isolation cannot be overstated. By isolating √[5x-5], we set the stage for squaring both sides of the equation in a way that minimizes complexity. If we were to square the original equation directly, we would end up with a term that still contains both radicals, making the simplification process significantly more challenging. This technique underscores a core principle in solving radical equations: simplify as much as possible at each step to prevent the problem from becoming unmanageable. Therefore, this preliminary step of isolating one of the square root terms is not just a matter of preference; it is a strategic necessity for an efficient solution.
Furthermore, the decision to isolate √[5x-5] rather than √[x+3] is somewhat arbitrary in this specific case, but in other equations, the choice of which radical to isolate first might be influenced by other factors, such as coefficients or the complexity of the expression under the radical. However, for this particular equation, either choice would lead to a correct solution if followed by the correct steps. Thus, the emphasis here is on the process of isolation itself, which is a universally applicable technique when solving equations involving radicals.
Step 2: Square Both Sides
Now that we have isolated one radical, √[5x-5] = 2 - √[x+3], we can eliminate it by squaring both sides of the equation. Squaring both sides gives us:
(√[5x-5])² = (2 - √[x+3])²
This simplifies to:
5x - 5 = (2 - √[x+3])(2 - √[x+3])
Expanding the right side, we get:
5x - 5 = 4 - 4√[x+3] + (x + 3)
The critical idea behind squaring both sides is to remove the square root, thereby converting the radical equation into a standard algebraic equation that is easier to manipulate. However, it's imperative to remember that squaring both sides can introduce extraneous solutions, which are solutions that satisfy the squared equation but not the original equation. This is why we must verify our solutions in the original equation later in the process. The expansion of (2 - √[x+3])² requires careful attention to the distributive property and the correct application of the binomial square formula, which is crucial for accurate algebraic manipulation.
When we expand (2 - √[x+3])², we are essentially multiplying the binomial (2 - √[x+3]) by itself. The result of this multiplication includes not only the squared terms (2² and (√[x+3])²) but also the cross-term -2 * 2 * √[x+3], which simplifies to -4√[x+3]. This cross-term is a direct consequence of the binomial expansion and is often a source of errors if not handled carefully. It's also important to realize that while squaring eliminates the square root on terms like (√[5x-5])², it also squares the entire expression on the other side, which can lead to a more complex expression as seen here.
The presence of the term -4√[x+3] indicates that we still have a radical in the equation, meaning we will need to repeat the process of isolating the radical and squaring both sides once more to fully eliminate the radicals. This is a common situation when dealing with equations containing multiple radicals. Each time we square, we reduce the number of radicals present, bringing us closer to a solution. This iterative approach is a key strategy in solving complex radical equations and highlights the methodical nature of algebraic problem-solving.
Step 3: Simplify and Isolate the Remaining Radical
Continuing from our previous step, we have:
5x - 5 = 4 - 4√[x+3] + (x + 3)
Combine like terms on the right side:
5x - 5 = x + 7 - 4√[x+3]
Now, isolate the remaining radical term. Subtract x and 7 from both sides:
4x - 12 = -4√[x+3]
We can simplify this equation by dividing both sides by -4:
-x + 3 = √[x+3]
Simplification at this stage is crucial for making the subsequent steps easier to handle. By combining like terms and isolating the remaining radical, we are setting ourselves up for another round of squaring, but with a simpler equation. This process exemplifies the strategy of reducing complexity incrementally in mathematical problem-solving. The algebraic manipulations involved here, such as combining like terms and dividing both sides of the equation by a constant, are fundamental skills that are applied across a wide range of mathematical contexts. Mastery of these techniques is essential for success in algebra and beyond.
The simplification step, specifically dividing both sides by -4, is a prime example of how strategic manipulation can lead to a more manageable equation. Without this division, the coefficients would be larger, and the subsequent squaring and simplification steps would be more cumbersome. This underscores the importance of looking for opportunities to simplify at every stage of the problem-solving process. Moreover, the resulting equation, -x + 3 = √[x+3], is now in a form that is much easier to square, further highlighting the effectiveness of the simplification.
The transition from 4x - 12 = -4√[x+3] to -x + 3 = √[x+3] not only simplifies the equation numerically but also reduces the potential for arithmetic errors in the following steps. When dealing with equations, especially those involving radicals, maintaining accuracy is paramount. Each simplification step is a safeguard against mistakes that can compound and lead to an incorrect final answer. Therefore, the emphasis on simplification is not just about making the problem easier; it's also about enhancing the reliability of the solution process.
Step 4: Square Both Sides Again
To eliminate the remaining radical, we square both sides of the equation -x + 3 = √[x+3]:
(-x + 3)² = (√[x+3])²
Expanding and simplifying:
x² - 6x + 9 = x + 3
This step is a direct repetition of the process used earlier, but now applied to a simpler equation. By squaring both sides, we successfully remove the remaining square root, transforming the radical equation into a quadratic equation, which we can then solve using standard algebraic methods. The quadratic equation is a common form that appears in many mathematical contexts, making the ability to solve quadratic equations a fundamental skill.
Expanding (-x + 3)² requires careful application of the binomial square formula or the distributive property. Common mistakes in this step involve incorrect signs or coefficients, highlighting the need for meticulous attention to detail. The expansion results in x² - 6x + 9, which is a standard quadratic expression. Setting up and solving the resulting quadratic equation is a routine process in algebra, but it is essential to perform each step accurately to avoid errors.
It is also worth noting that squaring both sides again reinforces the potential for introducing extraneous solutions. As we mentioned earlier, whenever we raise both sides of an equation to an even power, we must be vigilant about verifying our solutions in the original equation. This is because the squaring operation can mask sign differences that are crucial for the original equation. Therefore, the step of squaring both sides is a powerful tool for eliminating radicals, but it comes with the responsibility of thoroughly checking our final answers to ensure they are valid.
Step 5: Solve the Quadratic Equation
Now we have a quadratic equation: x² - 6x + 9 = x + 3. To solve it, we first set the equation equal to zero:
x² - 6x + 9 - x - 3 = 0
Combine like terms:
x² - 7x + 6 = 0
Factor the quadratic:
(x - 6)(x - 1) = 0
Set each factor equal to zero and solve for x:
x - 6 = 0 or x - 1 = 0
x = 6 or x = 1
Thus, we have two potential solutions: x = 6 and x = 1. The process of solving the quadratic equation involves several key steps: rearranging the equation into standard form, factoring the quadratic expression, and then applying the zero-product property. Each of these steps is a fundamental technique in algebra, and proficiency in these methods is essential for solving a wide range of mathematical problems.
Factoring the quadratic expression x² - 7x + 6 into (x - 6)(x - 1) is a critical step. It requires finding two numbers that multiply to 6 and add up to -7, which in this case are -6 and -1. Factoring is often the most efficient method for solving quadratic equations, but if factoring is not straightforward, the quadratic formula can also be used. The ability to recognize and apply the appropriate method for solving a quadratic equation is an important skill in algebra.
Finding the roots of the factored equation (x - 6)(x - 1) = 0 involves applying 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 allows us to set each factor equal to zero and solve for x, yielding the potential solutions x = 6 and x = 1. However, as we have emphasized, these are only potential solutions, and we must now verify them in the original radical equation to ensure they are not extraneous.
Step 6: Check for Extraneous Solutions
It is crucial to check our potential solutions in the original equation, √[5x-5] + √[x+3] = 2, because squaring both sides can introduce extraneous solutions.
Check x = 6:
√[5(6)-5] + √[6+3] = √[25] + √[9] = 5 + 3 = 8
Since 8 ≠ 2, x = 6 is an extraneous solution.
Check x = 1:
√[5(1)-5] + √[1+3] = √[0] + √[4] = 0 + 2 = 2
Since 2 = 2, x = 1 is a valid solution.
This step highlights a critical aspect of solving radical equations: the necessity of verification. Extraneous solutions arise because squaring both sides of an equation can transform it into a different equation with a larger solution set. The verification process involves substituting each potential solution back into the original equation to ensure that it satisfies the equation. If a potential solution does not satisfy the original equation, it is deemed extraneous and must be discarded.
The process of checking x = 6 involves substituting 6 for x in the original equation and simplifying. We find that the left side of the equation evaluates to 8, which is not equal to the right side (2). This discrepancy indicates that x = 6 is an extraneous solution and cannot be included in the final answer. This is a classic example of how squaring both sides can lead to a value that satisfies the transformed equation but not the original one.
Conversely, when we check x = 1, we substitute 1 for x in the original equation and find that the left side evaluates to 2, which is equal to the right side. This confirms that x = 1 is a valid solution and satisfies the original radical equation. This thorough verification process ensures that our final answer is correct and complete. The ability to identify and eliminate extraneous solutions is a key skill in solving radical equations and demonstrates a thorough understanding of the underlying mathematical principles.
Final Answer
Therefore, the only solution to the equation √[5x-5] + √[x+3] = 2 is x = 1. This solution was found through a careful process of isolating radicals, squaring both sides, simplifying the equation, and, most importantly, verifying the potential solutions in the original equation. The process illustrates the importance of systematic problem-solving and the necessity of checking for extraneous solutions in the context of radical equations. The final answer encapsulates the entire solution process, demonstrating the practical application of algebraic techniques to solve radical equations effectively.
In this comprehensive guide, we have walked through the detailed steps to solve the radical equation √[5x-5] + √[x+3] = 2. We emphasized the importance of isolating radicals, squaring both sides, simplifying equations, and verifying solutions to avoid extraneous results. By mastering these techniques, you can confidently approach and solve a wide variety of radical equations. Remember, the key to success in algebra lies in understanding the fundamental principles and applying them methodically. The skills developed in solving radical equations are invaluable and extend to many areas of mathematics and its applications. This structured approach not only leads to the correct answer but also reinforces the critical thinking and problem-solving skills that are essential in mathematical education. Therefore, practice and attention to detail are your best allies in solving radical equations and other algebraic challenges.
