Mastering E(x): Simplify, Calculate E(√3), Prove Divisibility By 3

Hey math enthusiasts! Are you ready to dive deep into the fascinating world of algebraic expressions? Today, we're tackling a super interesting problem that will put your simplification, evaluation, and critical thinking skills to the test. We're going to break down the expression E(x) = (x√3 + 2)² - (x√3 + 4)(x√3 - 4) + (2x√3 - 1), where x is a real number. Our mission, should we choose to accept it, involves two main parts:

• Part A: Calculating the value of E(√3).
• Part B: Proving that E(n) is divisible by 3 for any natural number n.

This isn't just about crunching numbers, guys; it's about understanding the underlying principles of algebra and number theory. We'll explore how careful algebraic manipulation can turn a complex-looking beast into a simple, elegant expression, making subsequent calculations a breeze. Then, we'll hit a bit of a curveball with the divisibility proof, which will give us a chance to really flex our critical thinking muscles and appreciate the nuances of mathematical problem-solving. So, buckle up, grab your virtual pencils, and let's embark on this mathematical adventure together!

Unpacking the Algebraic Expression E(x): A Deep Dive into Simplification

Alright, folks, before we can calculate anything or prove divisibility, the absolute first and most crucial step in any complex algebraic problem is to simplify the expression. Trust me, trying to plug values directly into that original 'monster' of an equation would be an unnecessary headache. Our goal is to transform E(x) = (x√3 + 2)² - (x√3 + 4)(x√3 - 4) + (2x√3 - 1) into its simplest, most elegant form. This involves recalling a couple of fundamental algebraic identities that are super handy in these situations. Let's break it down term by term, ensuring we don't miss any pesky signs or coefficients.

First, let's look at the term (x√3 + 2)². This is a classic example of squaring a binomial, which follows the formula (a + b)² = a² + 2ab + b². Here, 'a' is x√3 and 'b' is 2. So, expanding this, we get:

• (x√3)² = x² * (√3)² = x² * 3 = 3x²
• 2 * (x√3) * 2 = 4x√3
• 2² = 4

Combining these, the first part becomes 3x² + 4x√3 + 4. Easy peasy, right?

Next up, we have -(x√3 + 4)(x√3 - 4). This term is a perfect candidate for the 'difference of squares' formula, which is (a + b)(a - b) = a² - b². In this case, 'a' is x√3 and 'b' is 4. Remember, the negative sign in front of the whole product is super important, so we'll apply it after we've expanded the product.

• (x√3)² = 3x² (just like before)
• 4² = 16

So, (x√3 + 4)(x√3 - 4) simplifies to (3x² - 16). Now, applying that negative sign that was lurking outside, we get -(3x² - 16) = -3x² + 16. See how that changes everything? Missing a sign can totally derail your entire solution, so stay sharp!

Finally, we have the simplest term: +(2x√3 - 1). There's no fancy expansion here, we just drop the parentheses since there's a positive sign in front. So, this term remains +2x√3 - 1.

Now that we've broken down each piece, it's time to put them all back together and see what cancels out or combines. Let's gather our simplified terms:

E(x) = (3x² + 4x√3 + 4) + (-3x² + 16) + (2x√3 - 1)

Let's group the like terms together:

• x² terms: 3x² - 3x² = 0x² (Woohoo! The x² terms cancel out completely, making our expression much simpler!)
• x√3 terms: 4x√3 + 2x√3 = 6x√3
• Constant terms: 4 + 16 - 1 = 20 - 1 = 19

And just like that, our seemingly complex expression E(x) simplifies beautifully to: E(x) = 6x√3 + 19. This is our golden ticket, folks! This simplified form will make the next two parts of our problem incredibly straightforward. Always remember, a little effort in simplification goes a long way in preventing errors and making your life easier down the line!

Part A – Calculating E(√3): Plugging in the Numbers for a Clear Answer

Alright, with our beautifully simplified expression E(x) = 6x√3 + 19 in hand, Part A becomes an absolute breeze. The question asks us to calculate E(√3). This means we simply need to substitute √3 wherever we see x in our simplified expression. Think of it like replacing a placeholder with a specific value. If you had tried to substitute √3 into the original, unsimplified expression, you'd be looking at a much more cumbersome and error-prone calculation involving multiple squares and products of terms with square roots. But thanks to our meticulous simplification, this step is now quick and clean.

Let's go step-by-step:

We have E(x) = 6x√3 + 19.

Now, substitute x = √3:

E(√3) = 6(√3)√3 + 19

Remember that when you multiply a square root by itself, you get the number inside the root. So, √3 * √3 = 3. This is a fundamental property of square roots that makes working with them so much easier.

E(√3) = 6(3) + 19

Now, perform the multiplication:

E(√3) = 18 + 19

And finally, the addition:

E(√3) = 37

There you have it! The value of E(√3) is 37. See how crucial our initial simplification was? It transformed what could have been a lengthy and potentially frustrating calculation into a matter of a few simple arithmetic steps. This really underscores the power and efficiency that algebraic manipulation brings to problem-solving. By investing time upfront in careful simplification, we saved ourselves a ton of effort and minimized the chances of making mistakes in the subsequent evaluation. It's like sharpening your tools before you start building something magnificent; the end result is always better and achieved with less struggle. This confidence in your simplified expression allows you to move forward with certainty, tackling the next part of the problem with a clear head. Great job, everyone!

Part B – Proving Divisibility by 3 for E(n): A Deep Dive into a Curious Twist

Alright folks, now for the grand finale, Part B: proving that E(n) is divisible by 3 for any natural number n. This is where things get super interesting, and we might uncover a little twist that highlights the importance of precise definitions in mathematics! We're tasked with demonstrating that our expression E(n) is always a multiple of 3 when 'n' is a natural number. Natural numbers are typically defined as positive integers (1, 2, 3, ...), and sometimes include zero, depending on the context. For our purposes, let's consider n = 1, 2, 3, and so on.

Remember our beautifully simplified expression, E(x) = 6x√3 + 19. When we substitute 'n' (a natural number) for 'x', we get:

E(n) = 6n√3 + 19

Now, here's the kicker, guys. For E(n) to be 'divisible by 3' in the standard mathematical sense, E(n) itself needs to be an integer, and the result of E(n)/3 should also be an integer. This is the fundamental definition of divisibility that we apply to whole numbers. However, let's scrutinize our expression E(n) = 6n√3 + 19.

Since 'n' is a natural number (meaning n is an integer greater than or equal to 1), the term 6n will always be an integer (e.g., if n=1, 6n=6; if n=2, 6n=12, and so on). Now, consider the term 6n√3. Because √3 is an irrational number (it cannot be expressed as a simple fraction, and its decimal representation goes on forever without repeating), multiplying an integer (like 6n) by an irrational number (like √3) results in an irrational number (unless 6n is zero, which only happens if n=0, but even then, 19 is not divisible by 3). Furthermore, adding an integer (like 19) to an irrational number results in... you guessed it, an irrational number.

This means that E(n) = 6n√3 + 19 will always be an irrational number for any natural number n.

So, strictly speaking, an irrational number cannot be