Simplify The Radical Quotient

by Andrew McMorgan 30 views

Hey guys! Ever stumbled upon a math problem that looks like a jumble of roots and fractions and thought, "What in the world is this?" Well, you're in the right place! Today, we're diving deep into a problem that tests our skills in simplifying radical expressions. We've got this gnarly quotient: 6+115+3\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3}}. Our mission, should we choose to accept it, is to simplify this beast and find out which of the given options is the correct answer. Get ready to flex those mathematical muscles, because this is going to be a fun ride!

The Challenge: A Tangled Web of Roots

Let's break down the problem. We're asked to find the value of 6+115+3\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3}}. This isn't just about plugging numbers into a calculator; it's about understanding the properties of square roots and how to manipulate them. The presence of square roots in both the numerator and the denominator means we're likely looking at a problem that requires rationalizing the denominator. This is a standard technique when you have a radical in the denominator of a fraction, and it basically means we want to get rid of that radical down there to make the expression easier to work with and compare to our answer choices. Our goal is to transform the expression into a simpler form, ideally one that matches one of the options provided: A) 30+32+55+338\frac{\sqrt{30}+3 \sqrt{2}+\sqrt{55}+\sqrt{33}}{8}, B) 30βˆ’32+55βˆ’332\frac{\sqrt{30}-3 \sqrt{2}+\sqrt{55}-\sqrt{33}}{2}, C) 178\frac{17}{8}, or D) βˆ’52-\frac{5}{2}. Keep your eyes peeled, because we're going to go step-by-step, and by the end, we'll have this bad boy conquered.

Rationalizing the Denominator: Our Secret Weapon

Alright, so the main hurdle here is that pesky 5+3\sqrt{5}+\sqrt{3} in the denominator. To tackle this, we use the concept of conjugates. Remember that the conjugate of a binomial (a+b)(a+b) is (aβˆ’b)(a-b), and vice versa. The magic of conjugates is that when you multiply them, you get a difference of squares: (a+b)(aβˆ’b)=a2βˆ’b2(a+b)(a-b) = a^2 - b^2. In our case, the denominator is 5+3\sqrt{5}+\sqrt{3}. So, its conjugate is 5βˆ’3\sqrt{5}-\sqrt{3}. To keep our fraction balanced, whatever we multiply the denominator by, we must also multiply the numerator by. This ensures we're not changing the value of the original expression. So, let's multiply our fraction by 5βˆ’35βˆ’3\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}, which is just a fancy way of multiplying by 1.

Our expression now looks like this: 6+115+3Γ—5βˆ’35βˆ’3\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}.

Let's tackle the denominator first because that's where the magic of rationalization happens. Using the difference of squares formula, we get: (5+3)(5βˆ’3)=(5)2βˆ’(3)2=5βˆ’3=2(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2.

Boom! The denominator is now a nice, simple integer: 2. No more radicals down there, which is exactly what we wanted. This is a huge win, guys!

Conquering the Numerator: The FOIL Method in Action

Now, we need to deal with the numerator. This is where we multiply (6+11)(\sqrt{6}+\sqrt{11}) by (5βˆ’3)(\sqrt{5}-\sqrt{3}). This calls for the FOIL method (First, Outer, Inner, Last), which is just a systematic way of multiplying two binomials. Let's do it:

  • First: 6Γ—5=6Γ—5=30\sqrt{6} \times \sqrt{5} = \sqrt{6 \times 5} = \sqrt{30}
  • Outer: 6Γ—(βˆ’3)=βˆ’6Γ—3=βˆ’18\sqrt{6} \times (-\sqrt{3}) = -\sqrt{6 \times 3} = -\sqrt{18}
  • Inner: 11Γ—5=11Γ—5=55\sqrt{11} \times \sqrt{5} = \sqrt{11 \times 5} = \sqrt{55}
  • Last: 11Γ—(βˆ’3)=βˆ’11Γ—3=βˆ’33\sqrt{11} \times (-\sqrt{3}) = -\sqrt{11 \times 3} = -\sqrt{33}

Putting it all together, the numerator becomes: 30βˆ’18+55βˆ’33\sqrt{30} - \sqrt{18} + \sqrt{55} - \sqrt{33}.

Simplifying the Radicals: Don't Stop Now!

We're almost there, but we need to check if any of these radicals can be simplified further. Remember, a radical n\sqrt{n} can be simplified if nn has a perfect square factor. Let's look at each term:

  • 30\sqrt{30}: The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. None of these (other than 1) are perfect squares. So, 30\sqrt{30} is already in its simplest form.
  • 18\sqrt{18}: The factors of 18 are 1, 2, 3, 6, 9, 18. Aha! 9 is a perfect square (323^2). So, we can rewrite 18\sqrt{18} as 9Γ—2=9Γ—2=32\sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}.
  • 55\sqrt{55}: The factors of 55 are 1, 5, 11, 55. No perfect squares here. So, 55\sqrt{55} is simplified.
  • 33\sqrt{33}: The factors of 33 are 1, 3, 11, 33. No perfect squares. So, 33\sqrt{33} is simplified.

Now, let's substitute the simplified 18\sqrt{18} back into our numerator expression: 30βˆ’32+55βˆ’33\sqrt{30} - 3\sqrt{2} + \sqrt{55} - \sqrt{33}.

Putting It All Together: The Final Answer

We've successfully rationalized the denominator and simplified the numerator. Now we just need to combine our results. Our expression is:

SimplifiedΒ NumeratorSimplifiedΒ Denominator=30βˆ’32+55βˆ’332\frac{\text{Simplified Numerator}}{\text{Simplified Denominator}} = \frac{\sqrt{30} - 3\sqrt{2} + \sqrt{55} - \sqrt{33}}{2}.

Let's compare this to our answer choices:

A. 30+32+55+338\frac{\sqrt{30}+3 \sqrt{2}+\sqrt{55}+\sqrt{33}}{8} B. 30βˆ’32+55βˆ’332\frac{\sqrt{30}-3 \sqrt{2}+\sqrt{55}-\sqrt{33}}{2} C. 178\frac{17}{8} D. βˆ’52-\frac{5}{2}

It looks like our result, 30βˆ’32+55βˆ’332\frac{\sqrt{30} - 3\sqrt{2} + \sqrt{55} - \sqrt{33}}{2}, perfectly matches option B! We did it, team! High fives all around!

Why Rationalizing Matters: A Quick Recap

So, why go through all this trouble? Rationalizing the denominator might seem like an extra step, but it's crucial for a few reasons. First, it puts expressions into a standard form, making it easier to compare different results. Imagine trying to compare 12\frac{1}{\sqrt{2}} with 22\frac{\sqrt{2}}{2} if you didn't have a standard way to write them. Rationalizing helps us see that they are indeed the same. Second, it's often a requirement in mathematical contexts, like exams or when presenting solutions. Third, it can simplify calculations, especially when dealing with approximations or further algebraic manipulations. In this problem, rationalizing was the key to unlocking the simplified form that matched one of our options. Without it, we'd be stuck with radicals in the denominator, making it much harder to identify the correct answer. So, next time you see a radical in the denominator, remember the conjugate and the FOIL method – they're your best friends in the world of radical simplification!

This journey through simplifying radicals is a classic example of how fundamental algebraic techniques are applied to solve seemingly complex problems. By understanding concepts like conjugates and the properties of square roots, we can systematically break down challenging expressions into manageable parts. The problem required us to be meticulous with our calculations, especially when applying the FOIL method to the numerator and simplifying the terms afterward. The simplification of 18\sqrt{18} to 323\sqrt{2} was a key step that allowed our final expression to match option B. It’s a reminder that even small simplifications can make a big difference in the final form of an answer. So, keep practicing, stay curious, and don't shy away from those radical expressions – they're just waiting to be simplified!

Final Answer: The final answer is 30βˆ’32+55βˆ’332\boxed{\frac{\sqrt{30}-3 \sqrt{2}+\sqrt{55}-\sqrt{33}}{2}}