Solving (x+6)(x+2)=60 A Step By Step Guide

In the realm of algebra, solving equations is a fundamental skill. One common type of equation encountered is the quadratic equation, which takes the general form ax² + bx + c = 0. This article delves into the process of solving a specific quadratic equation, (x+6)(x+2)=60, providing a step-by-step guide and exploring the underlying concepts. We will transform the given equation into standard quadratic form, explore different solution methods, and arrive at the correct answers.

Understanding the Problem: (x+6)(x+2)=60

Before diving into the solution, it's crucial to understand the equation we are dealing with. The equation (x+6)(x+2) = 60 is a quadratic equation presented in factored form on the left side. Our goal is to find the values of x that satisfy this equation. These values are also known as the roots or solutions of the equation. The challenge here is that the equation isn't in the standard quadratic form (ax² + bx + c = 0), which makes it difficult to directly apply methods like the quadratic formula. Therefore, our first step involves expanding the factored form and rearranging the equation into the standard quadratic form. This transformation is essential for simplifying the problem and making it amenable to standard solution techniques. We aim to manipulate the equation without changing its inherent solutions, thus preserving the integrity of the mathematical statement. The process of expanding and simplifying will reveal the coefficients a, b, and c, which are crucial for subsequent steps in solving the equation. By carefully applying algebraic principles, we will ensure that the transformed equation is equivalent to the original one, allowing us to confidently proceed towards finding the values of x. This initial phase of understanding and transforming the equation is a critical foundation for solving quadratic equations effectively.

Step 1: Expanding and Simplifying

The first step in solving the equation (x+6)(x+2)=60 is to expand the left side. We can do this by using the distributive property (also known as the FOIL method): First, Outer, Inner, Last. This expansion is a critical step because it allows us to transform the equation from a factored form to a polynomial form, which is necessary for identifying the coefficients needed for solving the quadratic equation. When we multiply (x+6) by (x+2), we get:

• x * x = x²
• x * 2 = 2x
• 6 * x = 6x
• 6 * 2 = 12

Adding these terms together, we get x² + 2x + 6x + 12. Combining the like terms (2x and 6x), we simplify the expression to x² + 8x + 12. So, the left side of the equation, (x+6)(x+2), expands to x² + 8x + 12. Now, our equation looks like this: x² + 8x + 12 = 60. The next step in the simplification process involves moving the constant term from the right side of the equation to the left side. This is essential for setting the equation equal to zero, which is the standard form for a quadratic equation (ax² + bx + c = 0). To achieve this, we subtract 60 from both sides of the equation. This operation maintains the balance of the equation and ensures that the solutions remain the same. Subtracting 60 from both sides gives us: x² + 8x + 12 - 60 = 60 - 60, which simplifies to x² + 8x - 48 = 0. Now, the equation is in the standard quadratic form, where a = 1, b = 8, and c = -48. This form is crucial because it allows us to apply various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. The simplified equation, x² + 8x - 48 = 0, is the foundation upon which we will build our solution. It represents the same relationship as the original equation but in a form that is much easier to work with.

Step 2: Factoring the Quadratic Equation

Now that we have the quadratic equation in the standard form x² + 8x - 48 = 0, we can try to solve it by factoring. Factoring involves expressing the quadratic expression as a product of two binomials. This method relies on finding two numbers that multiply to the constant term (-48) and add up to the coefficient of the linear term (8). In this case, we need to find two numbers that, when multiplied, give us -48, and when added, give us 8. Let's consider the factors of -48. Some possible pairs are: (1, -48), (-1, 48), (2, -24), (-2, 24), (3, -16), (-3, 16), (4, -12), (-4, 12), (6, -8), and (-6, 8). Among these pairs, we are looking for the pair that adds up to 8. By examining the pairs, we can see that the numbers 12 and -4 satisfy these conditions: 12 * (-4) = -48 and 12 + (-4) = 8. Therefore, we can factor the quadratic expression x² + 8x - 48 as (x + 12)(x - 4). This factorization is a key step because it transforms the quadratic equation into a form where we can easily find the solutions. The equation x² + 8x - 48 = 0 can now be written as (x + 12)(x - 4) = 0. This factored form tells us that the product of two expressions is equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. This property is the cornerstone of solving equations by factoring. It allows us to break down a complex equation into simpler ones. In our case, the two factors are (x + 12) and (x - 4). Therefore, either (x + 12) = 0 or (x - 4) = 0. These are now simple linear equations that we can easily solve for x. Solving these equations will give us the values of x that make the original quadratic equation true. The success of factoring as a method depends on the ability to find the correct factors. In some cases, factoring might not be straightforward, and alternative methods, such as the quadratic formula, may be more appropriate. However, when factoring is possible, it is often the quickest and most efficient way to solve a quadratic equation.

Step 3: Solving for x

Now that we have the equation in factored form, (x + 12)(x - 4) = 0, we can apply the zero-product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. This principle is fundamental to solving equations by factoring and is a powerful tool in algebra. In our equation, the factors are (x + 12) and (x - 4). Therefore, we have two possibilities: either (x + 12) = 0 or (x - 4) = 0. These two equations represent separate scenarios, each of which can lead to a solution for x. To solve the first equation, (x + 12) = 0, we need to isolate x on one side of the equation. We can do this by subtracting 12 from both sides of the equation. This operation maintains the equality and moves us closer to finding the value of x. Subtracting 12 from both sides gives us: x + 12 - 12 = 0 - 12, which simplifies to x = -12. So, one solution to the quadratic equation is x = -12. This means that if we substitute -12 for x in the original equation, the equation will hold true. Similarly, to solve the second equation, (x - 4) = 0, we need to isolate x. We can do this by adding 4 to both sides of the equation. This is the inverse operation of subtraction and will effectively cancel out the -4 on the left side. Adding 4 to both sides gives us: x - 4 + 4 = 0 + 4, which simplifies to x = 4. Therefore, the other solution to the quadratic equation is x = 4. This means that if we substitute 4 for x in the original equation, it will also hold true. Thus, we have found two solutions for the equation (x + 6)(x + 2) = 60: x = -12 and x = 4. These solutions are the values of x that make the equation a true statement. They are also known as the roots of the quadratic equation. It's important to note that quadratic equations can have up to two real solutions, as we have seen in this case. The ability to solve for x in these situations is a cornerstone of algebraic problem-solving.

The Solutions: x = -12 and x = 4

We have successfully found the solutions to the equation (x+6)(x+2)=60. Through the process of expanding, simplifying, factoring, and applying the zero-product property, we have determined that the values of x that satisfy the equation are x = -12 and x = 4. These solutions are the roots of the quadratic equation, representing the points where the parabola defined by the equation intersects the x-axis. It is important to verify these solutions to ensure their accuracy. We can do this by substituting each value of x back into the original equation and checking if the equation holds true. Let's first verify x = -12. Substituting -12 for x in the original equation (x+6)(x+2)=60, we get: (-12 + 6)(-12 + 2) = (-6)(-10) = 60. Since 60 = 60, the solution x = -12 is correct. Now, let's verify x = 4. Substituting 4 for x in the original equation, we get: (4 + 6)(4 + 2) = (10)(6) = 60. Since 60 = 60, the solution x = 4 is also correct. Therefore, we have confirmed that both x = -12 and x = 4 are valid solutions to the equation (x+6)(x+2)=60. These solutions represent the values of x that make the equation a true statement. Understanding how to find and verify solutions is crucial in mathematics. It ensures that we have not only arrived at an answer but also that the answer is accurate and consistent with the given equation. The ability to verify solutions is a fundamental skill that strengthens our understanding of mathematical concepts and problem-solving techniques. This comprehensive process, from transforming the equation to verifying the solutions, demonstrates a thorough approach to solving quadratic equations.

Conclusion

In this article, we have explored the process of solving the quadratic equation (x+6)(x+2)=60. We began by understanding the equation and recognizing the need to transform it into the standard quadratic form. We then expanded and simplified the equation, which allowed us to factor it effectively. By applying the zero-product property, we were able to find the two solutions: x = -12 and x = 4. Finally, we verified these solutions to ensure their accuracy. This step-by-step approach demonstrates a systematic method for solving quadratic equations. The key takeaways from this exercise include the importance of transforming equations into standard forms, the power of factoring in simplifying expressions, the application of the zero-product property, and the necessity of verifying solutions. These principles are not only applicable to this specific equation but also to a wide range of algebraic problems. Mastering these techniques is crucial for success in mathematics. Furthermore, this process highlights the interconnectedness of different algebraic concepts. Expanding, simplifying, factoring, and solving are all interrelated skills that contribute to the overall problem-solving process. By understanding these connections, we can approach mathematical challenges with greater confidence and efficiency. The ability to solve quadratic equations is a fundamental skill in algebra and is essential for various applications in mathematics, science, and engineering. This article has provided a comprehensive guide to solving one such equation, and the principles discussed can be applied to many other quadratic equations. Understanding the underlying concepts and practicing these techniques will strengthen your problem-solving abilities and enhance your mathematical proficiency.