Solving Equations: How Many Solutions Exist?

by Dimemap Team 45 views

Hey guys! Let's dive into a cool algebra problem today. We're going to figure out how many solutions a tricky equation has and, more importantly, what those solutions actually are. Equations can seem intimidating at first, but breaking them down step-by-step makes the whole process much easier. So, grab your thinking caps, and let's get started!

Understanding the Equation

The equation we're tackling today is this beast:

x(x−1)(x−2)(x−3)(x−4)(x+1)(x−2)(x2−9)=2025(x−1)(x−2)(x−3)(x−4)(x+1)(x−2)(x2−9)\frac{x(x - 1)(x - 2)(x - 3)(x - 4)}{(x + 1)(x - 2)(x^2 - 9)} = \frac{2025(x - 1)(x - 2)(x - 3)(x - 4)}{(x + 1)(x - 2)(x^2 - 9)}

Okay, it looks pretty wild, right? But don't worry, we'll tame it. The first thing we need to do is really understand what we're looking at. We've got fractions, we've got 'x' all over the place, and we've got some parentheses action happening. Our main goal is to find out what values of 'x' will make this equation true. That's what solving an equation is all about!

Before we start crunching numbers, let's take a closer look at the structure of the equation. Notice that we have fractions on both sides. A key strategy when dealing with fractions in equations is to try and simplify them. This often involves canceling out common factors. But we need to be super careful when we do this, because sometimes canceling can hide solutions or create problems. We'll talk more about that in a bit.

Also, check out those parentheses! They're grouping terms together, which is a big hint that we might need to think about distribution or factoring. And that x^2 - 9 in the denominator? That should ring a bell – it looks like a difference of squares! Recognizing these patterns is a huge part of mastering algebra.

So, to recap, we're dealing with a fractional equation, we're looking for values of 'x' that make it true, and we've already spotted some opportunities for simplification. Now, let's get our hands dirty and start solving!

Simplifying the Equation

Alright, let's dive into simplifying this equation. Remember, our goal is to make it easier to work with while being careful not to lose any solutions along the way. The first thing that probably jumps out at you (and it should!) is that we've got a bunch of the same factors on both sides of the equation. Specifically, we see (x - 1), (x - 2), (x - 3), and (x - 4) all over the place. This is great news because it means we can simplify by dividing both sides by these common factors... with a major caveat.

Here's the catch: we can only divide by these factors if they're not equal to zero. Why? Because dividing by zero is a big no-no in math – it's undefined and will break our equation. So, before we go all-in on canceling, we need to consider what happens if any of those factors are zero. This is super important for finding all the solutions.

Let's think about each factor: (x - 1), (x - 2), (x - 3), and (x - 4). If any of these equals zero, then 'x' will have a specific value. For example, if (x - 1) = 0, then x = 1. Similarly, if (x - 2) = 0, then x = 2, and so on. These values (1, 2, 3, and 4) are potential solutions to our equation. We need to keep them in mind and check them later to see if they actually work.

Now, with that crucial point addressed, let's go ahead and divide both sides of the equation by those common factors. This gives us a much cleaner equation to work with:

x(x+1)(x2−9)=2025(x+1)(x2−9)\frac{x}{(x + 1)(x^2 - 9)} = \frac{2025}{(x + 1)(x^2 - 9)}

See? Much less scary! But we haven't forgotten about those potential solutions we identified earlier. They're still hanging out in the back of our minds.

Next up, let's tackle that (x^2 - 9) in the denominator. This is a classic difference of squares, which we can factor. Factoring is our friend when it comes to solving equations, so let's see how it helps us here.

Factoring and Further Simplification

Okay, let's continue simplifying our equation. We left off with:

x(x+1)(x2−9)=2025(x+1)(x2−9)\frac{x}{(x + 1)(x^2 - 9)} = \frac{2025}{(x + 1)(x^2 - 9)}

As we noted before, that (x^2 - 9) is begging to be factored. Remember the difference of squares pattern? It says that a^2 - b^2 can be factored into (a + b)(a - b). In our case, x^2 - 9 is the same as x^2 - 3^2, so we can factor it into (x + 3)(x - 3). Let's do that:

x(x+1)(x+3)(x−3)=2025(x+1)(x+3)(x−3)\frac{x}{(x + 1)(x + 3)(x - 3)} = \frac{2025}{(x + 1)(x + 3)(x - 3)}

Now our equation is looking even more interesting! We've got a fully factored denominator on both sides. Notice that we now have (x + 1), (x + 3), and (x - 3) in the denominator. Just like before, we need to be careful about these factors. They can't be equal to zero, because that would make our fractions undefined. So, let's figure out what values of 'x' would make them zero.

  • If (x + 1) = 0, then x = -1
  • If (x + 3) = 0, then x = -3
  • If (x - 3) = 0, then x = 3

These values (-1, -3, and 3) are also potential troublemakers. They could make the denominator zero, which means they can't be valid solutions to our equation. We'll need to keep them in mind as we move forward.

Now, let's get back to simplifying. We've got the same denominator on both sides of the equation, which is awesome! It means we can multiply both sides by that entire denominator to get rid of the fractions. This will make our equation much easier to handle. So, let's multiply both sides by (x + 1)(x + 3)(x - 3):

x=2025x = 2025

Wow! That's a huge simplification. We've gone from a complicated fractional equation to a simple statement: x = 2025. This looks like a solution, but we're not done yet. We need to remember all those potential troublemakers we identified earlier.

Checking for Extraneous Solutions

Alright, we've arrived at a possible solution: x = 2025. But in the world of equations, especially those with fractions, things aren't always as they seem. We need to do a crucial step called checking for extraneous solutions. Extraneous solutions are values that we find through our solving process, but they don't actually work in the original equation. They're like imposters pretending to be solutions!

Why do extraneous solutions happen? They often pop up when we perform operations that aren't reversible, like squaring both sides of an equation or, as in our case, multiplying or dividing by expressions that could be zero. Remember how we carefully noted the values that would make our denominators zero? This is where that pays off.

Let's recap the potential troublemakers we found:

  • x = 1, x = 2, x = 3, x = 4 (from canceling common factors in the numerator)
  • x = -1, x = -3, x = 3 (from the denominator)

We need to make sure that our solution, x = 2025, doesn't fall into this list of values that would make the original equation undefined. Thankfully, 2025 is nowhere near these values! It's much bigger and doesn't cause any problems with our denominators.

But what about those other values we found (1, 2, 3, 4, -1, -3)? These are the ones we really need to worry about. They came from factors that we canceled or denominators that could be zero. To determine if they are solutions, we would need to substitute each one back into the original equation. If they make the equation true, they're solutions. If they make the equation undefined (like by causing division by zero), they're not solutions.

In our case, we already know that x = 2, x = -1, x = 3 will lead to division by zero, because we had those factors in the denominator. So, those are definitely not solutions. We would need to test 1, 4, and -3 by plugging them into the original equation to see if they work.

Determining the Number of Solutions

Okay, we've done a lot of work! We simplified the equation, found a potential solution, and considered the possibility of extraneous solutions. Now, let's bring it all together and figure out how many solutions our equation actually has.

We found that x = 2025 is a valid solution because it doesn't make any of the denominators zero. That's one solution in the bag!

Then, we identified some values (1, 2, 3, 4, -1, -3) that we needed to check carefully. We already know that 2, -1, and 3 are not solutions because they make the denominator zero. We would need to plug in 1, 4, and -3 into the original equation to check if they are valid solutions.

Let's consider what happens if we plug in x = 1 into the original equation:

1(1−1)(1−2)(1−3)(1−4)(1+1)(1−2)(12−9)=2025(1−1)(1−2)(1−3)(1−4)(1+1)(1−2)(12−9)\frac{1(1 - 1)(1 - 2)(1 - 3)(1 - 4)}{(1 + 1)(1 - 2)(1^2 - 9)} = \frac{2025(1 - 1)(1 - 2)(1 - 3)(1 - 4)}{(1 + 1)(1 - 2)(1^2 - 9)}

Notice that we have a (1 - 1) term, which is zero, in both the numerator on the left and the numerator on the right. This means both sides of the equation become zero, so x = 1 is a solution!

Similarly, if we plug in x = 4, we get a (4-4) term, which makes both sides zero, so x = 4 is also a solution.

However, if we plug in x = -3, the denominator (x^2 - 9) becomes zero, so x = -3 is not a solution.

So, after all that work, we've found that our equation has three solutions: x = 1, x = 4, and x = 2025.

Conclusion

Woohoo! We made it through that equation! It was a bit of a journey, but we learned some super important things along the way. We saw how to simplify fractional equations, how to factor expressions, and, most importantly, how to check for those sneaky extraneous solutions. Remember, solving equations is like detective work – you need to be careful, methodical, and always double-check your answers!

So, the final answer is: the equation has three solutions, and they are x = 1, x = 4, and x = 2025. Great job, guys! Keep practicing, and you'll become equation-solving masters in no time!