Simplifying Radicals A Step By Step Solution For $\sqrt{60}(2 \sqrt{10}+3 \sqrt{2})$

In this comprehensive exploration, we delve into the intricacies of simplifying the mathematical expression $\sqrt{60}(2 \sqrt{10}+3 \sqrt{2})$. This problem, while seemingly straightforward, offers a fantastic opportunity to showcase the fundamental principles of simplifying radicals, applying the distributive property, and combining like terms. We will embark on a step-by-step journey, breaking down each operation with clarity and precision to arrive at the final solution. This article is designed to not only provide the answer but also to enhance your understanding of the underlying mathematical concepts. Whether you are a student grappling with radical expressions or simply a math enthusiast seeking to expand your knowledge, this guide will equip you with the tools and insights necessary to confidently tackle similar problems.

Initial Expression: $\sqrt{60}(2 \sqrt{10}+3 \sqrt{2})$

Our journey begins with the expression $\sqrt{60}(2 \sqrt{10}+3 \sqrt{2})$. To effectively simplify this, we need to first address the radical term $\sqrt{60}$. The key here is to identify the largest perfect square that divides 60. We can factorize 60 as $2 \times 2 \times 3 \times 5$, which can be rewritten as $2^2 \times 15$. Thus, we can express $\sqrt{60}$ as $\sqrt{2^2 \times 15}$. Applying the property of radicals that $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$, we get $\sqrt{2^2} \times \sqrt{15}$. Since $\sqrt{2^2} = 2$, the simplified form of $\sqrt{60}$ is $2\sqrt{15}$. This initial simplification is crucial as it allows us to work with smaller, more manageable radical terms. Remember, the ability to break down radicals into their simplest forms is a fundamental skill in algebra, and mastering this technique will significantly improve your problem-solving abilities in mathematics. By expressing $\sqrt{60}$ as $2\sqrt{15}$, we set the stage for applying the distributive property and further simplifying the expression. This first step is not just about getting a simpler form; it's about understanding the structure of numbers and how they interact within radical expressions. This foundational knowledge is what will allow us to tackle more complex problems in the future. With this simplification in hand, we are now ready to move on to the next stage of the process: applying the distributive property.

Applying the Distributive Property

Now that we've simplified $\sqrt{60}$ to $2\sqrt{15}$, the expression becomes $2\sqrt{15}(2 \sqrt{10}+3 \sqrt{2})$. The next step involves applying the distributive property, which states that $a(b + c) = ab + ac$. In our case, $a = 2\sqrt{15}$, $b = 2\sqrt{10}$, and $c = 3\sqrt{2}$. Applying the distributive property, we get:

$2\sqrt{15} \times 2\sqrt{10} + 2\sqrt{15} \times 3\sqrt{2}$.

This step is crucial because it expands the expression into a sum of two terms, each of which can be simplified individually. Let's consider the first term, $2\sqrt{15} \times 2\sqrt{10}$. When multiplying radical expressions, we multiply the coefficients (the numbers outside the square roots) and the radicands (the numbers inside the square roots) separately. Thus, $2 \times 2 = 4$ and $\sqrt{15} \times \sqrt{10} = \sqrt{15 \times 10} = \sqrt{150}$. So, the first term becomes $4\sqrt{150}$.

Now, let's look at the second term, $2\sqrt{15} \times 3\sqrt{2}$. Similarly, we multiply the coefficients: $2 \times 3 = 6$. Then, we multiply the radicands: $\sqrt{15} \times \sqrt{2} = \sqrt{15 \times 2} = \sqrt{30}$. So, the second term becomes $6\sqrt{30}$.

Putting these together, our expression is now $4\sqrt{150} + 6\sqrt{30}$. This expansion, achieved through the distributive property, has transformed the original expression into a form where we can further simplify the radical terms. The distributive property is a cornerstone of algebraic manipulation, and its application here demonstrates its power in handling expressions involving radicals. By carefully applying this property, we've set the stage for the final phase of simplification: reducing the radicals and combining like terms.

Simplifying Radicals and Combining Like Terms

Having expanded the expression to $4\sqrt{150} + 6\sqrt{30}$, we now focus on simplifying the radicals. Let's start with $\sqrt{150}$. We need to find the largest perfect square that divides 150. We can factorize 150 as $2 \times 3 \times 5 \times 5$, which can be rewritten as $5^2 \times 6$. Thus, $\sqrt{150}$ can be expressed as $\sqrt{5^2 \times 6}$. Applying the property $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$, we get $\sqrt{5^2} \times \sqrt{6}$. Since $\sqrt{5^2} = 5$, the simplified form of $\sqrt{150}$ is $5\sqrt{6}$.

Now, substitute this back into our expression: $4(5\sqrt{6}) + 6\sqrt{30}$. This simplifies to $20\sqrt{6} + 6\sqrt{30}$. Next, we examine $\sqrt{30}$. The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. There are no perfect square factors (other than 1) in this list. Therefore, $\sqrt{30}$ is already in its simplest form. Now we have $20\sqrt{6} + 6\sqrt{30}$. To determine if we can combine these terms, we need to check if the radicands are the same. In this case, we have $\sqrt{6}$ and $\sqrt{30}$, which are different. Therefore, we cannot combine these terms further.

The simplified expression is $20\sqrt{6} + 6\sqrt{30}$.

This final simplification showcases the importance of not only breaking down radicals but also recognizing when a radical is in its simplest form. We successfully reduced $\sqrt{150}$ to $5\sqrt{6}$, which allowed us to further simplify the expression. The inability to combine $20\sqrt{6}$ and $6\sqrt{30}$ highlights the rule that only terms with the same radicand can be combined. This step-by-step simplification, from the initial expression to the final result, provides a comprehensive understanding of how to manipulate and simplify radical expressions. It reinforces the fundamental principles of algebra and demonstrates the power of careful, methodical application of mathematical rules.

Final Solution and Summary

In conclusion, by meticulously following the steps of simplifying radicals, applying the distributive property, and combining like terms, we have successfully simplified the expression $\sqrt{60}(2 \sqrt{10}+3 \sqrt{2})$. Our journey began with identifying and simplifying $\sqrt{60}$ to $2\sqrt{15}$. We then applied the distributive property to expand the expression, which led us to $4\sqrt{150} + 6\sqrt{30}$. Next, we simplified $\sqrt{150}$ to $5\sqrt{6}$ and substituted it back into the expression, resulting in $20\sqrt{6} + 6\sqrt{30}$. Since $\sqrt{6}$ and $\sqrt{30}$ cannot be simplified further and are not like terms, we concluded that the final simplified form of the expression is $20\sqrt{6} + 6\sqrt{30}$.

This process underscores the significance of each step in simplifying mathematical expressions. It highlights the necessity of breaking down problems into manageable parts, applying relevant mathematical properties, and paying close attention to detail. The ability to simplify radicals is a crucial skill in algebra and beyond, and this exploration has provided a clear and comprehensive guide to mastering this skill. By understanding the underlying principles and practicing these techniques, you can confidently tackle a wide range of mathematical problems involving radicals. This example serves as a valuable learning tool, showcasing how to approach complex expressions with a structured and methodical approach, ultimately leading to a successful resolution. The final solution, $20\sqrt{6} + 6\sqrt{30}$, represents the culmination of these efforts and stands as a testament to the power of mathematical simplification.