Simplifying Radicals: A Step-by-Step Guide

by Andrew McMorgan 43 views

Hey guys! Today, we're diving into a fun math problem that's all about simplifying radicals. Specifically, we're tackling the expression $-10 \sqrt{12} \cdot 3 \sqrt{45}$. It might look a little intimidating at first, but trust me, breaking it down into smaller, manageable steps makes it a breeze. Ready to flex those math muscles? Let's jump in! We'll explore how to simplify radicals like a pro, making sure you understand every step of the process. This isn't just about finding the right answer; it's about understanding why that answer is correct. By the end of this guide, you'll be confident in your ability to simplify radical expressions, and maybe even impress your friends with your newfound skills. Let's get started!

Understanding the Basics of Simplifying Radicals

Before we get our hands dirty with the specific problem, let's quickly recap what simplifying radicals is all about. At its core, simplifying a radical means rewriting it in a form where the number under the radical sign (the radicand) is as small as possible, and there are no radicals left in the denominator. This usually involves finding perfect square factors within the radicand and extracting their square roots. For instance, consider 20\sqrt{20}. We can rewrite this as 4â‹…5\sqrt{4 \cdot 5}. Since 4 is a perfect square (4=2\sqrt{4} = 2), we can simplify this to 252\sqrt{5}. The key here is to look for perfect square factors like 4, 9, 16, 25, and so on. Also, remember that a radical is simply another way of expressing a fractional exponent. For instance, x\sqrt{x} is the same as x12x^{\frac{1}{2}}. Keeping these basics in mind will make the process of simplifying our original expression much smoother.

Now, let's break down the problem step-by-step. The initial expression is $-10 \sqrt{12} \cdot 3 \sqrt{45}$. The first thing we need to do is to focus on simplifying the radicals individually. Let's look at 12\sqrt{12} first. We can rewrite 12 as 4â‹…34 \cdot 3, and since 4 is a perfect square, we can rewrite 12\sqrt{12} as 4â‹…3=23\sqrt{4 \cdot 3} = 2\sqrt{3}. Next, let's simplify 45\sqrt{45}. We can rewrite 45 as 9â‹…59 \cdot 5, and since 9 is a perfect square, we can rewrite 45\sqrt{45} as 9â‹…5=35\sqrt{9 \cdot 5} = 3\sqrt{5}. So far, so good. We've simplified both radicals individually, and we're ready for the next step. Remember, the goal is to make these expressions as simple as possible before we combine everything together.

Step-by-Step Simplification of the Expression

Alright, let's get down to the nitty-gritty and work through the simplification step by step. We have the expression $-10 \sqrt{12} \cdot 3 \sqrt{45}$. As we discussed, the first step is to simplify the individual radicals. We found that 12=23\sqrt{12} = 2\sqrt{3} and 45=35\sqrt{45} = 3\sqrt{5}. Now we'll substitute these simplified forms back into the original expression. This gives us:

−10⋅(23)⋅3⋅(35)-10 \cdot (2\sqrt{3}) \cdot 3 \cdot (3\sqrt{5})

Next, we'll multiply the numbers outside the radicals together. That means we multiply -10, 2, and 3, and then multiply the terms involving the square roots. Multiplying the numbers, we get $-10 \cdot 2 \cdot 3 = -60$. Then, we multiply the remaining square roots, 3â‹…5\sqrt{3} \cdot \sqrt{5}. When multiplying square roots, you can combine them under a single radical sign. Thus, 3â‹…5=3â‹…5=15\sqrt{3} \cdot \sqrt{5} = \sqrt{3 \cdot 5} = \sqrt{15}. So, after multiplying, we have:

−6015-60 \sqrt{15}

This gives us our simplified expression. Now that we have simplified the expression, let's look at the multiple-choice options to find the correct answer.

Finding the Correct Answer Among the Options

We have simplified the original expression $-10 \sqrt{12} \cdot 3 \sqrt{45}$ to $-60 \sqrt{15}$. Now, we need to find the option that matches this result. Let's take a look at the given options:

A. $-180 \sqrt{15}$ B. $-7 \sqrt{33}$ C. $-1,080 \sqrt{15}$ D. $-30 \sqrt{15}$

Comparing our simplified answer ( $-60 \sqrt15}$) with the options, we can see that none of the given options exactly match our simplified result. However, there might have been a mistake during our calculations. Let's go back and carefully check the steps. We started with $-10 \sqrt{12} \cdot 3 \sqrt{45}$. First, simplify the radicals $\sqrt{12 = 2\sqrt3}$ and 45=35\sqrt{45} = 3\sqrt{5}. Substitute these values back into the expression $-10 \cdot 2\sqrt{3 \cdot 3 \cdot 3\sqrt5}$. Then, multiply the numbers outside the radicals $-10 \cdot 2 \cdot 3 \cdot 3 = -180$. Finally, multiply the radicals: $\sqrt{3 \cdot \sqrt{5} = \sqrt{15}$. Thus, the final simplified expression is $-180\sqrt{15}$. Now let's compare with the given options to find which one is equal to $-180\sqrt{15}$.

Comparing our revised simplified answer ( $-180 \sqrt{15}$) with the options, we can see that option A matches our final result. Therefore, the correct answer is option A, which is $-180 \sqrt{15}$. Therefore, the correct answer is A. It's crucial to go back and double-check each step. Careful calculation is key in these types of problems!

Conclusion

And that's a wrap, folks! We've successfully simplified the radical expression $-10 \sqrt{12} \cdot 3 \sqrt{45}$ and found the correct answer. The key takeaways from this problem are:

  • Simplify Radicals Individually: Always start by simplifying each radical separately by finding the largest perfect square factor.
  • Combine Numbers: After simplifying the radicals, multiply the numbers outside the radicals together.
  • Combine Radicals: Multiply the remaining radicals by combining them under a single radical sign.
  • Double-Check: Always double-check your work to avoid any silly mistakes. This is especially important during exams or quizzes.

By following these steps, you'll be well on your way to mastering radical simplification. Keep practicing, and you'll find that these problems become easier and more intuitive over time. Remember, the more you practice, the more comfortable you'll become. So, keep up the great work, and happy simplifying!

I hope this step-by-step guide has been helpful. Keep an eye out for more math tips and tricks! Bye for now!