Rationalize The Denominator: 2/√3 Simplified!

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Alright, guys, let's dive into a common mathematical problem: rationalizing the denominator. Specifically, we're going to tackle the expression 23\frac{2}{\sqrt{3}}. You might be wondering, "What does 'rationalizing the denominator' even mean?" Well, it's all about getting rid of any square roots (or other radicals) from the bottom of a fraction. It's a standard practice in math to present our answers in a clean and simplified form, and having a radical in the denominator is generally frowned upon. So, how do we do it?

The main idea behind rationalizing the denominator is to multiply the fraction by a clever form of '1'. This means we multiply both the numerator (the top part of the fraction) and the denominator (the bottom part) by the same value. The value we choose is specifically designed to eliminate the radical in the denominator. In our case, we have 3\sqrt{3} in the denominator. To get rid of this, we'll multiply both the numerator and the denominator by 3\sqrt{3}. This works because 3∗3=3\sqrt{3} * \sqrt{3} = 3, which is a nice, rational number (no square root!). Let's walk through the steps:

  1. Start with the original expression: 23\frac{2}{\sqrt{3}}
  2. Multiply the numerator and denominator by 3\sqrt{3}: 23∗33\frac{2}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}}
  3. Multiply the numerators: 2∗3=232 * \sqrt{3} = 2\sqrt{3}
  4. Multiply the denominators: 3∗3=3\sqrt{3} * \sqrt{3} = 3
  5. Combine the results: 233\frac{2\sqrt{3}}{3}

And there you have it! The expression 23\frac{2}{\sqrt{3}} has been rationalized to 233\frac{2\sqrt{3}}{3}. Notice that the denominator is now a rational number (3), and we've successfully eliminated the square root from the bottom of the fraction. This is the simplified form we're looking for. Remember, the key is to identify the radical in the denominator and multiply both the numerator and denominator by that radical. This technique works for simple square roots, but what about more complex denominators? Don't worry; we'll get to those in a bit!

Advanced Rationalization Techniques

Okay, so you've mastered the basic rationalization. What if the denominator is a bit more complicated? What if it involves a sum or difference with a square root? For example, consider something like 11+2\frac{1}{1 + \sqrt{2}}. We can't just multiply by 2\sqrt{2} in this case, because that would only get rid of the square root term; it wouldn't eliminate the radical completely from the denominator. This is where the concept of a conjugate comes in handy. The conjugate of an expression like a+ba + b is simply a−ba - b. Similarly, the conjugate of a−ba - b is a+ba + b. The magic of conjugates is that when you multiply an expression by its conjugate, you eliminate the square root. Let’s see how this works with our example:

  1. Start with the expression: 11+2\frac{1}{1 + \sqrt{2}}
  2. Identify the conjugate of the denominator: The conjugate of 1+21 + \sqrt{2} is 1−21 - \sqrt{2}
  3. Multiply the numerator and denominator by the conjugate: 11+2∗1−21−2\frac{1}{1 + \sqrt{2}} * \frac{1 - \sqrt{2}}{1 - \sqrt{2}}
  4. Multiply the numerators: 1∗(1−2)=1−21 * (1 - \sqrt{2}) = 1 - \sqrt{2}
  5. Multiply the denominators: (1+2)∗(1−2)=1∗1−1∗2+2∗1−2∗2=1−2+2−2=1−2=−1(1 + \sqrt{2}) * (1 - \sqrt{2}) = 1*1 - 1*\sqrt{2} + \sqrt{2}*1 - \sqrt{2}*\sqrt{2} = 1 - \sqrt{2} + \sqrt{2} - 2 = 1 - 2 = -1
  6. Combine the results: 1−2−1\frac{1 - \sqrt{2}}{-1}
  7. Simplify: −(1−2)=−1+2=2−1-(1 - \sqrt{2}) = -1 + \sqrt{2} = \sqrt{2} - 1

So, 11+2\frac{1}{1 + \sqrt{2}} rationalizes to 2−1\sqrt{2} - 1. Notice how multiplying by the conjugate eliminated the square root from the denominator, leaving us with a rational number (-1, which we then used to simplify the entire expression).

Let's do another example to solidify this concept. Suppose we have 32−5\frac{3}{2 - \sqrt{5}}.

  1. Start with the expression: 32−5\frac{3}{2 - \sqrt{5}}
  2. Identify the conjugate of the denominator: The conjugate of 2−52 - \sqrt{5} is 2+52 + \sqrt{5}
  3. Multiply the numerator and denominator by the conjugate: 32−5∗2+52+5\frac{3}{2 - \sqrt{5}} * \frac{2 + \sqrt{5}}{2 + \sqrt{5}}
  4. Multiply the numerators: 3∗(2+5)=6+353 * (2 + \sqrt{5}) = 6 + 3\sqrt{5}
  5. Multiply the denominators: (2−5)∗(2+5)=2∗2+2∗5−5∗2−5∗5=4+25−25−5=4−5=−1(2 - \sqrt{5}) * (2 + \sqrt{5}) = 2*2 + 2*\sqrt{5} - \sqrt{5}*2 - \sqrt{5}*\sqrt{5} = 4 + 2\sqrt{5} - 2\sqrt{5} - 5 = 4 - 5 = -1
  6. Combine the results: 6+35−1\frac{6 + 3\sqrt{5}}{-1}
  7. Simplify: −(6+35)=−6−35-(6 + 3\sqrt{5}) = -6 - 3\sqrt{5}

Thus, 32−5\frac{3}{2 - \sqrt{5}} rationalizes to −6−35-6 - 3\sqrt{5}.

When Rationalizing Matters

Now, you might be thinking, "Okay, I know how to rationalize, but why do I need to know this?" That's a valid question! Rationalizing the denominator is important for several reasons:

  • Simplifying Expressions: It presents mathematical expressions in a cleaner, more standard form, making them easier to understand and work with.
  • Combining Fractions: When adding or subtracting fractions, having rational denominators makes it much simpler to find a common denominator.
  • Evaluating Expressions: It can make it easier to approximate the value of an expression. For example, it's easier to estimate 22\frac{\sqrt{2}}{2} than 12\frac{1}{\sqrt{2}}.
  • Further Calculations: In more advanced mathematics, rationalizing the denominator can be a necessary step in solving equations or simplifying complex formulas.

Practice Problems

To really master rationalizing the denominator, practice is key! Here are a few problems for you to try:

  1. 57\frac{5}{\sqrt{7}}
  2. 111\frac{1}{\sqrt{11}}
  3. 43+2\frac{4}{3 + \sqrt{2}}
  4. 25−3\frac{2}{5 - \sqrt{3}}
  5. 35−2\frac{\sqrt{3}}{\sqrt{5} - \sqrt{2}}

Solutions to Practice Problems

Here are the solutions to the practice problems:

  1. 57=577\frac{5}{\sqrt{7}} = \frac{5\sqrt{7}}{7}
  2. 111=1111\frac{1}{\sqrt{11}} = \frac{\sqrt{11}}{11}
  3. 43+2=4(3−2)(3+2)(3−2)=12−429−2=12−427\frac{4}{3 + \sqrt{2}} = \frac{4(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} = \frac{12 - 4\sqrt{2}}{9 - 2} = \frac{12 - 4\sqrt{2}}{7}
  4. 25−3=2(5+3)(5−3)(5+3)=10+2325−3=10+2322=5+311\frac{2}{5 - \sqrt{3}} = \frac{2(5 + \sqrt{3})}{(5 - \sqrt{3})(5 + \sqrt{3})} = \frac{10 + 2\sqrt{3}}{25 - 3} = \frac{10 + 2\sqrt{3}}{22} = \frac{5 + \sqrt{3}}{11}
  5. 35−2=3(5+2)(5−2)(5+2)=15+65−2=15+63\frac{\sqrt{3}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{3}(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{\sqrt{15} + \sqrt{6}}{5 - 2} = \frac{\sqrt{15} + \sqrt{6}}{3}

Keep practicing, and you'll become a pro at rationalizing denominators in no time! Remember, the goal is to eliminate those radicals from the bottom of your fractions and present your answers in the simplest, cleanest form possible. Good luck, and happy calculating!