Finding K For Unique Solution Of Quadratic System | Math Problem
In this article, we delve into the fascinating world of quadratic equations and systems of equations. Specifically, we will explore how to determine the value of a constant, denoted as k, in a system of equations such that the system has exactly one real solution. This involves understanding the nature of quadratic equations, their graphical representations, and the conditions under which they intersect with linear equations at a single point. We'll dissect the given system:
y = x^2 + 4x + 14
y = 2k
and employ algebraic techniques to find the value of k that satisfies the given condition. This problem is a great example of how algebraic concepts can be applied to solve real-world problems. The core concept revolves around the discriminant of a quadratic equation, which provides crucial information about the nature of the roots (solutions) of the equation. By setting up the equations correctly and analyzing the discriminant, we can pinpoint the value of k that results in a single, unique solution.
To begin, let's break down the problem statement. We are given a system of two equations. The first equation, y = x^2 + 4x + 14, is a quadratic equation, which, when graphed, represents a parabola. The second equation, y = 2k, is a linear equation that represents a horizontal line. The problem asks us to find the value of the constant k such that the system of equations has exactly one real solution. Graphically, this means we are looking for the horizontal line that touches the parabola at only one point. This point of contact represents the unique solution to the system. A deep dive into quadratic equations is essential, as they form the backbone of this problem. Quadratic equations, characterized by their x^2 term, exhibit a parabolic curve when plotted on a graph. This curve can open upwards or downwards, depending on the coefficient of the x^2 term. The vertex of the parabola, the point where the curve changes direction, plays a crucial role in determining the number of solutions the quadratic equation has in conjunction with a linear equation. Understanding the discriminant, a key component of the quadratic formula, is paramount. The discriminant, denoted as Δ, is given by the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. The value of the discriminant dictates the nature of the roots: if Δ > 0, there are two distinct real roots; if Δ = 0, there is exactly one real root (a repeated root); and if Δ < 0, there are no real roots (two complex roots). For our problem, we need the condition where the discriminant is equal to zero, as this corresponds to the case where the horizontal line y = 2k intersects the parabola at a single point, indicating a unique solution. This condition forms the basis of our approach to solving for the value of k.
To solve the system, we can use the substitution method. Since both equations are expressed in terms of y, we can set them equal to each other:
x^2 + 4x + 14 = 2k
Now, rearrange the equation to form a standard quadratic equation:
x^2 + 4x + (14 - 2k) = 0
For the quadratic equation to have exactly one real solution, the discriminant must be equal to zero. The discriminant, denoted as Δ, is given by the formula:
Δ = b^2 - 4ac
In our equation, a = 1, b = 4, and c = 14 - 2k. Plugging these values into the discriminant formula, we get:
Δ = 4^2 - 4(1)(14 - 2k)
Setting the discriminant equal to zero:
0 = 16 - 4(14 - 2k)
Now, we solve for k:
0 = 16 - 56 + 8k
0 = -40 + 8k
8k = 40
k = 5
Therefore, the value of the constant k that makes the system of equations have exactly one real solution is 5. Let's delve deeper into the steps taken to arrive at this solution, emphasizing the importance of each algebraic manipulation. The substitution method, where we equated the two expressions for y, is a fundamental technique in solving systems of equations. This step allows us to eliminate one variable, y, and obtain a single equation in terms of x. Rearranging the equation into the standard quadratic form, ax^2 + bx + c = 0, is crucial because it allows us to directly apply the discriminant formula. Identifying the coefficients a, b, and c correctly is paramount for accurate calculation of the discriminant. The discriminant, Δ = b^2 - 4ac, is the key to determining the nature of the roots. Setting Δ = 0 is the cornerstone of our approach, as it represents the condition for a unique real solution. The subsequent algebraic steps, including expanding, simplifying, and isolating k, are standard procedures in solving linear equations. The final result, k = 5, is the value that ensures the parabola and the horizontal line intersect at exactly one point, satisfying the problem's condition. A thorough understanding of these steps and the underlying principles is essential for tackling similar problems involving quadratic equations and systems of equations.
To verify our solution, we can substitute k = 5 back into the original system of equations:
y = x^2 + 4x + 14
y = 2(5) = 10
Now, set the equations equal to each other:
x^2 + 4x + 14 = 10
Rearrange the equation:
x^2 + 4x + 4 = 0
This quadratic equation can be factored as:
(x + 2)^2 = 0
This equation has one real solution, x = -2. This confirms that when k = 5, the system of equations has exactly one real solution. Verifying the solution is an essential step in problem-solving, as it ensures the accuracy of the calculated value and provides confidence in the final answer. Substituting the obtained value of k back into the original system allows us to check if the resulting equation indeed satisfies the condition of having exactly one real solution. In this case, substituting k = 5 leads to the equation x^2 + 4x + 4 = 0, which, upon factoring, yields (x + 2)^2 = 0. This equation has a single repeated root, x = -2, confirming that our solution is correct. Furthermore, we can find the corresponding y-value by substituting x = -2 into either of the original equations. Using y = 2k, we get y = 2(5) = 10. Thus, the point of intersection is (-2, 10), which represents the unique solution to the system when k = 5. This verification process not only validates our answer but also reinforces our understanding of the problem and the concepts involved. It highlights the connection between the algebraic solution and the graphical representation of the equations, where the parabola and the horizontal line intersect at a single point. The ability to verify solutions is a valuable skill in mathematics, as it promotes accuracy and helps in identifying potential errors in the problem-solving process.
In conclusion, we have successfully determined the value of the constant k for which the given system of equations has exactly one real solution. By setting the discriminant of the quadratic equation to zero, we found that k = 5. This problem demonstrates the importance of understanding the properties of quadratic equations and their graphical representations. The discriminant, a powerful tool, allows us to analyze the nature of the roots and determine the conditions for unique solutions. The ability to solve systems of equations is a fundamental skill in mathematics, with applications in various fields such as physics, engineering, and economics. The process of solving this problem involved several key steps, including setting up the equations, applying the discriminant formula, and solving for the unknown variable. Each step required a solid understanding of algebraic principles and techniques. The verification process further solidified our understanding and confirmed the accuracy of our solution. Furthermore, this problem underscores the significance of connecting algebraic concepts with their graphical interpretations. The condition of having exactly one real solution corresponds to the geometric scenario where a parabola and a horizontal line intersect at a single point. This connection between algebra and geometry enhances our problem-solving abilities and provides a deeper appreciation for the beauty and elegance of mathematics. Mastering these concepts and techniques will undoubtedly prove beneficial in tackling more complex mathematical challenges in the future. The exploration of quadratic equations and their solutions is a cornerstone of mathematical education, and this problem serves as a valuable exercise in reinforcing these foundational concepts.
