Finding K Value: Solving Complex Equations
Hey guys! Today, let's dive into a fun math problem where we need to find the value of k
in a somewhat complex equation. Don't worry, we'll break it down step by step so it's super easy to follow. Our equation looks like this:
(y^2 - y)^2 + (y + 1)^2 + 2(1 - k) = y^4 - 2y^3 + (y + 1)^2 + y^2 - k
It might seem intimidating at first, but trust me, it's all about simplifying and rearranging terms. So, grab your pencils, and let's get started!
Simplifying the Equation
The first thing we're going to do is simplify both sides of the equation. This involves expanding the squared terms and combining like terms. Remember, the goal here is to make the equation less cluttered so we can isolate k
. Let’s begin by expanding (y^2 - y)^2
. This term can be expanded as (y^2 - y) * (y^2 - y)
. When we multiply this out, we get:
y^4 - y^3 - y^3 + y^2 = y^4 - 2y^3 + y^2
Now, let's look at the term (y + 1)^2
. Expanding this is more straightforward:
(y + 1)^2 = (y + 1) * (y + 1) = y^2 + y + y + 1 = y^2 + 2y + 1
Great! We've expanded the squared terms. Now, let's plug these expansions back into our original equation. This gives us:
y^4 - 2y^3 + y^2 + y^2 + 2y + 1 + 2(1 - k) = y^4 - 2y^3 + y^2 + 2y + 1 + y^2 - k
Notice that we’ve replaced (y^2 - y)^2
with y^4 - 2y^3 + y^2
and (y + 1)^2
with y^2 + 2y + 1
on both sides of the equation. This substitution makes the equation much clearer and sets us up for the next simplification steps. Guys, don't worry if it looks a bit long now; the magic happens when we start canceling out terms!
Now, let’s distribute the 2
in the term 2(1 - k)
on the left side of the equation. This gives us:
2 * 1 - 2 * k = 2 - 2k
So, our equation now looks like this:
y^4 - 2y^3 + y^2 + y^2 + 2y + 1 + 2 - 2k = y^4 - 2y^3 + y^2 + 2y + 1 + y^2 - k
Next, we'll combine like terms on both sides to further simplify the equation. On the left side, we can combine the y^2
terms and the constant terms. This gives us:
y^4 - 2y^3 + 2y^2 + 2y + 3 - 2k
On the right side, we also combine the y^2
terms, which gives us:
y^4 - 2y^3 + 2y^2 + 2y + 1 - k
So, our simplified equation is now:
y^4 - 2y^3 + 2y^2 + 2y + 3 - 2k = y^4 - 2y^3 + 2y^2 + 2y + 1 - k
This looks much cleaner, right? We've managed to get rid of the parentheses and group similar terms together. This simplified form will make it much easier to isolate k
and find its value. Remember, the key to solving complex equations is to take it one step at a time, and that’s exactly what we’re doing here!
Isolating k
Alright, now that we have a simplified equation, the next step is to isolate k
. This means we want to get all the terms involving k
on one side of the equation and everything else on the other side. Looking at our equation:
y^4 - 2y^3 + 2y^2 + 2y + 3 - 2k = y^4 - 2y^3 + 2y^2 + 2y + 1 - k
We can see that there are several terms that appear on both sides of the equation. These terms can be canceled out, which will make our equation even simpler. Let's start by canceling out the common terms. We have y^4
on both sides, so we can subtract y^4
from both sides. Similarly, -2y^3
appears on both sides, so we can add 2y^3
to both sides. The term 2y^2
also appears on both sides, so we subtract 2y^2
. Lastly, 2y
appears on both sides, so we subtract 2y
. After canceling these terms, our equation becomes:
3 - 2k = 1 - k
Wow, that’s a much simpler equation to work with! Now, let’s move the terms involving k
to one side of the equation. We can do this by adding 2k
to both sides:
3 - 2k + 2k = 1 - k + 2k
This simplifies to:
3 = 1 + k
Now, we want to isolate k
, so we need to get rid of the 1
on the right side. We can do this by subtracting 1
from both sides:
3 - 1 = 1 + k - 1
This simplifies to:
2 = k
So, we’ve found that k = 2
! That wasn't so bad, was it? By systematically simplifying the equation and isolating k
, we were able to find its value. Remember, the key to solving these types of problems is to take it one step at a time and stay organized. Now, let's verify our solution to make sure we've got it right.
Verifying the Solution
It's always a good idea to double-check our work, especially in math. To verify our solution, we'll substitute k = 2
back into the original equation and see if both sides of the equation are equal. Our original equation was:
(y^2 - y)^2 + (y + 1)^2 + 2(1 - k) = y^4 - 2y^3 + (y + 1)^2 + y^2 - k
Now, let's substitute k = 2
into the equation:
(y^2 - y)^2 + (y + 1)^2 + 2(1 - 2) = y^4 - 2y^3 + (y + 1)^2 + y^2 - 2
Let's simplify the left side first. We have 2(1 - 2)
, which is 2(-1) = -2
. So the left side becomes:
(y^2 - y)^2 + (y + 1)^2 - 2
We already know from our simplification steps that (y^2 - y)^2 = y^4 - 2y^3 + y^2
and (y + 1)^2 = y^2 + 2y + 1
. So, substituting these back in, we get:
y^4 - 2y^3 + y^2 + y^2 + 2y + 1 - 2
Combining like terms on the left side, we have:
y^4 - 2y^3 + 2y^2 + 2y - 1
Now let's simplify the right side of the equation. We have:
y^4 - 2y^3 + (y + 1)^2 + y^2 - 2
Again, we substitute (y + 1)^2 = y^2 + 2y + 1
:
y^4 - 2y^3 + y^2 + 2y + 1 + y^2 - 2
Combining like terms on the right side, we get:
y^4 - 2y^3 + 2y^2 + 2y - 1
Now, let's compare the simplified left side and the simplified right side:
Left side: y^4 - 2y^3 + 2y^2 + 2y - 1
Right side: y^4 - 2y^3 + 2y^2 + 2y - 1
They are equal! This confirms that our solution k = 2
is correct. Guys, remember, verifying your solution is a crucial step to ensure accuracy. It gives you confidence that you've solved the problem correctly and helps catch any potential mistakes.
Conclusion
So, there you have it! We successfully found the value of k
in the equation and verified our solution. The key takeaways from this problem are:
- Simplify: Always simplify the equation by expanding terms and combining like terms.
- Isolate: Isolate the variable you're trying to find by moving terms around.
- Verify: Always verify your solution by substituting it back into the original equation.
Math problems might seem daunting at first, but by breaking them down into smaller, manageable steps, they become much easier to tackle. Keep practicing, and you'll become a pro at solving these types of equations. Great job, guys! Keep up the awesome work!