Radical Calculations: Finding 'n' In Complex Expressions
Hey guys! Today, we're diving into the fascinating world of radical expressions and calculations. We've got some pretty interesting problems here where we need to calculate the value of 'n' in different expressions. These expressions involve radicals, and it's going to be a fun journey to simplify them and find our answers. So, let's buckle up and get started!
Understanding the Basics of Radical Expressions
Before we jump into the calculations, let's quickly recap what radical expressions are all about. At their core, radicals involve roots β square roots, cube roots, and so on. When we see a radical symbol (β), it means we're looking for a number that, when multiplied by itself (for a square root), or by itself twice (for a cube root), gives us the number under the radical. Understanding this fundamental concept is crucial for tackling the problems ahead. We need to be comfortable simplifying radicals, combining like terms, and performing arithmetic operations with them. Think of it as building the foundation for a mathematical skyscraper β a solid base ensures we can go high!
Furthermore, simplifying radicals often involves breaking down the number under the root into its prime factors. This allows us to identify perfect squares (or cubes, etc.) that can be taken out of the radical, making the expression simpler and easier to work with. For instance, β8 can be simplified to 2β2 because 8 can be written as 4 * 2, and β4 is 2. Mastering this technique is like learning a secret code that unlocks the true potential of radical expressions. Without this key, the problems can seem daunting, but with it, we can conquer even the most complex-looking radicals. We will apply these simplification techniques extensively in the following calculations.
In the context of our problems, we'll encounter expressions where radicals are added, subtracted, and multiplied. The order of operations (PEMDAS/BODMAS) is our trusty guide here, ensuring we perform the operations in the correct sequence. Remember, we can only add or subtract radicals if they have the same number under the radical sign (they are βlike termsβ). Think of it like adding apples and oranges β we canβt directly add them until we express them in a common unit (like βfruitsβ). Similarly, with radicals, we need to simplify them first and then combine the like terms. This careful, step-by-step approach is what will lead us to the correct solutions. So, letβs put on our mathematical detective hats and start unraveling these radical mysteries!
a) Calculating n = 2x - 5y
Breaking Down x = 2β192 + 4β48 - 5β147
Let's start with the first part of our problem, which is finding the value of 'x'. We're given that x = 2β192 + 4β48 - 5β147. The first thing we need to do is simplify these radicals. Remember our earlier discussion about breaking down numbers into their prime factors? That's exactly what we'll do here. Think of each radical as a puzzle waiting to be solved. We need to find the perfect squares hidden within these numbers.
For β192, we can break 192 down into 64 * 3. And guess what? 64 is a perfect square (8 * 8). So, β192 becomes β(64 * 3) = 8β3. See how we're already making progress? It's like peeling away the layers of an onion, revealing the simpler core. Next, let's tackle β48. We can break 48 into 16 * 3, and 16 is another perfect square (4 * 4). So, β48 becomes β(16 * 3) = 4β3. We're on a roll here! Finally, let's look at β147. This one can be broken down into 49 * 3, and 49 is a perfect square (7 * 7). Thus, β147 becomes β(49 * 3) = 7β3.
Now that we've simplified the radicals, we can rewrite our expression for 'x' as: x = 2(8β3) + 4(4β3) - 5(7β3). Notice how all the terms now have β3? This is fantastic because it means we can combine them. Think of β3 as a common unit, like apples. We can easily add and subtract βappleβ terms. So, letβs do the math: x = 16β3 + 16β3 - 35β3. Combining the terms, we get x = (16 + 16 - 35)β3 = -3β3. Awesome! We've found the simplified value of 'x'. It's like finding the missing piece of a jigsaw puzzle, and it fits perfectly!
Simplifying y = 2β108 - 3β75 + 4β27 - 5β12
Now, let's move on to the second part of this problem and find the value of 'y'. We're given that y = 2β108 - 3β75 + 4β27 - 5β12. Just like we did with 'x', our first step here is to simplify each of these radicals. Let's put on our simplifying hats again and get to work!
Starting with β108, we can break 108 down into 36 * 3, and 36 is a perfect square (6 * 6). So, β108 becomes β(36 * 3) = 6β3. Weβre getting good at this, aren't we? Next up is β75. We can break 75 into 25 * 3, and 25 is a perfect square (5 * 5). Therefore, β75 becomes β(25 * 3) = 5β3. Moving along to β27, we can break it down into 9 * 3, and 9 is a perfect square (3 * 3). So, β27 becomes β(9 * 3) = 3β3. Almost there! Lastly, let's tackle β12. We can break 12 into 4 * 3, and 4 is a perfect square (2 * 2). Thus, β12 becomes β(4 * 3) = 2β3.
Now that we've simplified all the radicals, we can rewrite the expression for 'y' as: y = 2(6β3) - 3(5β3) + 4(3β3) - 5(2β3). Just like with 'x', we notice that all terms have β3, which means we can combine them. This is like a symphony of simplification, where all the individual parts come together to create a harmonious whole. Let's do the math: y = 12β3 - 15β3 + 12β3 - 10β3. Combining the terms, we get y = (12 - 15 + 12 - 10)β3 = -β3. Fantastic! We've successfully found the simplified value of 'y'. It feels like weβve climbed another mountain and reached the summit!
Calculating n = 2x - 5y
Now that we've found the simplified values of both 'x' and 'y', we can finally calculate 'n'. We know that n = 2x - 5y, and we've determined that x = -3β3 and y = -β3. So, let's plug these values into the expression for 'n'. Itβs like fitting the last pieces of the puzzle together to reveal the complete picture.
Substituting the values, we get: n = 2(-3β3) - 5(-β3). Remember, a negative times a negative is a positive, so this becomes: n = -6β3 + 5β3. Now, we can combine the terms, just like we did before: n = (-6 + 5)β3 = -β3. And there you have it! We've successfully calculated the value of 'n' for the first part of our problem. Itβs a moment of mathematical triumph, like reaching the end of a challenging quest and claiming the treasure!
b) Calculating n = 5x - 3y
Simplifying x = 2β128 - 3β72 - 5β8 + 4β32
Let's move on to the second part of our problem, where we need to calculate 'n' using different expressions for 'x' and 'y'. This time, we're given that n = 5x - 3y, and we need to start by simplifying x = 2β128 - 3β72 - 5β8 + 4β32. Just like before, our first task is to simplify each of these radical terms. Think of it as a mathematical spring cleaning β we're decluttering the expressions to make them more manageable.
Let's start with β128. We can break 128 down into 64 * 2, and 64 is a perfect square (8 * 8). So, β128 becomes β(64 * 2) = 8β2. Next up, we have β72. We can break 72 into 36 * 2, and 36 is a perfect square (6 * 6). So, β72 becomes β(36 * 2) = 6β2. Moving on to β8, we can break it down into 4 * 2, and 4 is a perfect square (2 * 2). Thus, β8 becomes β(4 * 2) = 2β2. Finally, let's simplify β32. We can break 32 into 16 * 2, and 16 is a perfect square (4 * 4). So, β32 becomes β(16 * 2) = 4β2.
Now that we've simplified all the radicals, we can rewrite the expression for 'x' as: x = 2(8β2) - 3(6β2) - 5(2β2) + 4(4β2). Notice how all terms now have β2? This means we can combine them, just like we did in the previous problem. It's like organizing a messy drawer β once everything is sorted, it's much easier to work with. Let's do the math: x = 16β2 - 18β2 - 10β2 + 16β2. Combining the terms, we get x = (16 - 18 - 10 + 16)β2 = 4β2. Great! We've found the simplified value of 'x'. Itβs like discovering a hidden treasure map that leads us to the solution!
Simplifying y = 2β98 + 3β32 - 6β18
Next, we need to simplify the expression for 'y', which is given as y = 2β98 + 3β32 - 6β18. We'll use the same techniques we've been using β breaking down the numbers under the radicals to find perfect squares. Think of this as a mathematical workout, strengthening our simplification muscles.
Let's start with β98. We can break 98 down into 49 * 2, and 49 is a perfect square (7 * 7). So, β98 becomes β(49 * 2) = 7β2. Moving on to β32, we've already simplified this in the previous step (when we were working on 'x'), and we know that β32 = 4β2. Lastly, let's simplify β18. We can break 18 into 9 * 2, and 9 is a perfect square (3 * 3). Thus, β18 becomes β(9 * 2) = 3β2.
Now that we've simplified all the radicals, we can rewrite the expression for 'y' as: y = 2(7β2) + 3(4β2) - 6(3β2). All the terms have β2, so we can combine them. It's like assembling a team of radicals, each playing its part to reach the final answer. Let's do the math: y = 14β2 + 12β2 - 18β2. Combining the terms, we get y = (14 + 12 - 18)β2 = 8β2. Excellent! We've found the simplified value of 'y'. It feels like we've cracked another code, revealing the hidden value.
Calculating n = 5x - 3y
Now that we have the simplified values of 'x' and 'y', we can calculate 'n'. We know that n = 5x - 3y, and we've found that x = 4β2 and y = 8β2. Let's plug these values into the expression for 'n'. It's like putting the finishing touches on a masterpiece, bringing all the elements together.
Substituting the values, we get: n = 5(4β2) - 3(8β2). Multiplying, we have: n = 20β2 - 24β2. Now, we can combine the terms: n = (20 - 24)β2 = -4β2. And there we have it! We've successfully calculated the value of 'n' for the second part of our problem. Itβs a satisfying moment, like solving a challenging riddle and finally understanding the answer!
c) Calculating n = 8x + 5y
Simplifying x = 3β384 - 2β216 - 4β24
For the third part of our problem, we need to calculate 'n' where n = 8x + 5y. Let's start by simplifying the expression for 'x', which is given as x = 3β384 - 2β216 - 4β24. Our familiar strategy applies here: we'll break down the numbers under the radicals to find perfect square factors. Think of this as a mathematical treasure hunt β we're searching for the hidden perfect squares.
First, let's simplify β384. We can break 384 down into 64 * 6, and 64 is a perfect square (8 * 8). So, β384 becomes β(64 * 6) = 8β6. Moving on, let's simplify β216. We can break 216 down into 36 * 6, and 36 is a perfect square (6 * 6). So, β216 becomes β(36 * 6) = 6β6. Now, let's tackle β24. We can break 24 down into 4 * 6, and 4 is a perfect square (2 * 2). Thus, β24 becomes β(4 * 6) = 2β6.
Now that we've simplified the individual radicals, we can rewrite the expression for 'x' as: x = 3(8β6) - 2(6β6) - 4(2β6). Notice that all terms now have β6, which means we can combine them. Itβs like having a common currency β we can easily add and subtract these terms. Let's do the math: x = 24β6 - 12β6 - 8β6. Combining the terms, we get x = (24 - 12 - 8)β6 = 4β6. Excellent! We've found the simplified value of 'x'. Itβs like unlocking a secret level in a game, revealing new challenges and rewards.
The Missing 'y' and the Conclusion
Oops! It seems thereβs a slight problem here. The expression for 'y' is missing in the original problem statement for part (c). Without knowing the expression for 'y', we can't calculate the final value of 'n'. Itβs like having a recipe with a missing ingredient β we can't complete the dish without it.
However, we've successfully simplified the expression for 'x', which is a significant step forward. If we had the expression for 'y', we would follow the same process: simplify the radicals, combine like terms, and then substitute the values of 'x' and 'y' into the equation n = 8x + 5y. We've demonstrated the techniques and the logical flow required to solve these types of problems, so you'd be well-equipped to tackle the rest once you have the full information.
Final Thoughts
So, guys, we've journeyed through some complex radical calculations today! We've learned how to simplify radicals, combine like terms, and use these skills to find the value of 'n' in different expressions. Remember, the key is to break down the problems into smaller, manageable steps. Simplify each radical individually, and then combine the terms. It's like building a house β brick by brick, we create a solid structure. I hope this explanation has been helpful, and remember, practice makes perfect! Keep exploring the fascinating world of mathematics, and you'll be amazed at what you can achieve. If you have any questions or want to try more examples, feel free to ask. Keep up the great work, and see you next time!