Solving Expressions With Radicals: A Step-by-Step Guide

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Hey guys! Today, we're diving into the world of radical expressions and tackling some problems that might seem tricky at first glance. But don't worry, we'll break them down step by step so you can master them. We'll be working through three expressions involving multiplication and division of radicals. Let's get started!

Understanding Radical Expressions

Before we jump into solving, let's make sure we're all on the same page about radical expressions. A radical expression is simply an expression that contains a square root, cube root, or any other root. The key to simplifying these expressions lies in understanding how to manipulate the radicals themselves. Remember, the goal is often to express the radical in its simplest form, meaning that the number under the root has no perfect square factors (for square roots), perfect cube factors (for cube roots), and so on.

When you're dealing with radical expressions, it's super important to remember the rules for multiplying and dividing them. Here’s the gist: you can multiply or divide the numbers inside the square roots, but you gotta keep them separate from the numbers outside the square roots. Think of it like this: you're grouping similar things together. This helps in simplifying the expressions and getting to the right answer without a headache. Also, always try to simplify each radical before you start multiplying or dividing; it'll make the whole process way easier and less prone to mistakes. Trust me, your future self will thank you!

Key Concepts

  • Simplifying Radicals: Before performing any operations, simplify each radical as much as possible.
  • Multiplication: aΓ—b=aΓ—b\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}
  • Division: aΓ·b=ab\sqrt{a} \div \sqrt{b} = \sqrt{\frac{a}{b}}

Now, let's apply these concepts to our expressions.

Problem a: (56)Γ—(215)Γ·(155)(5 \sqrt{6}) \times (2 \sqrt{15}) \div (15 \sqrt{5})

Let's break down this expression step by step. Our main goal here is to simplify things by using the rules of radicals we just talked about. We'll start by focusing on multiplying the terms and then handle the division to get to our final, simplified answer. Keep an eye on how we combine the numbers inside and outside the square rootsβ€”that's where the magic happens!

Step 1: Multiply the First Two Terms

We'll start by multiplying (56)(5 \sqrt{6}) and (215)(2 \sqrt{15}). Remember, we multiply the numbers outside the radical and the numbers inside the radical separately.

So, we have:

(56)Γ—(215)=5Γ—2Γ—6Γ—15(5 \sqrt{6}) \times (2 \sqrt{15}) = 5 \times 2 \times \sqrt{6} \times \sqrt{15}

This simplifies to:

106Γ—15=109010 \sqrt{6 \times 15} = 10 \sqrt{90}

Step 2: Simplify the Radical

Now we need to simplify 90\sqrt{90}. We're looking for the largest perfect square that divides 90. We know that 90=9Γ—1090 = 9 \times 10, and 9 is a perfect square (323^2). So, we can rewrite the radical as:

90=9Γ—10=9Γ—10=310\sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \times \sqrt{10} = 3 \sqrt{10}

Substituting this back into our expression, we get:

1090=10Γ—310=301010 \sqrt{90} = 10 \times 3 \sqrt{10} = 30 \sqrt{10}

Step 3: Divide by the Last Term

Now we'll divide our result by (155)(15 \sqrt{5}):

3010155=3015Γ—105\frac{30 \sqrt{10}}{15 \sqrt{5}} = \frac{30}{15} \times \frac{\sqrt{10}}{\sqrt{5}}

This simplifies to:

2Γ—105=222 \times \sqrt{\frac{10}{5}} = 2 \sqrt{2}

Final Answer for Problem a

Therefore, (56)Γ—(215)Γ·(155)=22(5 \sqrt{6}) \times (2 \sqrt{15}) \div (15 \sqrt{5}) = 2 \sqrt{2}.

Problem b: (312)Γ—(βˆ’8)Γ·(66)(3 \sqrt{12}) \times (-\sqrt{8}) \div (6 \sqrt{6})

Okay, let's tackle the next one! This problem looks a little more complex, but we'll use the same methods. The key here is to stay organized and take things one step at a time. First up, we’re going to simplify the radicals where we can, then handle the multiplication and division. Remember, a negative sign in front of a radical or number just means we’re dealing with a negative value, so we need to keep track of those signs as we go.

Step 1: Simplify the Radicals

Before we multiply or divide, let's simplify 12\sqrt{12} and 8\sqrt{8}.

  • 12=4Γ—3=4Γ—3=23\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}
  • 8=4Γ—2=4Γ—2=22\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}

Now we can rewrite the expression as:

(3Γ—23)Γ—(βˆ’22)Γ·(66)=(63)Γ—(βˆ’22)Γ·(66)(3 \times 2 \sqrt{3}) \times (-2 \sqrt{2}) \div (6 \sqrt{6}) = (6 \sqrt{3}) \times (-2 \sqrt{2}) \div (6 \sqrt{6})

Step 2: Multiply the First Two Terms

Multiply (63)(6 \sqrt{3}) by (βˆ’22)(-2 \sqrt{2}):

(63)Γ—(βˆ’22)=6Γ—βˆ’2Γ—3Γ—2=βˆ’126(6 \sqrt{3}) \times (-2 \sqrt{2}) = 6 \times -2 \times \sqrt{3} \times \sqrt{2} = -12 \sqrt{6}

Step 3: Divide by the Last Term

Now divide our result by (66)(6 \sqrt{6}):

βˆ’12666=βˆ’126Γ—66\frac{-12 \sqrt{6}}{6 \sqrt{6}} = \frac{-12}{6} \times \frac{\sqrt{6}}{\sqrt{6}}

This simplifies to:

βˆ’2Γ—1=βˆ’2-2 \times 1 = -2

Final Answer for Problem b

Therefore, (312)Γ—(βˆ’8)Γ·(66)=βˆ’2(3 \sqrt{12}) \times (-\sqrt{8}) \div (6 \sqrt{6}) = -2.

Problem c: (βˆ’26)Γ—(148)Γ·(3412)(-2 \sqrt{6}) \times (\frac{1}{4} \sqrt{8}) \div (\frac{3}{4} \sqrt{12})

Alright, let's jump into our last problem! This one’s got some fractions in the mix, but don't sweat it, we'll handle them just like the radicals. We're sticking to our plan: simplify those radicals first, then multiply, and finally divide. Fractions might seem a bit scary at first, but they're just numbers too, and we've got this! Remember, keeping track of all the parts and staying organized is key to nailing these kinds of problems.

Step 1: Simplify the Radicals

Let's simplify 8\sqrt{8} and 12\sqrt{12}:

  • 8=4Γ—2=4Γ—2=22\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}
  • 12=4Γ—3=4Γ—3=23\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}

Now we can rewrite the expression as:

(βˆ’26)Γ—(14Γ—22)Γ·(34Γ—23)=(βˆ’26)Γ—(122)Γ·(323)(-2 \sqrt{6}) \times (\frac{1}{4} \times 2 \sqrt{2}) \div (\frac{3}{4} \times 2 \sqrt{3}) = (-2 \sqrt{6}) \times (\frac{1}{2} \sqrt{2}) \div (\frac{3}{2} \sqrt{3})

Step 2: Multiply the First Two Terms

Multiply (βˆ’26)(-2 \sqrt{6}) by (122)(\frac{1}{2} \sqrt{2}):

(βˆ’26)Γ—(122)=βˆ’2Γ—12Γ—6Γ—2=βˆ’1Γ—12(-2 \sqrt{6}) \times (\frac{1}{2} \sqrt{2}) = -2 \times \frac{1}{2} \times \sqrt{6} \times \sqrt{2} = -1 \times \sqrt{12}

Simplify 12\sqrt{12}:

12=4Γ—3=23\sqrt{12} = \sqrt{4 \times 3} = 2 \sqrt{3}

So our expression becomes:

βˆ’1Γ—23=βˆ’23-1 \times 2 \sqrt{3} = -2 \sqrt{3}

Step 3: Divide by the Last Term

Now divide our result by (323)(\frac{3}{2} \sqrt{3}):

βˆ’23323=βˆ’2Γ—23Γ—33\frac{-2 \sqrt{3}}{\frac{3}{2} \sqrt{3}} = -2 \times \frac{2}{3} \times \frac{\sqrt{3}}{\sqrt{3}}

This simplifies to:

βˆ’43Γ—1=βˆ’43-\frac{4}{3} \times 1 = -\frac{4}{3}

Final Answer for Problem c

Therefore, (βˆ’26)Γ—(148)Γ·(3412)=βˆ’43(-2 \sqrt{6}) \times (\frac{1}{4} \sqrt{8}) \div (\frac{3}{4} \sqrt{12}) = -\frac{4}{3}.

Final Thoughts

So there you have it! We've worked through three problems involving radical expressions, and hopefully, you're feeling a bit more confident about tackling these types of questions. Remember, the key is to simplify, multiply, and divide step by step. Keep practicing, and you'll become a pro in no time!