Finding The Value Of G In The Equation (x+7)(x-4) = X^2 + Gx - 28
Introduction: Unraveling the Algebraic Puzzle
In the realm of mathematics, algebraic equations often present themselves as puzzles waiting to be solved. One such puzzle involves finding the value of an unknown coefficient, denoted here as 'g', within a quadratic equation. The equation at hand is (x+7)(x-4) = x^2 + gx - 28. To determine the correct value of 'g', we must delve into the principles of algebraic expansion and equation solving. This article will explore a step-by-step approach to solving this equation, ensuring a clear understanding of the underlying concepts. We will focus on expanding the left side of the equation, comparing coefficients, and ultimately isolating 'g' to find its numerical value. This process not only provides the answer but also reinforces fundamental algebraic techniques applicable to a wide range of mathematical problems. Understanding how to manipulate and solve such equations is crucial for students and anyone involved in fields that require analytical thinking and problem-solving skills. Let's embark on this mathematical journey to uncover the value of 'g' and enhance our algebraic prowess. By carefully examining each step, we can demystify the process and gain confidence in tackling similar algebraic challenges.
Expanding the Left Side of the Equation: A Crucial First Step
The initial step in solving the equation (x+7)(x-4) = x^2 + gx - 28 involves expanding the left side. This expansion is a critical process that transforms the product of two binomials into a standard quadratic expression. To achieve this, we employ the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last). This method ensures that each term in the first binomial is multiplied by each term in the second binomial. Let's break down the expansion:
- First: Multiply the first terms of each binomial: x * x = x^2
- Outer: Multiply the outer terms of the binomials: x * -4 = -4x
- Inner: Multiply the inner terms of the binomials: 7 * x = 7x
- Last: Multiply the last terms of each binomial: 7 * -4 = -28
Combining these products, we get x^2 - 4x + 7x - 28. The next step is to simplify this expression by combining like terms. Specifically, we combine the '-4x' and '7x' terms. Adding these together, -4x + 7x, we arrive at 3x. Thus, the expanded form of the left side of the equation is x^2 + 3x - 28. This expansion is a fundamental algebraic technique, and mastering it is essential for solving various equations and simplifying expressions. The expanded form now allows us to directly compare the left side of the equation with the right side, which is a crucial step in isolating 'g'. Understanding this process is not only vital for this particular problem but also for a broad spectrum of mathematical challenges involving polynomial expressions. This careful expansion lays the groundwork for the subsequent steps in solving for the unknown variable, 'g'.
Comparing Coefficients: Unveiling the Value of 'g'
With the left side of the equation expanded to x^2 + 3x - 28, we can now directly compare it to the right side, which is x^2 + gx - 28. This comparison is the key to unveiling the value of 'g'. By aligning the two sides of the equation, we can identify corresponding terms and their coefficients. The equation now stands as: x^2 + 3x - 28 = x^2 + gx - 28. Notice that the x^2 terms and the constant terms (-28) are already equal on both sides. This simplifies our task, allowing us to focus solely on the 'x' terms. On the left side, the coefficient of the 'x' term is 3, while on the right side, the coefficient of the 'x' term is 'g'. For the equation to hold true, the coefficients of the corresponding terms must be equal. Therefore, we can directly equate the coefficients of the 'x' terms: 3 = g. This straightforward comparison leads us to the solution: g = 3. This method of comparing coefficients is a powerful technique in algebra, especially when dealing with polynomials. It allows us to isolate unknown variables by focusing on the terms that contain them. The simplicity of this step highlights the importance of the initial expansion; without it, this direct comparison would not be possible. Understanding and applying this method provides a clear and efficient way to solve for unknowns in various algebraic contexts. It's a technique that underscores the elegance and precision of mathematical problem-solving.
The Solution: Isolating 'g' to Solve the Puzzle
After expanding the left side of the equation and comparing coefficients, we have arrived at the solution for 'g'. The equation x^2 + 3x - 28 = x^2 + gx - 28 clearly shows that the coefficient of the 'x' term on the left side, which is 3, must be equal to the coefficient of the 'x' term on the right side, which is 'g'. Therefore, g = 3. This value makes the equation true, as substituting 3 for 'g' results in identical quadratic expressions on both sides of the equation. To verify this, we can substitute g = 3 back into the original equation: (x+7)(x-4) = x^2 + 3x - 28. We already know that the left side expands to x^2 + 3x - 28, so the equation holds true. This process of isolating the unknown variable is fundamental in algebra. By systematically manipulating the equation through expansion and comparison, we were able to pinpoint the value of 'g' with certainty. The solution g = 3 not only satisfies the equation but also demonstrates the power of algebraic techniques in solving mathematical puzzles. This exercise reinforces the importance of careful expansion, coefficient comparison, and logical deduction in arriving at the correct answer. Understanding how to isolate variables in equations is a core skill in mathematics, applicable across various fields and problems. The satisfaction of solving for 'g' highlights the beauty and precision of algebraic problem-solving.
Conclusion: Mastering Algebraic Equations
In conclusion, solving the equation (x+7)(x-4) = x^2 + gx - 28 to find the value of 'g' has been a journey through fundamental algebraic principles. We began by expanding the left side of the equation using the distributive property, a critical step in transforming the equation into a comparable form. This expansion yielded x^2 + 3x - 28, setting the stage for the next phase. Next, we compared the coefficients of the expanded left side with the right side of the equation. This direct comparison allowed us to equate the coefficients of the 'x' terms, leading to the solution g = 3. The process of isolating 'g' highlighted the power of algebraic manipulation and the importance of precision in mathematical operations. The solution not only answers the specific question but also reinforces essential skills in algebra, including expansion, simplification, and coefficient comparison. These skills are transferable and applicable to a wide range of mathematical problems. Understanding how to approach and solve algebraic equations is crucial for students and professionals alike. It fosters analytical thinking, problem-solving abilities, and a deeper appreciation for the elegance of mathematics. The journey of finding 'g' in this equation serves as a valuable lesson in the systematic approach to solving mathematical puzzles, emphasizing the importance of each step in the process. By mastering these techniques, we empower ourselves to tackle more complex challenges in the world of mathematics and beyond. The value of 'g' may be just one solution, but the skills gained in finding it are invaluable.
