Solutions To The Quadratic Equation (5y+6)^2=24 A Detailed Guide
Introduction: Understanding Quadratic Equations
In the realm of mathematics, quadratic equations hold a position of fundamental importance. These equations, characterized by their squared variable terms, appear in various scientific and engineering applications. A quintessential example of such an equation is the one we aim to solve in this comprehensive guide: (5y + 6)² = 24. To effectively navigate the solution process, it's crucial to grasp the inherent nature of quadratic equations and the array of methodologies available for their resolution. Quadratic equations, defined as polynomial equations of the second degree, assume the general form of ax² + bx + c = 0, where 'a,' 'b,' and 'c' represent constants, and 'x' symbolizes the variable to be determined. The hallmark of a quadratic equation lies in the presence of the squared term (x²), distinguishing it from linear equations where the highest power of the variable is one. Solving quadratic equations involves finding the values of the variable that satisfy the equation, commonly known as the roots or solutions. These roots represent the points where the quadratic function intersects the x-axis on a graph. There are several methods to solve quadratic equations, each with its own strengths and applicability. Factoring, completing the square, and the quadratic formula are among the most widely used techniques. Factoring involves expressing the quadratic equation as a product of two linear factors, allowing for easy determination of the roots. Completing the square transforms the equation into a perfect square trinomial, facilitating the isolation of the variable. The quadratic formula provides a universal solution, applicable to any quadratic equation regardless of its factorability. Understanding the characteristics of quadratic equations and the available solution methods is paramount for successfully tackling a wide range of mathematical problems. In the subsequent sections, we will delve into the step-by-step process of solving the specific equation (5y + 6)² = 24, employing appropriate techniques and elucidating the underlying principles.
Problem Statement: Deconstructing (5y+6)^2=24
The quadratic equation at the heart of our exploration is (5y + 6)² = 24. This particular equation presents a unique structure that requires careful consideration. Unlike the standard form ax² + bx + c = 0, this equation is expressed in a squared binomial form. To effectively solve it, we must first unravel its structure and transform it into a more manageable format. The presence of the squared binomial (5y + 6)² indicates that we are dealing with a perfect square trinomial. Expanding this binomial is a crucial first step in simplifying the equation. This expansion involves applying the distributive property or using the formula (a + b)² = a² + 2ab + b². Once expanded, the equation can be rearranged into the standard quadratic form, allowing us to apply familiar solution methods. However, before embarking on the expansion, it's beneficial to consider alternative approaches that might streamline the solution process. One such approach involves directly taking the square root of both sides of the equation. This method leverages the inverse relationship between squaring and taking the square root, potentially simplifying the equation without the need for full expansion. It's important to note that when taking the square root, we must account for both positive and negative roots, as both can satisfy the original equation. By carefully analyzing the structure of the equation and considering different solution strategies, we can choose the most efficient path to determine the values of 'y' that satisfy the given condition. In the following sections, we will explore both the expansion method and the square root method, comparing their advantages and disadvantages in the context of this specific problem. Understanding the nuances of each approach will empower us to solve similar quadratic equations with greater confidence and proficiency. The key lies in recognizing the underlying principles and applying them strategically to simplify the problem and arrive at the correct solutions. Our journey through this equation will not only provide the answers but also enhance our understanding of quadratic equation-solving techniques.
Method 1: Solving by Taking the Square Root
The first method we'll employ to solve the quadratic equation (5y + 6)² = 24 involves directly taking the square root of both sides. This approach leverages the inverse relationship between squaring and taking the square root, potentially simplifying the equation without the need for full expansion. By directly addressing the squared term, we can isolate the binomial (5y + 6) and proceed towards solving for 'y.' The initial step in this method is to apply the square root operation to both sides of the equation. This yields √(5y + 6)² = ±√24. It's crucial to include both positive and negative square roots of 24, as both values, when squared, will result in 24. This consideration of both positive and negative roots is a fundamental aspect of solving quadratic equations. The square root of (5y + 6)² simplifies to (5y + 6), while √24 can be simplified by factoring out the perfect square factor. In this case, 24 can be expressed as 4 * 6, where 4 is a perfect square. Therefore, √24 becomes √(4 * 6) = √4 * √6 = 2√6. Substituting this simplified value back into the equation, we get 5y + 6 = ±2√6. Now, we have two separate equations to solve: 5y + 6 = 2√6 and 5y + 6 = -2√6. Solving each equation involves isolating 'y' by performing algebraic operations. First, we subtract 6 from both sides of each equation: 5y = -6 + 2√6 and 5y = -6 - 2√6. Finally, we divide both sides of each equation by 5 to obtain the solutions for 'y': y = (-6 + 2√6) / 5 and y = (-6 - 2√6) / 5. These two values represent the roots of the quadratic equation (5y + 6)² = 24. By directly taking the square root, we bypassed the need for expanding the binomial, streamlining the solution process. This method highlights the importance of recognizing the structure of the equation and choosing the most efficient approach. The square root method is particularly effective when dealing with equations in the form of a squared expression equal to a constant. In the next section, we will explore an alternative method – solving by expansion – and compare its advantages and disadvantages to the square root method.
Method 2: Solving by Expansion and Factoring (or Quadratic Formula)
An alternative approach to solving the quadratic equation (5y + 6)² = 24 involves expanding the squared binomial and rearranging the equation into the standard quadratic form (ay² + by + c = 0). This method, while potentially more laborious than directly taking the square root, provides a comprehensive understanding of the equation's structure and allows for the application of various solution techniques, such as factoring or the quadratic formula. The initial step in this method is to expand the binomial (5y + 6)². Using the formula (a + b)² = a² + 2ab + b², we get (5y + 6)² = (5y)² + 2(5y)(6) + 6² = 25y² + 60y + 36. Substituting this expansion back into the original equation, we have 25y² + 60y + 36 = 24. Now, we need to rearrange the equation into the standard quadratic form by subtracting 24 from both sides: 25y² + 60y + 36 - 24 = 0, which simplifies to 25y² + 60y + 12 = 0. At this point, we have a quadratic equation in the standard form. We can attempt to solve it by factoring, but in this case, factoring might not be straightforward. Therefore, the quadratic formula provides a reliable alternative. The quadratic formula states that for an equation of the form ay² + by + c = 0, the solutions for 'y' are given by y = (-b ± √(b² - 4ac)) / (2a). In our equation, a = 25, b = 60, and c = 12. Substituting these values into the quadratic formula, we get y = (-60 ± √(60² - 4 * 25 * 12)) / (2 * 25). Simplifying the expression under the square root, we have 60² - 4 * 25 * 12 = 3600 - 1200 = 2400. Thus, the equation becomes y = (-60 ± √2400) / 50. We can simplify √2400 by factoring out the perfect square factor. 2400 can be expressed as 400 * 6, where 400 is a perfect square. Therefore, √2400 = √(400 * 6) = √400 * √6 = 20√6. Substituting this back into the equation, we get y = (-60 ± 20√6) / 50. Now, we can simplify the expression by dividing both the numerator and denominator by 10: y = (-6 ± 2√6) / 5. This gives us the two solutions for 'y': y = (-6 + 2√6) / 5 and y = (-6 - 2√6) / 5. These solutions are identical to those obtained using the square root method, confirming the consistency of both approaches. Solving by expansion and using the quadratic formula provides a robust method for tackling quadratic equations, even when factoring is not readily apparent. It demonstrates the power of the quadratic formula as a universal solution tool. In the next section, we will compare the two methods and discuss their relative advantages and disadvantages.
Comparing the Two Methods: Square Root vs. Expansion
Having explored two distinct methods for solving the quadratic equation (5y + 6)² = 24, it's crucial to compare their relative advantages and disadvantages. This comparison will provide valuable insights into choosing the most efficient approach for similar problems in the future. The first method, solving by taking the square root, offers a streamlined solution when the equation is presented in the form of a squared expression equal to a constant. This approach directly addresses the squared term, bypassing the need for expansion and potentially simplifying the algebraic manipulations. The key advantage of this method lies in its efficiency. By directly taking the square root of both sides, we reduce the number of steps required to isolate the variable 'y.' This can save time and effort, especially when dealing with more complex equations. However, the square root method is most effective when the equation is already in the appropriate form. If the equation is not presented as a squared expression equal to a constant, rearranging it into this form might require additional steps. The second method, solving by expansion and using the quadratic formula, provides a more general approach that can be applied to any quadratic equation, regardless of its initial form. This method involves expanding the squared binomial, rearranging the equation into the standard quadratic form (ay² + by + c = 0), and then applying the quadratic formula. The quadratic formula provides a universal solution for quadratic equations, ensuring that we can always find the roots, even when factoring is not feasible. The advantage of this method lies in its robustness and applicability to a wide range of quadratic equations. However, it can be more time-consuming than the square root method, especially when the expansion and simplification steps are lengthy. Additionally, the quadratic formula involves multiple calculations, increasing the potential for errors. In the case of the equation (5y + 6)² = 24, the square root method proves to be more efficient. It directly addresses the structure of the equation, leading to a quicker solution. However, if we were to encounter a quadratic equation that is not easily expressed in the squared form, the expansion method and the quadratic formula would be a more reliable choice. Ultimately, the best method depends on the specific characteristics of the equation and the solver's familiarity with each technique. Proficiency in both methods allows for flexibility and adaptability in solving quadratic equations. In the concluding section, we will summarize our findings and provide the final solutions to the equation.
Final Solutions and Conclusion
In this comprehensive exploration, we embarked on a journey to solve the quadratic equation (5y + 6)² = 24. We employed two distinct methods – solving by taking the square root and solving by expansion and using the quadratic formula – each offering unique insights into the problem. Through the square root method, we directly addressed the squared term, simplifying the equation and efficiently isolating the variable 'y.' This approach highlighted the importance of recognizing the structure of the equation and leveraging inverse operations to streamline the solution process. Alternatively, the expansion method provided a more general approach, applicable to a wider range of quadratic equations. By expanding the binomial and rearranging the equation into the standard quadratic form, we demonstrated the power of the quadratic formula as a universal solution tool. Both methods ultimately led us to the same solutions, confirming the consistency of our approaches. The solutions to the quadratic equation (5y + 6)² = 24 are: y = (-6 + 2√6) / 5 and y = (-6 - 2√6) / 5. These values represent the roots of the equation, the points where the corresponding quadratic function intersects the x-axis. In conclusion, solving quadratic equations requires a multifaceted understanding of algebraic principles and techniques. The choice of method depends on the specific characteristics of the equation and the solver's preferences. Proficiency in both the square root method and the expansion method, along with the quadratic formula, equips us with the necessary tools to tackle a wide range of quadratic equation problems. This exploration not only provided the solutions to a specific equation but also enhanced our understanding of quadratic equation-solving strategies. By mastering these techniques, we gain confidence in our ability to navigate mathematical challenges and apply these principles to real-world applications. The journey through this equation serves as a testament to the power and elegance of mathematical problem-solving.
Final Answer: The final answer is
