Simplify Radical Expressions: Step-by-Step Guide

by Kenji Nakamura 49 views

Hey there, math enthusiasts! Let's dive into simplifying this radical expression: 128x5y62x7y5\sqrt{\frac{128 x^5 y^6}{2 x^7 y^5}}. Our goal is to find an equivalent expression from the given options. Don't worry, we'll break it down step by step to make sure everyone's on board. Before we start, remember the golden rules: x>0x > 0 and y>0y > 0. This means we're dealing with positive numbers, which simplifies things quite a bit.

Step 1: Simplify the Fraction Inside the Square Root

The first thing we need to tackle is the fraction inside the square root. We've got 128x5y62x7y5\frac{128 x^5 y^6}{2 x^7 y^5}. Let's simplify the numerical part and then handle the variables. So, when you first look at the expression, it might seem a bit intimidating with all those exponents and variables. But trust me, we're going to break it down into manageable chunks, and by the end, you'll be like, "Oh, that's all there is to it?" We'll start by simplifying the numerical part of the fraction, which is 1282\frac{128}{2}. What's 128 divided by 2, guys? That's right, it's 64. So, we can rewrite our fraction as 64x5y6x7y5\frac{64 x^5 y^6}{x^7 y^5}. Now, let's move on to the variables. Remember the rule for dividing exponents with the same base? You subtract the exponents. For xx, we have x5x^5 in the numerator and x7x^7 in the denominator. So, we subtract the exponents: 5−7=−25 - 7 = -2. That means we'll have x−2x^{-2}, which is the same as 1x2\frac{1}{x^2}. For yy, we have y6y^6 in the numerator and y5y^5 in the denominator. Subtracting the exponents, we get 6−5=16 - 5 = 1. So, we have y1y^1, which is just yy. Putting it all together, we've simplified the fraction inside the square root to 64yx2\frac{64y}{x^2}. See? We're making progress already!

Numerical Part

Let's start with the numbers: 1282\frac{128}{2}. This simplifies to 64. Easy peasy!

Variable Part: x

Now, let's look at the xx terms: x5x7\frac{x^5}{x^7}. Remember the rule for dividing exponents with the same base? You subtract the exponents: x5−7=x−2x^{5-7} = x^{-2}. A negative exponent means we can rewrite this as 1x2\frac{1}{x^2}. So, when you see those exponents, don't sweat it! Just remember the rules. Dividing exponents with the same base? Subtract 'em. A negative exponent? Flip it and make it positive. These are the little tricks that make simplifying expressions like this so much easier. And once you get the hang of it, you'll be cruising through these problems like a pro. We've tackled the numerical part and the x part, and now we're onto the y's. We're building up our simplified fraction piece by piece, and it's all coming together nicely. So, let's keep the momentum going and see what we can do with those y's!

Variable Part: y

Next up, the yy terms: y6y5\frac{y^6}{y^5}. Again, subtract the exponents: y6−5=y1y^{6-5} = y^1, which is just yy. Fantastic!

Putting It Together

Combining these simplifications, our fraction becomes:

64yx2\frac{64y}{x^2}

Step 2: Apply the Square Root

Now we have 64yx2\sqrt{\frac{64 y}{x^2}}. Remember, the square root applies to both the numerator and the denominator. So, we can think of this as 64yx2\frac{\sqrt{64y}}{\sqrt{x^2}}. Applying the square root might seem like the trickiest part, but it's really just about knowing your squares and using a few key properties of square roots. First, let's think about the numerator: 64y\sqrt{64y}. We can actually split this up into 64â‹…y\sqrt{64} \cdot \sqrt{y}. Why? Because the square root of a product is the product of the square roots. This is a super handy trick to remember. Now, what's the square root of 64? If you're thinking 8, you're spot on! So, 64\sqrt{64} is 8, and we have 8y8\sqrt{y} in the numerator. Next, let's look at the denominator: x2\sqrt{x^2}. This one's pretty straightforward. The square root of x2x^2 is just xx, since the square root and the square cancel each other out. So, now we have 8yx\frac{8\sqrt{y}}{x}. We've simplified the expression inside the square root, and we've applied the square root to both the numerator and the denominator. We're almost there, guys! We just need to take a look at our answer choices and see which one matches up with what we've got. So, let's do that now.

Square Root of the Numerator

Let's break down the numerator: 64y\sqrt{64y}. We can split this into 64â‹…y\sqrt{64} \cdot \sqrt{y}. The square root of 64 is 8, so we have 8y8\sqrt{y}. Remember that aâ‹…b=aâ‹…b\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}. This property is super useful!

Square Root of the Denominator

The denominator is x2\sqrt{x^2}. The square root and the square cancel each other out, leaving us with xx. So, whenever you see a variable squared inside a square root, you can just take the variable itself as the result. It's like a magic trick, but it's math!

Putting It Together Again

Our expression now looks like this:

8yx\frac{8 \sqrt{y}}{x}

Step 3: Match the Result with the Options

Our simplified expression is 8yx\frac{8 \sqrt{y}}{x}. Looking at the options, we can see that this matches option D.

So, let's recap what we've done. We started with a seemingly complex expression, 128x5y62x7y5\sqrt{\frac{128 x^5 y^6}{2 x^7 y^5}}, and we systematically simplified it step by step. We divided the numerical coefficients, subtracted the exponents for the variables, and then applied the square root to both the numerator and the denominator. And all that hard work paid off, because we arrived at the simplified expression 8yx\frac{8 \sqrt{y}}{x}, which matches option D. Remember, guys, math isn't about memorizing formulas or blindly following procedures. It's about understanding the underlying principles and applying them in a logical way. So, keep practicing, keep exploring, and keep having fun with it!

Final Answer

The equivalent expression is:

D. 8yx\frac{8 \sqrt{y}}{x}

So, there you have it! We've successfully navigated through the simplification of this radical expression. Remember, the key is to break it down into smaller, manageable steps. Simplify the fraction first, then apply the square root, and finally, match your result with the given options. You've got this!