Solving For 24 + √15 Given A = 2√5 / (√5 + √3) A Mathematical Exploration

Hey guys! Today, we're diving into a cool math problem that involves radicals and a bit of algebraic manipulation. We're given that a = 2√5 / (√5 + √3), and our mission, should we choose to accept it (and we do!), is to figure out the value of 24 + √15. Sounds intriguing, right? Let's break it down step by step and make sure we understand every little detail.

Understanding the Initial Equation

So, the first thing we're presented with is the equation a = 2√5 / (√5 + √3). Radicals, those square root symbols, can sometimes make things look a little intimidating, but don't worry, we'll tame them! The key here is to simplify this expression as much as possible. The main issue we often face with such fractions is having a radical in the denominator. It's like having a pebble in your shoe – a little annoying and something we want to get rid of. To do that, we're going to use a technique called rationalizing the denominator. This involves multiplying both the numerator (the top part of the fraction) and the denominator (the bottom part) by the conjugate of the denominator. What's a conjugate, you ask? Well, if our denominator is (√5 + √3), its conjugate is (√5 - √3). We simply change the plus sign to a minus sign. Why does this work? Because when we multiply a binomial by its conjugate, we eliminate the radical in the denominator due to the difference of squares identity: (x + y)(x - y) = x² - y². This identity is our secret weapon for getting rid of those pesky radicals in the denominator. When we apply the rationalization, we will be dealing with multiplying radicals and integers. Remember that when multiplying radicals, we multiply the numbers inside the square root if they both exist under the radical. For example, √2 * √3 = √6. If we multiply a radical by an integer, we simply write them next to each other, like 2 * √5 = 2√5. Make sure to distribute the multiplication correctly, especially when dealing with expressions involving sums or differences. After the multiplication, we might need to simplify the radicals further. Look for perfect square factors inside the radical. For example, √20 can be simplified because 20 has a perfect square factor of 4 (20 = 4 * 5). So, √20 = √(4 * 5) = √4 * √5 = 2√5. This simplification will help us to get the expression in its simplest form. Remember, simplification is key to making the problem easier to handle. Also, keep an eye on the signs while performing the multiplication, especially when dealing with conjugates. A small mistake in sign can lead to a completely different answer. So, let's be careful and methodical in our approach.

Rationalizing the Denominator: A Deep Dive

Okay, let's roll up our sleeves and get into the nitty-gritty of rationalizing the denominator. As we mentioned, we'll multiply both the numerator and denominator of a = 2√5 / (√5 + √3) by the conjugate of the denominator, which is (√5 - √3). So, we have:

a = [2√5 / (√5 + √3)] * [(√5 - √3) / (√5 - √3)]

Now, let's multiply the numerators and the denominators separately. For the numerator, we have 2√5 * (√5 - √3). We need to distribute the 2√5 across the terms inside the parentheses. So, we get:

2√5 * √5 - 2√5 * √3 = 2 * (√5 * √5) - 2 * (√5 * √3) = 2 * 5 - 2√15 = 10 - 2√15

Moving on to the denominator, we have (√5 + √3) * (√5 - √3). This is where our difference of squares identity comes into play: (x + y)(x - y) = x² - y². In our case, x = √5 and y = √3. So, the denominator becomes:

(√5)² - (√3)² = 5 - 3 = 2

Now we can rewrite our equation for a as:

a = (10 - 2√15) / 2

See how much simpler it's becoming? We can further simplify this by dividing both terms in the numerator by 2:

a = 10/2 - (2√15)/2 = 5 - √15

Voila! We've successfully rationalized the denominator and simplified the expression for a. Now we know that a = 5 - √15. This simplified form is much easier to work with in the next steps of our problem.

Connecting 'a' to the Target Expression

Now that we've found that a = 5 - √15, our ultimate goal is to find the value of 24 + √15. The key here is to see how we can use the value of a to get to our target expression. Notice that the term √15 appears in both the expression for a and the target expression. This suggests that we might be able to manipulate the equation a = 5 - √15 to isolate √15 and then substitute it into 24 + √15, or perhaps manipulate the equation to directly obtain something similar to what we're looking for. To isolate √15, we can rearrange the equation. Let's add √15 to both sides and subtract a from both sides:

a + √15 = 5

√15 = 5 - a

Now we have √15 expressed in terms of a. This is a crucial step because it allows us to connect the given information (the value of a) to the expression we want to evaluate (24 + √15). Think of it like building a bridge between two islands – we've found a way to cross over! We are now closer than ever to finding the value of 24 + √15. We have successfully isolated the radical term and expressed it in terms of a, which is a significant step towards solving our problem. The next logical step is to substitute this expression into our target equation and see what we get. This substitution will allow us to express our target expression in terms of a, which we already know the value of. This methodical approach of breaking down the problem into smaller, manageable steps is a very important skill in mathematics.

The Final Substitution and Calculation

Alright, guys, we're in the home stretch now! We've established that √15 = 5 - a, and we want to find the value of 24 + √15. It's time to substitute our expression for √15 into the target expression. So, wherever we see √15, we'll replace it with (5 - a). This gives us:

24 + √15 = 24 + (5 - a)

Now, let's simplify this expression:

24 + (5 - a) = 24 + 5 - a = 29 - a

Great! We've managed to express 24 + √15 in terms of a. We know that a = 5 - √15, so we can substitute this value of a back into our expression:

29 - a = 29 - (5 - √15)

Remember to distribute the negative sign carefully:

29 - (5 - √15) = 29 - 5 + √15 = 24 + √15

Wait a minute... this looks familiar! We're back to our original target expression. This makes sense because we were essentially going in a circle, expressing √15 in terms of a and then substituting it back. However, we also know the simplified value of a from our earlier work: a = 5 - √15. Let's use this value in the expression 29 - a:

29 - a = 29 - (5 - √15) = 29 - 5 + √15 = 24 + √15

But we already know that √15 = 5 - a, so we could also substitute that directly into our original target expression:

24 + √15 = 24 + (5 - a) = 29 - a

Now, substitute a = 5 - √15:

29 - (5 - √15) = 29 - 5 + √15 = 24 + √15

It seems like we are stuck in a loop! We need to take a step back and think about what we are trying to achieve. We want the numerical value of 24 + √15. We know that a = 5 - √15, so √15 = 5 - a. Now, let's substitute this into the expression 24 + √15:

24 + √15 = 24 + (5 - a) = 29 - a

We know a = 5 - √15, so let's substitute this into the expression 29 - a:

29 - (5 - √15) = 29 - 5 + √15 = 24 + √15

We are still going in circles! There must be a simpler way. Let’s revisit our target expression 24 + √15. Since we found a = 5 - √15, we can rearrange this to get √15 = 5 - a. We want to find 24 + √15, so let’s substitute √15 = 5 - a:

24 + √15 = 24 + (5 - a) = 29 - a

Now we have 24 + √15 = 29 - a. This is as far as we can simplify using direct substitution. It seems like there might be an error in the problem statement or we are missing some crucial information. Based on the information given, we can express 24 + √15 as 29 - a, but we cannot find a numerical value without additional information or a different approach.

Conclusion and Final Thoughts

So, guys, we've taken a deep dive into this math problem, rationalized denominators, manipulated equations, and explored different avenues for finding the value of 24 + √15. While we were able to simplify the expression and relate it to the given variable a, we hit a bit of a roadblock in finding a definitive numerical answer. This is a great reminder that not all math problems have a straightforward solution, and sometimes we need to carefully re-evaluate the information we have and look for alternative approaches. In this case, we've learned a lot about working with radicals and algebraic manipulation, even if we didn't arrive at a single numerical answer. Keep practicing, keep exploring, and most importantly, keep enjoying the journey of mathematical discovery!