Solve For A, B, C In X = 1 + (2x + 2)^2

Hey math whizzes! Ever stumbled upon an equation that looks a bit wild and wondered how to tame it into something simpler? Well, guys, today we're diving deep into an expression that looks a bit tricky at first glance: $x = 1 + (2x + 2)^2$. Our mission, should we choose to accept it (and we totally should!), is to figure out what this beast is equivalent to in the standard quadratic form, $ax^2 + bx + c$. We're not just going to slap an answer down; we're going to break it down, show our work, and make sure everyone understands how we got there. So, buckle up, grab your calculators (or your trusty pens and paper), and let's get this done!

Unpacking the Expression: What's Really Going On?

Alright, let's talk about the initial expression: $x = 1 + (2x + 2)^2$. At first glance, it might seem like a simple linear equation because of the '$x$' on the left side. But then you spot that squared term, $(2x + 2)^2$, and you know things are about to get interesting. This squared binomial is the key player here. Remember your algebra basics, people! When you see something squared, it means you're multiplying it by itself. So, $(2x + 2)^2$ isn't just $4x^2 + 4$, oh no. We need to expand it properly using the FOIL method or the binomial expansion formula $(a+b)^2 = a^2 + 2ab + b^2$. Let's go with the expansion formula because it's super clean. Here, $a = 2x$ and $b = 2$.

Applying the formula, we get: $(2x + 2)^2 = (2x)^2 + 2(2x)(2) + (2)^2$

Let's break that down piece by piece:

• $(2x)^2$: This means $2x$ multiplied by $2x$. So, $(2 imes 2) imes (x imes x) = 4x^2$. Easy peasy!
• $2(2x)(2)$: This is just multiplication. $2 imes 2x = 4x$. Then, $4x imes 2 = 8x$. Got it!
• $(2)^2$: This is simply $2 imes 2 = 4$. Straightforward.

So, when we put it all together, $(2x + 2)^2$ expands to $4x^2 + 8x + 4$. Pretty neat, huh? Now, we just need to plug this back into our original equation. Remember, our original equation was $x = 1 + (2x + 2)^2$. We've just figured out that $(2x + 2)^2$ is $4x^2 + 8x + 4$. So, substitute that in:

$x = 1 + (4x^2 + 8x + 4)$

Now, we can simplify the right side by combining the constant terms. We have a '$1$' and a '$+ 4$'. That gives us '$+ 5$'. So the equation becomes:

$x = 4x^2 + 8x + 5$

We're getting closer, guys! The equation is starting to look more like a quadratic. But we're not quite at the $ax^2 + bx + c$ form yet. To get there, we need to have zero on one side of the equation. Usually, we aim to get all terms on the right side to keep the $x^2$ term positive, which is generally preferred. So, we need to move that '$x$' from the left side over to the right.

How do we do that? We subtract '$x$' from both sides of the equation. This is a fundamental rule in algebra: whatever you do to one side, you must do to the other to maintain equality.

So, let's subtract '$x$' from both sides:

$x - x = 4x^2 + 8x + 5 - x$

On the left side, $x - x$ is simply $0$. On the right side, we combine the '$x$' terms. We have '$8x$' and we're subtracting '$x$' (which is like subtracting $1x$). So, $8x - 1x = 7x$. The '$4x^2$' and the '$+ 5$' remain as they are because they don't have any other like terms to combine with.

This leaves us with:

$0 = 4x^2 + 7x + 5$

And there you have it! We've successfully transformed the original expression into the standard quadratic form $ax^2 + bx + c = 0$. The equation is now $4x^2 + 7x + 5 = 0$. This is equivalent to the original expression for all values of $x$.

Identifying the Coefficients: a, b, and c

Now that we've got our quadratic equation in the standard form $ax^2 + bx + c = 0$, the next step is super straightforward: identify the values of $a$, $b$, and $c$. These are simply the coefficients of the $x^2$ term, the $x$ term, and the constant term, respectively. It's like matching puzzle pieces, guys!

Our equation is: $4x^2 + 7x + 5 = 0$

Let's compare this to the general form $ax^2 + bx + c = 0$:

• The coefficient of the $x^2$ term ($a$): In our equation, the number multiplying $x^2$ is $4$. So, $a = 4$. This tells us that our quadratic opens upwards, if we were to graph it.
• The coefficient of the $x$ term ($b$): The number multiplying $x$ is $7$. So, $b = 7$. This coefficient influences the position and steepness of the parabola.
• The constant term ($c$): The term without any '$x$' is $5$. So, $c = 5$. This is the y-intercept of the parabola when graphed.

So, the values are $a=4$, $b=7$, and $c=5$. We did it! We took a seemingly complex equation and broke it down into its fundamental quadratic components. This process is super useful in algebra, whether you're trying to solve for the roots of an equation, understand the graph of a parabola, or simplify complex mathematical statements. Remember, persistence and breaking down problems into smaller steps are your best friends in math!

Why This Matters: The Power of Equivalence

Understanding that $x = 1 + (2x + 2)^2$ is equivalent to $4x^2 + 7x + 5 = 0$ is a big deal in mathematics, guys. Equivalence means that both expressions represent the exact same relationship between $x$ and the operations performed on it. They will yield the same solutions for $x$ if we were to solve them. This is the magic of algebraic manipulation. We can transform equations into different forms without changing their underlying truth or the values of the variables that satisfy them.

Why would we want to do this? Well, the original form $x = 1 + (2x + 2)^2$ isn't very helpful if you're trying to find specific values of $x$ that make the statement true. You can't easily isolate $x$, and it's not immediately obvious if it's a linear, quadratic, or some other type of equation. However, once we transform it into the standard quadratic form $ax^2 + bx + c = 0$, we unlock a whole toolbox of methods for solving it.

Think about it: with $4x^2 + 7x + 5 = 0$, we can now:

1. Use the Quadratic Formula: x = rac{-b itle{ ext{plus or minus }} itle{ ext{sqrt}}(b^2 - 4ac)}{2a}. This formula is a universal solver for any quadratic equation and will give us the exact values of $x$ that satisfy the equation.
2. Factor the Quadratic: If the quadratic is factorable (which this one isn't, easily), we could express it as $(px+q)(rx+s)=0$ and then solve for $x$ by setting each factor to zero.
3. Complete the Square: Another method to solve for $x$ or to rewrite the quadratic in vertex form, which reveals the parabola's vertex.
4. Graphing: We can graph the function $y = 4x^2 + 7x + 5$ and find where it intersects the x-axis (the roots or solutions).

So, the ability to convert an expression into a standard form like $ax^2 + bx + c$ is not just an academic exercise; it's a practical skill that makes complex problems solvable. It’s like translating a foreign language into your native tongue – suddenly, everything makes sense and you can interact with it more effectively. The coefficients $a, b,$ and $c$ are the essential parameters that define the behavior of the quadratic function, and knowing them is the first step to unlocking its secrets. So, remember this process, guys – it’s a fundamental building block for so much more in math!