Completing The Square Find The Intermediate Step

by Mei Lin 49 views

Hey guys! Today, we're diving deep into a fundamental concept in algebra: completing the square. Specifically, we're going to tackle the task of transforming a quadratic equation into the form (x+a)2=b(x+a)^2 = b. This form is super useful because it allows us to easily solve for x. We'll break down each step in detail, making sure you understand not just how to do it, but why it works. So, grab your pencils and paper, and let's get started!

Understanding the Goal: (x+a)2=b(x+a)^2 = b

Before we jump into the nitty-gritty, let's make sure we're all on the same page about our goal. The form (x+a)2=b(x+a)^2 = b represents a perfect square trinomial. Expanding the left side, we get:

(x+a)2=(x+a)(x+a)=x2+2ax+a2(x+a)^2 = (x+a)(x+a) = x^2 + 2ax + a^2

Notice the pattern here. The coefficient of the x term is 2a, and the constant term is a². This relationship is key to completing the square. Our objective is to manipulate the given quadratic equation so that the left side matches this perfect square pattern. Once we achieve this form, solving for x becomes a breeze – just take the square root of both sides and isolate x. We can visualize this process as fitting the pieces of a puzzle together, where the perfect square trinomial is the complete picture we're aiming for. This method isn't just a trick; it's a powerful technique rooted in algebraic principles that allows us to transform and solve equations in a systematic way. Understanding this underlying structure is crucial for mastering not only this specific type of problem but also more advanced algebraic concepts later on.

The Problem: x2+2x−108=−10x+3x^2 + 2x - 108 = -10x + 3

Okay, let's get to the problem at hand: x2+2x−108=−10x+3x^2 + 2x - 108 = -10x + 3. This equation is a classic example where completing the square comes in handy. Our first task is to rearrange the equation so that all the x terms are on one side and the constant terms are on the other. This is a crucial first step because it sets the stage for creating our perfect square trinomial. By grouping the x terms together, we can then focus on identifying the missing piece needed to complete the square. Similarly, isolating the constant terms on the other side simplifies the subsequent steps and makes the equation easier to manipulate. It's like organizing your workspace before starting a project – it makes the whole process smoother and more efficient. The goal here is to get the equation into a form where we can easily see the quadratic expression on one side, ready for transformation. This initial rearrangement is not just a matter of aesthetics; it's a strategic move that simplifies the algebraic manipulations that follow.

To do this, we'll add 10x10x to both sides and add 108108 to both sides. This gives us:

x2+2x+10x−108+108=−10x+10x+3+108x^2 + 2x + 10x - 108 + 108 = -10x + 10x + 3 + 108

Simplifying, we get:

x2+12x=111x^2 + 12x = 111

Now, we have a quadratic equation in a more manageable form. The x² and x terms are nicely grouped on the left, and the constant term is isolated on the right. This sets us up perfectly for the next step: completing the square!

Completing the Square: The Core Technique

Now for the main event: completing the square. This technique is where the magic happens, and it's all about turning the left side of our equation into a perfect square trinomial. Remember that perfect square trinomial pattern we talked about earlier? x2+2ax+a2=(x+a)2x^2 + 2ax + a^2 = (x+a)^2. We need to figure out what constant term will fit into our equation, x2+12x=111x^2 + 12x = 111, to make it match this pattern. The key lies in the coefficient of our x term, which is 12 in this case. We take half of this coefficient (12 / 2 = 6), and then square it (6² = 36). This number, 36, is the magic constant we need to add to both sides of the equation.

Why does this work? Well, by adding 36, we're essentially creating the a² term in our perfect square trinomial. In this case, a is 6, so a² is indeed 36. This ensures that the left side can be factored into the form (x+a)2(x + a)², which is exactly what we're aiming for. It's like adding the missing puzzle piece that completes the picture. By understanding the relationship between the coefficient of the x term and the constant term in a perfect square trinomial, we can systematically determine the value needed to complete the square. This technique is not just a mathematical trick; it's a method based on algebraic principles that allow us to transform and solve equations effectively.

Adding 36 to both sides, we get:

x2+12x+36=111+36x^2 + 12x + 36 = 111 + 36

x2+12x+36=147x^2 + 12x + 36 = 147

The Intermediate Step: (x+6)2=147(x+6)^2 = 147

And here it is, the intermediate step we've been looking for! The left side of the equation, x2+12x+36x^2 + 12x + 36, is now a perfect square trinomial. We can factor it as (x+6)2(x + 6)^2. So, our equation becomes:

(x+6)2=147(x + 6)^2 = 147

This is the desired form, (x+a)2=b(x+a)^2 = b, where a is 6 and b is 147. We've successfully transformed the original equation into a form that's much easier to solve. This step is crucial because it bridges the gap between the original quadratic equation and its solution. By completing the square, we've essentially rewritten the equation in a way that isolates the variable x, making it straightforward to apply the square root property and find the solutions. The beauty of this method lies in its ability to convert a complex equation into a simpler, more manageable form. From here, solving for x is just a matter of taking the square root of both sides and performing a few basic algebraic manipulations.

Solving for x (The Final Steps)

While the problem asked for the intermediate step, let's go ahead and solve for x to see how this form is useful. To solve (x+6)2=147(x + 6)^2 = 147, we take the square root of both sides:

(x+6)2=±147\sqrt{(x + 6)^2} = \pm \sqrt{147}

x+6=±147x + 6 = \pm \sqrt{147}

Now, we subtract 6 from both sides:

x=−6±147x = -6 \pm \sqrt{147}

We can simplify 147\sqrt{147} by factoring out the largest perfect square, which is 49:

147=49⋅3=49⋅3=73\sqrt{147} = \sqrt{49 \cdot 3} = \sqrt{49} \cdot \sqrt{3} = 7\sqrt{3}

So, our solutions are:

x=−6+73x = -6 + 7\sqrt{3} and x=−6−73x = -6 - 7\sqrt{3}

These are the two values of x that satisfy the original equation. Notice how easily we were able to find these solutions once we had the equation in the completed square form. This highlights the power and elegance of the completing the square technique. By transforming the equation into this specific format, we've essentially unlocked a direct pathway to the solutions. The process of taking the square root and isolating x becomes straightforward and efficient. This method is not just about finding the answers; it's about understanding the structure of the equation and using algebraic manipulations to reveal its underlying solutions. It's a testament to the beauty and power of mathematical transformations.

Conclusion: Mastering Completing the Square

So, there you have it! We've successfully found the intermediate step in completing the square for the equation x2+2x−108=−10x+3x^2 + 2x - 108 = -10x + 3, which is (x+6)2=147(x + 6)^2 = 147. We've also gone the extra mile and solved for x, demonstrating the utility of this form. Completing the square is a fundamental technique in algebra, and mastering it will open doors to solving a wide range of quadratic equations and other mathematical problems. Remember, the key is to understand the pattern of a perfect square trinomial and how to manipulate equations to fit that pattern. With practice, you'll become a pro at completing the square!

This technique is not just a tool for solving equations; it's a cornerstone of algebraic thinking. It teaches us how to transform expressions, manipulate equations, and reveal hidden structures. These skills are invaluable not only in mathematics but also in various other fields that rely on logical reasoning and problem-solving. As you continue your mathematical journey, you'll find that completing the square is a skill that you'll use again and again, in various contexts and applications. So, embrace the challenge, practice diligently, and you'll be well on your way to mastering this essential technique.

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