Solving Geometric Problems A Detailed Exercise On Squares And Equations

Introduction: Understanding the Significance of Exercise in Mathematics

In the realm of mathematics, exercise isn't just a physical activity; it's a mental workout that strengthens our problem-solving skills and deepens our understanding of concepts. Mathematical exercises often involve applying theorems, formulas, and logical reasoning to solve problems, and these challenges are crucial for developing a strong foundation in the subject. This article delves into a specific exercise involving geometric shapes and algebraic equations, demonstrating how seemingly disparate mathematical concepts intertwine to form elegant solutions. We'll be exploring how to translate word problems into mathematical equations, manipulate those equations to reveal underlying relationships, and ultimately arrive at concrete answers. Through this exercise, we aim to not only solve the problem at hand but also to illustrate the broader importance of practice and problem-solving in mastering mathematics. Remember, each problem solved is a step forward in your mathematical journey.

Problem Statement: Unraveling the Mystery of Squares

The heart of our mathematical journey today lies within a geometric puzzle involving two squares. Imagine two squares existing side by side, each with its unique dimensions. The first square boasts sides measuring x cm, while the second square is slightly more complex, with sides extending to (2x-3) cm. These dimensions, while seemingly simple, hold the key to a fascinating relationship between the two squares. The problem reveals that the areas of these squares differ by a significant 24 cm². This seemingly small piece of information is the cornerstone of our challenge. Our mission is twofold: firstly, to demonstrate how this information elegantly translates into the algebraic equation x² - 4x - 5 = 0. Secondly, we must embark on a quest to unveil the actual dimensions of these enigmatic squares. This problem seamlessly blends geometry and algebra, requiring us to utilize our knowledge of area calculations and equation solving techniques. The journey ahead promises to be both challenging and rewarding, as we unravel the mystery of these squares.

Setting Up the Equation: Bridging Geometry and Algebra

To embark on our mathematical quest, we need to translate the given word problem into a language that mathematics understands – algebraic equations. This involves carefully dissecting the information provided and representing it using variables and symbols. The core concept here is the area of a square, which is calculated by squaring the length of its side. The first square, with sides of x cm, has an area of x² cm². Similarly, the second square, with sides of (2x-3) cm, has an area of (2x-3)² cm². The problem states that the difference in their areas is 24 cm². This key piece of information allows us to construct an equation. We can express the difference in areas as either (2x-3)² - x² = 24 or x² - (2x-3)² = 24. However, since the side length (2x-3) could potentially be smaller than x, we'll consider the absolute value of the difference to ensure we're dealing with a positive value. Therefore, the fundamental equation we'll work with is: |(2x-3)² - x²| = 24. This equation forms the bridge between the geometric description of the squares and the algebraic tools we'll use to solve for their dimensions. In the next steps, we'll simplify and manipulate this equation to reveal the quadratic equation that will lead us to the solution.

Expanding and Simplifying: Transforming the Equation

Having established the fundamental equation |(2x-3)² - x²| = 24, our next step involves expanding and simplifying it to reveal its underlying structure. This process is crucial for transforming the equation into a more manageable form, specifically a quadratic equation. Let's begin by expanding the term (2x-3)². Using the binomial expansion or the FOIL method, we get (2x-3)² = (2x-3)(2x-3) = 4x² - 12x + 9. Now, substituting this back into our equation, we have |(4x² - 12x + 9) - x²| = 24. Next, we simplify the expression inside the absolute value: |3x² - 12x + 9| = 24. At this stage, we need to consider two possibilities due to the absolute value: either 3x² - 12x + 9 = 24 or 3x² - 12x + 9 = -24. Let's analyze the first case, 3x² - 12x + 9 = 24. Subtracting 24 from both sides, we get 3x² - 12x - 15 = 0. Dividing the entire equation by 3, we arrive at x² - 4x - 5 = 0. This is the quadratic equation we were tasked with demonstrating. Now, let's briefly consider the second case, 3x² - 12x + 9 = -24. Adding 24 to both sides, we get 3x² - 12x + 33 = 0. Dividing by 3, we have x² - 4x + 11 = 0. We will address this case later when we discuss solving the quadratic equations. For now, our primary focus is on the equation x² - 4x - 5 = 0, which we have successfully derived from the given information. This equation encapsulates the relationship between the sides of the squares and their area difference, paving the way for us to determine their dimensions.

Solving the Quadratic Equation: Unveiling the Value of x

Now that we've successfully derived the quadratic equation x² - 4x - 5 = 0, the next critical step is to solve it. Solving this equation will reveal the possible values of x, which represents the side length of one of the squares. Quadratic equations can be solved using various methods, including factoring, completing the square, or using the quadratic formula. In this case, factoring seems to be the most straightforward approach. We need to find two numbers that multiply to -5 and add up to -4. These numbers are -5 and 1. Therefore, we can factor the quadratic equation as (x - 5)(x + 1) = 0. This factored form allows us to easily identify the solutions for x. For the product of two factors to be zero, at least one of them must be zero. So, either x - 5 = 0 or x + 1 = 0. Solving these linear equations, we find two possible values for x: x = 5 or x = -1. However, in the context of our problem, x represents the side length of a square, which cannot be negative. Therefore, we discard the solution x = -1. This leaves us with a single valid solution: x = 5. This means the side length of the first square is 5 cm. Let's now turn our attention to the other potential quadratic equation we identified earlier, x² - 4x + 11 = 0. This equation does not factor easily, so we can use the quadratic formula to find its solutions. The quadratic formula is given by x = [-b ± √(b² - 4ac)] / 2a, where a = 1, b = -4, and c = 11 in our case. Substituting these values, we get x = [4 ± √((-4)² - 4 * 1 * 11)] / 2 * 1 = [4 ± √(-28)] / 2. Since the discriminant (the value inside the square root) is negative, this equation has no real solutions. This confirms that x = 5 is the only valid solution for our problem. With the value of x determined, we are now poised to calculate the dimensions of both squares.

Determining the Dimensions: Unveiling the Squares

Having triumphantly solved the quadratic equation and found x = 5 cm, we can now determine the dimensions of both squares. The first square has sides of x cm, so its sides are simply 5 cm. The second square has sides of (2x - 3) cm. Substituting x = 5 into this expression, we get (2 * 5 - 3) cm = (10 - 3) cm = 7 cm. Therefore, the sides of the second square are 7 cm. We have successfully unveiled the dimensions of both squares: one with sides of 5 cm and the other with sides of 7 cm. To further solidify our understanding, let's calculate the areas of these squares and verify that their difference is indeed 24 cm². The area of the first square is 5² cm² = 25 cm². The area of the second square is 7² cm² = 49 cm². The difference in their areas is 49 cm² - 25 cm² = 24 cm², which matches the information given in the problem statement. This confirms the accuracy of our solution. We have not only found the dimensions of the squares but also validated our results, showcasing the power of mathematical problem-solving.

Conclusion: The Elegance of Mathematical Solutions

In conclusion, this exercise involving two squares has been a captivating journey through the realms of geometry and algebra. We successfully translated a word problem into a mathematical equation, navigated through the intricacies of quadratic equations, and ultimately unveiled the dimensions of the squares. The process involved setting up an equation based on the area difference, simplifying and expanding the equation, solving the resulting quadratic equation, and interpreting the solutions within the context of the problem. This exercise highlights the interconnectedness of mathematical concepts and the importance of problem-solving skills. It also showcases the elegance and precision of mathematical solutions. The key takeaways from this exercise are: the ability to translate word problems into algebraic equations, the mastery of solving quadratic equations, and the importance of verifying solutions. Each step, from formulating the equation to validating the results, reinforces the fundamental principles of mathematics and fosters a deeper appreciation for the subject. Remember, mathematics is not just about formulas and calculations; it's about logical reasoning, critical thinking, and the joy of discovery. Through exercises like this, we not only hone our mathematical skills but also cultivate a mindset of problem-solving that extends far beyond the classroom.