Simplifying Complex Radical Expressions: A Step-by-Step Guide

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Hey guys! In this guide, we're diving deep into the world of radical expressions. You know, those expressions with square roots, cube roots, and all sorts of roots! We're going to break down how to simplify some pretty complex-looking ones. If you've ever felt a little intimidated by these, don't worry – we'll take it step by step and make it super clear. So, grab your pencils, and let's get started!

Part 1: Simplifying (a) ((√3-1)/(√3+1) - (√3+1)/(√3-1)) ÷ √3

First up, we have this expression: ((√3-1)/(√3+1) - (√3+1)/(√3-1)) Γ· √3. It looks a bit scary, right? But trust me, we can handle this. The key here is to tackle the fraction within the parentheses first. We need to subtract these two fractions, which means finding a common denominator. Think of it like adding or subtracting regular fractions – same principle applies!

Finding a Common Denominator

The denominators we're dealing with are (√3 + 1) and (√3 - 1). The easiest way to find a common denominator is to multiply these two together. So, our common denominator will be (√3 + 1)(√3 - 1). Now, we need to rewrite each fraction with this new denominator. Remember, whatever we multiply the bottom by, we also have to multiply the top by – keeps things balanced, you know?

Let's rewrite the first fraction, (√3 - 1)/(√3 + 1). We multiply both the numerator and the denominator by (√3 - 1):

((√3 - 1) / (√3 + 1)) * ((√3 - 1) / (√3 - 1)) = ((√3 - 1)²) / ((√3 + 1)(√3 - 1))

Now, let's do the same for the second fraction, (√3 + 1)/(√3 - 1). We multiply both the numerator and the denominator by (√3 + 1):

((√3 + 1) / (√3 - 1)) * ((√3 + 1) / (√3 + 1)) = ((√3 + 1)²) / ((√3 - 1)(√3 + 1))

Expanding and Simplifying

Okay, now we have the fractions with a common denominator. Let's expand those numerators and denominators. Remember the formulas (a - b)Β² = aΒ² - 2ab + bΒ² and (a + b)Β² = aΒ² + 2ab + bΒ², and also the difference of squares: (a + b)(a - b) = aΒ² - bΒ².

For the first numerator, (√3 - 1)²:

(√3 - 1)² = (√3)² - 2(√3)(1) + 1² = 3 - 2√3 + 1 = 4 - 2√3

For the second numerator, (√3 + 1)²:

(√3 + 1)² = (√3)² + 2(√3)(1) + 1² = 3 + 2√3 + 1 = 4 + 2√3

For the common denominator, (√3 + 1)(√3 - 1):

(√3 + 1)(√3 - 1) = (√3)² - 1² = 3 - 1 = 2

Putting It All Together

So, now our expression looks like this:

((4 - 2√3) / 2) - ((4 + 2√3) / 2)

Since they have the same denominator, we can subtract the numerators:

(4 - 2√3) - (4 + 2√3) = 4 - 2√3 - 4 - 2√3 = -4√3

Now, we divide this by the common denominator, which is 2:

(-4√3) / 2 = -2√3

Don't Forget the Final Division!

We're not quite done yet! Remember, the original expression had a division by √3 at the end:

-2√3 ÷ √3 = -2

So, the simplified form of the first expression is -2. Woohoo! We did it!

Part 2: Simplifying (b) βˆ›(√2-1) Γ— ∜(3+2√2)

Next up, we've got this little number: βˆ›(√2-1) Γ— ⎬(3+2√2). This one involves cube roots and sixth roots, but don't let that scare you. The key here is to recognize patterns and see if we can rewrite the expressions inside the roots in a simpler form. Sometimes, what looks complicated can actually be simplified with a little algebraic magic.

Spotting the Pattern

Let's focus on the (3 + 2√2) part. Does that look familiar to anyone? It might not jump out at you right away, but this expression is actually a perfect square in disguise! We can rewrite it as (√2 + 1)². How cool is that?

To see why, let's expand (√2 + 1)²:

(√2 + 1)² = (√2)² + 2(√2)(1) + 1² = 2 + 2√2 + 1 = 3 + 2√2

See? It matches! This is a crucial step because it allows us to simplify the sixth root.

Rewriting the Expression

Now that we know 3 + 2√2 = (√2 + 1)², we can rewrite the original expression:

βˆ›(√2 - 1) Γ— ⎬((√2 + 1)Β²)

Now, let's deal with that sixth root. Remember, a sixth root is the same as raising something to the power of 1/6. So, we can rewrite the second term as:

((√2 + 1)²)^(1/6)

Using the power of a power rule (which says (am)n = a^(m*n)), we get:

(√2 + 1)^(2 * (1/6)) = (√2 + 1)^(1/3) = βˆ›(√2 + 1)

Putting It Together and Simplifying

Now our expression looks much cleaner:

βˆ›(√2 - 1) Γ— βˆ›(√2 + 1)

Since both terms are cube roots, we can combine them under a single cube root:

βˆ›((√2 - 1)(√2 + 1))

Now, we have that familiar difference of squares pattern again! Remember, (a - b)(a + b) = aΒ² - bΒ².

So, (√2 - 1)(√2 + 1) = (√2)² - 1² = 2 - 1 = 1

Therefore, our expression simplifies to:

βˆ›(1)

And the cube root of 1 is simply 1. So, the final simplified answer for this part is 1. Awesome!

Part 3: Simplifying (c) 1/(√11-√120) - 3/(√7-2√10)

Last but not least, we have this beauty: 1/(√11-√120) - 3/(√7-2√10). This one looks tricky because we have fractions with radicals in the denominator. But don't worry, we have a technique for this: it's called rationalizing the denominator. Sounds fancy, but it's really just a way of getting rid of those radicals in the bottom of the fraction.

Rationalizing the Denominator

The trick to rationalizing the denominator is to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate is just the same expression but with the opposite sign in the middle. For example, the conjugate of (a + b) is (a - b), and vice versa.

Let's start with the first fraction, 1/(√11 - √120). The conjugate of (√11 - √120) is (√11 + √120). So, we multiply both the top and bottom by (√11 + √120):

(1 / (√11 - √120)) * ((√11 + √120) / (√11 + √120)) = (√11 + √120) / ((√11 - √120)(√11 + √120))

Now, let's do the same for the second fraction, 3/(√7 - 2√10). The conjugate of (√7 - 2√10) is (√7 + 2√10). So, we multiply both the top and bottom by (√7 + 2√10):

(3 / (√7 - 2√10)) * ((√7 + 2√10) / (√7 + 2√10)) = (3(√7 + 2√10)) / ((√7 - 2√10)(√7 + 2√10))

Simplifying the Denominators

Now, let's simplify those denominators using the difference of squares pattern, (a - b)(a + b) = aΒ² - bΒ².

For the first fraction:

(√11 - √120)(√11 + √120) = (√11)² - (√120)² = 11 - 120 = -109

For the second fraction:

(√7 - 2√10)(√7 + 2√10) = (√7)² - (2√10)² = 7 - 4(10) = 7 - 40 = -33

Simplifying the Numerators

Now, let's simplify the numerators. The first numerator is already simple: (√11 + √120). For the second numerator, we just distribute the 3:

3(√7 + 2√10) = 3√7 + 6√10

Putting It All Together

Now our expression looks like this:

(√11 + √120) / (-109) - (3√7 + 6√10) / (-33)

This still looks a bit messy, but we're getting there! Let's rewrite the fractions to get rid of the negative signs in the denominators:

-(√11 + √120) / 109 + (3√7 + 6√10) / 33

Simplifying Radicals and Finding a Common Denominator

Before we combine these fractions, let's see if we can simplify any of the radicals. Notice that √120 can be simplified:

√120 = √(4 * 30) = 2√30

So, the first fraction becomes:

-(√11 + 2√30) / 109

Now, we need to find a common denominator for the two fractions. The least common multiple of 109 and 33 is 3597 (109 * 33). So, we need to rewrite each fraction with this new denominator.

For the first fraction, we multiply both the numerator and denominator by 33:

(-(√11 + 2√30) / 109) * (33/33) = (-33√11 - 66√30) / 3597

For the second fraction, we multiply both the numerator and denominator by 109:

((3√7 + 6√10) / 33) * (109/109) = (327√7 + 654√10) / 3597

Combining the Fractions

Now we can combine the fractions:

(-33√11 - 66√30 + 327√7 + 654√10) / 3597

Unfortunately, we can't simplify this expression any further because the radicals are all different. So, this is our final simplified answer:

(-33√11 - 66√30 + 327√7 + 654√10) / 3597

Conclusion

And there you have it! We've successfully simplified three pretty complex radical expressions. Remember, the key is to take it step by step, look for patterns, and don't be afraid to use those algebraic tricks we've learned. I hope this guide has been helpful, and now you feel a little more confident tackling those radical expressions. Keep practicing, and you'll become a pro in no time! Keep your math skills sharp, guys! You rock! πŸš€