Solving For X In Y = (x - 6)^2 - 10 Expressed As X = [?] ± √[?]
In mathematics, solving equations is a fundamental skill. In this comprehensive guide, we will delve into the process of finding the values of x in the equation y = (x - 6)^2 - 10. This equation represents a parabola, and understanding how to solve for x allows us to analyze its properties and behavior. We will break down the steps involved, explain the underlying concepts, and provide clear examples to illustrate the process. By the end of this guide, you'll be equipped with the knowledge and skills to confidently solve similar equations. This method involves isolating x and expressing it in the form x = [?] ± √[?], which is crucial for determining the roots or x-intercepts of the parabolic function. Understanding this process is essential not only for academic success but also for practical applications in various fields such as physics, engineering, and economics, where parabolic relationships are frequently encountered. The ability to manipulate equations and solve for specific variables is a cornerstone of mathematical proficiency, enabling us to model and understand real-world phenomena more effectively. This exploration will cover the algebraic manipulations necessary to isolate the variable x, ensuring a thorough understanding of each step involved. Through this detailed explanation, you will gain a solid foundation in solving quadratic equations and interpreting their solutions. The techniques discussed here are applicable to a wide range of problems, making this a valuable skill for any student or professional working with mathematical models.
Understanding the Equation
The given equation, y = (x - 6)^2 - 10, is a quadratic equation in vertex form. This form provides valuable information about the parabola it represents. The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. In our case, a = 1, h = 6, and k = -10. Therefore, the vertex of the parabola is at the point (6, -10). The vertex represents the minimum or maximum point of the parabola, depending on the sign of a. Since a = 1, which is positive, the parabola opens upwards, and the vertex is the minimum point. Understanding the vertex form allows us to quickly identify key features of the parabola, such as its vertex and direction of opening. This understanding is crucial for graphing the parabola and interpreting its behavior. The equation also reveals the axis of symmetry, which is a vertical line passing through the vertex. In this case, the axis of symmetry is x = 6. The parabola is symmetric about this line, meaning that the graph on one side of the line is a mirror image of the graph on the other side. Recognizing these properties helps us visualize the parabola and predict its behavior. Furthermore, the constant term -10 indicates the vertical shift of the parabola compared to the basic parabola y = x^2. The parabola has been shifted 6 units to the right and 10 units downwards. This transformation affects the position of the parabola in the coordinate plane. By analyzing the equation in vertex form, we gain a comprehensive understanding of the parabola's characteristics, including its vertex, axis of symmetry, and vertical shift, which are essential for solving the equation and interpreting its solutions.
Isolating the Squared Term
Our first goal in solving for x is to isolate the squared term, (x - 6)^2. To do this, we need to eliminate the constant term -10 from the right side of the equation. We achieve this by adding 10 to both sides of the equation. This maintains the equality and moves us closer to isolating the squared term. The equation then becomes: y + 10 = (x - 6)^2. This step is a fundamental algebraic manipulation technique used to simplify equations. By adding the same value to both sides, we ensure that the equation remains balanced, and the solutions remain unchanged. Isolating the squared term is crucial because it allows us to apply the square root property, which is the next step in solving for x. This step sets the stage for undoing the squaring operation and ultimately finding the values of x that satisfy the equation. The principle behind this manipulation is based on the properties of equality, which state that performing the same operation on both sides of an equation preserves the equality. This is a cornerstone of algebraic problem-solving and is used extensively in various mathematical contexts. By isolating the squared term, we create a situation where we can directly apply the inverse operation of squaring, which is taking the square root. This process is essential for solving quadratic equations and finding the values of the variable that make the equation true. Moreover, this step simplifies the equation, making it easier to work with and understand. It helps to streamline the solution process and reduces the complexity of the problem, allowing us to focus on the remaining steps required to find the values of x. The clarity achieved by isolating the squared term is vital for ensuring accuracy and efficiency in solving the equation.
Taking the Square Root
Now that we have isolated the squared term, (x - 6)^2 = y + 10, we can take the square root of both sides of the equation. This is a crucial step in solving for x, as it undoes the squaring operation. When we take the square root, we must remember to consider both the positive and negative roots. This is because both the positive and negative square roots, when squared, will result in the same positive value. Therefore, we have: √(x - 6)^2 = ±√(y + 10), which simplifies to x - 6 = ±√(y + 10). This step introduces the ± symbol, which is essential for capturing both possible solutions for x. Failing to consider both roots would lead to an incomplete solution, as we would miss one of the values of x that satisfy the equation. The square root property is a fundamental concept in algebra, and it is particularly important when solving quadratic equations. It allows us to effectively deal with squared terms and find the values of the variable that make the equation true. Understanding this property is crucial for mastering algebraic problem-solving techniques. The inclusion of both positive and negative roots reflects the symmetry inherent in quadratic equations. The parabola represented by the equation has two x-intercepts (or roots), which are equidistant from the axis of symmetry. These roots correspond to the two solutions obtained when taking the square root. Therefore, this step is not just a mathematical manipulation but also a reflection of the underlying geometric properties of the parabola. By carefully considering both roots, we ensure a comprehensive and accurate solution to the equation. This thoroughness is vital for both mathematical rigor and practical applications where missing a solution could have significant consequences.
Solving for x
To finally solve for x, we need to isolate x on one side of the equation. We have x - 6 = ±√(y + 10). To isolate x, we add 6 to both sides of the equation. This gives us: x = 6 ± √(y + 10). This is the solution for x in the required form, x = [?] ± √[?]. The [?] before the ± is 6, and the [?] inside the square root is y + 10. This form of the solution is particularly useful because it expresses x explicitly in terms of y. This allows us to easily find the values of x for any given value of y. For example, if we want to find the x-intercepts of the parabola, we would set y = 0 and solve for x. This would give us the points where the parabola intersects the x-axis. The solution x = 6 ± √(y + 10) also reveals the symmetry of the parabola. The two solutions for x are equidistant from the line x = 6, which is the axis of symmetry of the parabola. This symmetry is a characteristic feature of quadratic equations and their parabolic graphs. Understanding this symmetry can help us visualize the solutions and check our work. Moreover, this form of the solution highlights the relationship between the x and y values of the parabola. It shows how the x values vary as y changes, providing a deeper understanding of the function's behavior. The ability to express x in terms of y is a powerful tool in mathematical analysis and is essential for solving a variety of problems. By isolating x and expressing it in this form, we have completed the process of solving the equation and gained valuable insights into the properties of the parabola.
Expressing the Solution
The solution we found, x = 6 ± √(y + 10), expresses x in the form requested: x = [?] ± √[?]. Here, the value before the ± is 6, and the expression inside the square root is y + 10. This form is crucial for understanding the behavior of the equation and for finding specific values of x given a value of y. This representation of the solution allows us to easily calculate the two possible values of x for any given y. The ± sign indicates that there are two solutions: one where we add the square root and one where we subtract it. These two solutions correspond to the two points on the parabola that have the same y-coordinate. Understanding this duality is essential for interpreting the solutions correctly. The form x = 6 ± √(y + 10) also provides insights into the transformations applied to the basic square root function. The +10 inside the square root represents a vertical shift, and the +6 outside the square root represents a horizontal shift. These transformations affect the position and shape of the graph of the function. Moreover, this representation is particularly useful for finding the x-intercepts of the parabola. To find the x-intercepts, we set y = 0 and solve for x*, which gives us x = 6 ± √10. These two values of x are the points where the parabola intersects the x-axis. The ability to express the solution in this form is a testament to the power of algebraic manipulation and its ability to reveal the underlying structure of mathematical relationships. By expressing x in terms of y, we gain a comprehensive understanding of the equation and its properties, which is essential for both theoretical analysis and practical applications. This form is not just a final answer but a tool for further exploration and understanding.
Practical Applications and Further Exploration
Understanding how to solve for x in equations like y = (x - 6)^2 - 10 has numerous practical applications. These types of equations are commonly encountered in physics, engineering, and economics, where parabolic relationships are used to model various phenomena. For example, the trajectory of a projectile, the shape of a suspension bridge cable, and the cost-benefit analysis in economics can all be modeled using parabolic equations. The ability to solve for x allows us to determine key parameters such as the range of a projectile, the minimum cost, or the break-even point. In physics, projectile motion is a classic example where parabolic equations are used. The equation describes the height of a projectile as a function of its horizontal distance, and solving for x can help determine the range of the projectile. In engineering, parabolic shapes are used in the design of reflectors, antennas, and bridges. Understanding the properties of parabolas is crucial for optimizing the performance of these structures. In economics, quadratic functions are used to model cost, revenue, and profit. Solving for x can help determine the optimal production level or the break-even point for a business. Furthermore, the techniques discussed in this guide can be extended to solve more complex equations and systems of equations. Understanding the underlying principles of algebraic manipulation and problem-solving is essential for tackling a wide range of mathematical challenges. Further exploration of quadratic equations can involve analyzing their graphs, finding their vertices and axes of symmetry, and determining their maximum or minimum values. These concepts are fundamental to understanding the behavior of quadratic functions and their applications in various fields. Additionally, exploring different methods for solving quadratic equations, such as factoring, completing the square, and using the quadratic formula, can provide a deeper understanding of the topic. By continuing to explore these concepts, you can build a strong foundation in algebra and develop the skills necessary to solve a wide range of mathematical problems.
By following these steps, we have successfully solved for x in the equation y = (x - 6)^2 - 10 and expressed the solution in the required form. This process demonstrates the importance of algebraic manipulation and the application of fundamental mathematical principles.
