Finding 'a' For Equivalent Radical Expressions
Let's dive into the world of radical expressions and figure out how to find the value of 'a' that makes two expressions equivalent. This is a common type of problem in mathematics, and understanding the steps involved can help you tackle similar challenges with confidence. We'll break down the problem step-by-step, making sure to explain the reasoning behind each operation. This problem assumes that x > 0 and y ≥ 0. We aim to find the value of 'a' that satisfies the equation √((126xy⁵)/(32x³)) = √((63y⁵)/(ax⁶)).
Understanding the Problem
In this mathematical puzzle, we're presented with two radical expressions that are claimed to be equivalent. Our mission, should we choose to accept it, is to pinpoint the exact value of 'a' that makes this equivalence a reality. The expressions in question involve square roots, fractions, variables (x and y), and exponents – a delightful mix of mathematical concepts! To conquer this challenge, we'll need to roll up our sleeves and simplify these expressions. Simplifying expressions is not just about making them look neater; it's about revealing their underlying structure and making comparisons easier. Think of it as decluttering a room – once everything is in its place, it's much easier to find what you're looking for. In our case, we're looking for the value of 'a', and simplification will help us isolate and identify it. Simplifying radical expressions often involves identifying perfect squares (or cubes, or higher powers, depending on the root) within the radicand (the expression under the radical symbol). We can then extract these perfect squares, which simplifies the overall expression. For example, √(9x²) can be simplified to 3x because 9 is a perfect square (3²) and x² is a perfect square (x*x). We'll also be using the properties of exponents to help us simplify. Remember, when dividing terms with the same base, we subtract the exponents (e.g., x⁵ / x² = x³). These exponent rules are crucial for maneuvering variables and their powers within our expressions. Before we jump into the nitty-gritty calculations, it's important to have a plan. Our strategy will be to first simplify each radical expression individually. Then, we'll compare the simplified forms and solve for 'a'. It's like having a roadmap for our mathematical journey – it helps us stay on track and avoid getting lost in the complexities of the problem. With our plan in place and our tools at the ready, let's begin the simplification process! We'll start with the left-hand side of the equation, carefully dissecting it and applying our mathematical skills to bring it into its simplest form. By simplifying each side and then comparing, we'll reveal the hidden value of 'a' and solve this intriguing equation.
Simplifying the Left-Hand Side
Let's begin by tackling the left-hand side of our equation: √((126xy⁵)/(32x³)). Our goal is to simplify this expression as much as possible, making it easier to compare with the right-hand side. The first step in simplifying a radical expression is often to look for ways to simplify the fraction inside the radical. This means we want to reduce the fraction to its lowest terms by finding common factors in the numerator and denominator. Looking at the numbers 126 and 32, we can see that both are divisible by 2. Dividing both by 2 gives us 63 in the numerator and 16 in the denominator. So, our expression now looks like this: √((63xy⁵)/(16x³)). Next, let's consider the variables. We have 'x' in both the numerator and the denominator. Remember the rule for dividing exponents with the same base: we subtract the exponents. In this case, we have x¹ in the numerator and x³ in the denominator. So, x¹ / x³ = x^(1-3) = x⁻² which is equivalent to 1/x². Now our expression looks like: √((63y⁵)/(16x²)). We're making good progress! Now, we need to think about the square root. Remember, √(a/b) = √a / √b. This means we can split the radical into two separate radicals: √(63y⁵) / √(16x²). The denominator is looking promising. 16 is a perfect square (4²), and x² is also a perfect square. So, √(16x²) simplifies to 4x. The numerator needs a bit more work. Let's focus on √(63y⁵). We need to look for perfect square factors within 63 and y⁵. For 63, we can break it down into 9 * 7. 9 is a perfect square (3²), so we can rewrite √63 as √(9 * 7) = √9 * √7 = 3√7. Now let's think about y⁵. We can rewrite y⁵ as y⁴ * y. y⁴ is a perfect square because it's (y²)². So, √(y⁵) = √(y⁴ * y) = √(y⁴) * √y = y²√y. Putting it all together, √(63y⁵) becomes 3y²√(7y). So, our fully simplified left-hand side is (3y²√(7y)) / (4x). This is much cleaner and easier to work with than where we started! We've successfully simplified the left-hand side by reducing the fraction, applying exponent rules, and extracting perfect squares from the radical. Now, let's turn our attention to the right-hand side and see if we can simplify it in a similar way. By simplifying both sides, we'll be in a prime position to compare them and solve for the elusive value of 'a'.
Simplifying the Right-Hand Side
Now, let's simplify the right-hand side of the equation: √((63y⁵)/(ax⁶)). Similar to our approach with the left-hand side, we'll look for ways to break down the expression and identify any perfect squares. Our goal here is to get the right-hand side into its simplest form so we can compare it with the simplified left-hand side and ultimately solve for 'a'. The first thing we can do is split the radical, just like we did before, using the property √(a/b) = √a / √b. This gives us √(63y⁵) / √(ax⁶). We've already simplified √(63y⁵) when we worked on the left-hand side. We found that √(63y⁵) = 3y²√(7y). So, we can substitute that in, and our expression becomes (3y²√(7y)) / √(ax⁶). Now, let's focus on the denominator, √(ax⁶). We need to think about how to extract any perfect squares from this expression. We know that x⁶ is a perfect square because it's (x³)² (remember, (xm)n = x^(m*n)). So, √(x⁶) = x³. This simplifies our denominator to √(a) * x³. Now, our entire right-hand side expression looks like this: (3y²√(7y)) / (√(a)x³). We're getting closer to a simplified form! Notice that we have an x³ in the denominator, while on the left-hand side, we ended up with an x in the denominator. This difference in the power of x is something we'll need to address when we equate the two sides. The key to further simplification lies in understanding the role of 'a'. The value of 'a' will determine what, if any, perfect square factors we can extract from the radical in the denominator. We've successfully simplified the right-hand side as much as possible without knowing the specific value of 'a'. We've split the radical, simplified the known terms, and isolated the √(a) term. Now, we have the left-hand side simplified to (3y²√(7y)) / (4x) and the right-hand side simplified to (3y²√(7y)) / (√(a)x³). The next step is crucial: we'll equate these two simplified expressions. By setting them equal to each other, we create an equation that we can solve for 'a'. It's like connecting two pieces of a puzzle – once they're joined, the bigger picture starts to emerge. So, let's take the leap and equate the two sides. This will set us on the path to uncovering the numerical value of 'a'.
Equating and Solving for 'a'
Now comes the crucial step: equating the simplified expressions and solving for 'a'. We've worked diligently to simplify both sides of our original equation, and now we're ready to bring them together. We have the left-hand side simplified to (3y²√(7y)) / (4x) and the right-hand side simplified to (3y²√(7y)) / (√(a)x³). To find the value of 'a', we set these two expressions equal to each other:
(3y²√(7y)) / (4x) = (3y²√(7y)) / (√(a)x³)
This equation might look a bit intimidating at first glance, but don't worry, we can simplify it step by step. The first thing to notice is that we have the same term, 3y²√(7y), in the numerator of both fractions. This means we can cancel this term out from both sides of the equation, as long as it's not equal to zero. Since the problem states that y ≥ 0, we need to consider the case where y = 0. If y = 0, then both sides of the original equation would be zero, and any value of 'a' would technically work. However, let's assume y > 0 for now, so we can safely cancel out the term. After canceling the common term, our equation simplifies to:
1 / (4x) = 1 / (√(a)x³)
Now, we have a much cleaner equation to work with. To solve for 'a', we can cross-multiply. This means multiplying the numerator of the left side by the denominator of the right side, and vice versa. This gives us:
√(a)x³ = 4x
Our goal is to isolate 'a', so let's divide both sides of the equation by x. Remember, the problem states that x > 0, so we don't need to worry about dividing by zero. Dividing both sides by x gives us:
√(a)x² = 4
Now, we want to get √(a) by itself. Let's divide both sides by x²:
√(a) = 4 / x²
We're almost there! To get 'a' by itself, we need to square both sides of the equation. Squaring √(a) gives us 'a', and squaring 4/x² gives us (4/x²)² = 16/x⁴. So, we have:
a = 16 / x⁴
This is our solution for 'a'! The value of 'a' that makes the two original expressions equivalent is 16/x⁴. We've successfully navigated through the simplification, equation solving, and algebraic manipulation to arrive at our answer. It's like reaching the summit of a challenging climb – the view from the top is always worth the effort. Now, let's take a moment to reflect on the steps we took and the concepts we used along the way. We'll also discuss how we can verify our solution to ensure that it's correct.
Verifying the Solution
It's always a good practice in mathematics to verify your solution. Think of it as double-checking your work or proofreading an important document. Verifying our solution helps us catch any potential errors and gives us confidence that our answer is correct. In our case, we found that a = 16/x⁴. To verify this, we can substitute this value back into the original equation and see if both sides are indeed equal. Our original equation was:
√((126xy⁵)/(32x³)) = √((63y⁵)/(ax⁶))
Let's substitute a = 16/x⁴ into the right-hand side:
√((63y⁵)/((16/x⁴)x⁶))
Now, we need to simplify this expression. Dividing by a fraction is the same as multiplying by its reciprocal, so we can rewrite the denominator as:
(16/x⁴)x⁶ = 16x⁶/x⁴
Using the rule for dividing exponents (x^m / x^n = x^(m-n)), we get:
16x⁶/x⁴ = 16x²
So, the right-hand side now looks like:
√((63y⁵)/(16x²))
We've already simplified the left-hand side of the equation, and we found it to be:
(3y²√(7y)) / (4x)
Let's see if we can simplify the right-hand side to match this. We can split the radical:
√(63y⁵) / √(16x²)
We know that √(63y⁵) = 3y²√(7y) and √(16x²) = 4x. So, the right-hand side becomes:
(3y²√(7y)) / (4x)
Lo and behold! This is exactly the same as the simplified left-hand side. This confirms that our solution, a = 16/x⁴, is indeed correct. We've successfully verified our answer by substituting it back into the original equation and showing that both sides are equal. This gives us a high degree of confidence in our solution. In summary, to verify our solution, we substituted the value we found for 'a' back into the original equation. We then simplified both sides of the equation separately. If the simplified expressions are the same, then our solution is verified. If they are different, it means we made a mistake somewhere along the way, and we need to go back and check our work. Verification is a crucial step in problem-solving, especially in mathematics. It helps us ensure the accuracy of our results and deepens our understanding of the concepts involved. By taking the time to verify our solutions, we become more confident and proficient mathematicians.
Conclusion
In conclusion, we successfully determined the value of 'a' that makes the two radical expressions equivalent. We achieved this by systematically simplifying both expressions, equating them, and then solving for 'a'. We found that a = 16/x⁴. We also verified our solution by substituting it back into the original equation and confirming that both sides were equal. This problem highlights the importance of several key mathematical concepts, including simplifying radical expressions, using exponent rules, and solving equations. Mastering these skills is essential for success in algebra and beyond. Remember, simplification is often the key to solving complex problems. By breaking down expressions into their simplest forms, we can reveal their underlying structure and make comparisons easier. Exponent rules are powerful tools for manipulating variables and their powers. Understanding these rules allows us to combine or separate terms, making it easier to isolate the variable we're trying to solve for. Solving equations is a fundamental skill in mathematics. It involves using algebraic techniques to isolate the unknown variable and find its value. Practice is essential for mastering these skills. The more problems you solve, the more comfortable you'll become with the different techniques and strategies involved. Don't be afraid to make mistakes – they're a natural part of the learning process. When you make a mistake, take the time to understand why it happened and how to avoid it in the future. Mathematics is a journey of discovery, and each problem you solve is a step forward. Keep exploring, keep questioning, and keep practicing! And for further exploration of radical expressions and equation solving, check out resources like Khan Academy for additional practice problems and video explanations. Happy problem-solving!
