System Of Equations Y=6x^2+1 And Y=x^2+4 Two Solutions Explained

Jul 14, 2025 by Admin 65 views

System of Equations with Two Solutions: A Comprehensive Analysis

Introduction

Understanding systems of equations is a fundamental concept in mathematics, particularly in algebra. A system of equations involves two or more equations with the same set of variables. The solutions to a system of equations are the values that satisfy all the equations simultaneously. In this article, we will delve into a specific system of equations and explore why it has two solutions. We will focus on the graphical representation of these equations, which provides a visual understanding of the solutions. Specifically, we will analyze the system:

y = 6x^2 + 1
y = x^2 + 4

This system consists of two quadratic equations, each representing a parabola. Our objective is to explain why this system possesses two distinct solutions. The key to this explanation lies in understanding how the graphs of these equations intersect. The points of intersection represent the solutions to the system because these are the points where the y-values are equal for the same x-values in both equations. We will explore the properties of parabolas, such as their vertices and shapes, to illustrate why these two parabolas intersect at two points, leading to two solutions. This analysis will provide a clear understanding of the relationship between the algebraic representation of the system and its graphical interpretation, highlighting the significance of graphical methods in solving systems of equations. Furthermore, we will consider and refute alternative explanations, such as the misconception that each vertex represents a solution, to reinforce the correct understanding of the solution set. By the end of this discussion, you will have a thorough grasp of the underlying principles that govern the solutions of this specific system and systems of equations in general.

Understanding the Equations

To understand why the system of equations has two solutions, it's crucial to first break down the individual equations and their graphical representations. The equations given are:

y = 6x^2 + 1
y = x^2 + 4

Both equations are quadratic equations, which means they represent parabolas when graphed. A parabola is a U-shaped curve, and its properties are determined by the coefficients in the quadratic equation. Let's analyze each equation separately to understand their individual characteristics and how they contribute to the overall system.

The first equation, y = 6x^2 + 1, is a parabola that opens upwards because the coefficient of the x^2 term (6) is positive. The coefficient of x^2 also affects the width of the parabola. A larger coefficient, like 6 in this case, makes the parabola narrower compared to a parabola with a smaller coefficient. The constant term, +1, indicates the vertical shift of the parabola. This parabola's vertex, which is the lowest point on the curve, is at the point (0, 1). This is because when x is 0, y is at its minimum value of 1. The parabola extends upwards from this vertex, becoming steeper as x moves away from 0 in either the positive or negative direction. Understanding these features is crucial for visualizing the parabola and predicting its interactions with other curves.

The second equation, y = x^2 + 4, is also a parabola that opens upwards, as the coefficient of the x^2 term (1) is positive. However, the coefficient is smaller than the first equation, which means this parabola is wider. The constant term, +4, indicates that the vertex of this parabola is at the point (0, 4). This parabola is shifted upwards by 4 units compared to the standard parabola y = x^2. The broader shape and higher vertex position are key characteristics that differentiate it from the first parabola. Visualizing both parabolas on the same coordinate plane allows us to anticipate potential intersection points, which correspond to the solutions of the system. The difference in their shapes and vertical positions suggests that they will intersect at two points, which we will explore further in the subsequent sections. By analyzing these individual characteristics, we gain a foundational understanding of the equations and their graphical behavior.

Graphical Representation and Intersection Points

The key to understanding why the system of equations has two solutions lies in the graphical representation of the two parabolas and their intersection points. By plotting the graphs of the equations y = 6x^2 + 1 and y = x^2 + 4, we can visually identify the solutions.

As discussed earlier, the first equation, y = 6x^2 + 1, represents a narrow parabola with its vertex at (0, 1). The second equation, y = x^2 + 4, represents a wider parabola with its vertex at (0, 4). When these two parabolas are graphed on the same coordinate plane, they intersect at two distinct points. These points of intersection are the solutions to the system of equations because they represent the (x, y) coordinates that satisfy both equations simultaneously.

The intersection points are where the y-values of both parabolas are equal for the same x-values. To find these points algebraically, we set the two equations equal to each other:

6x^2 + 1 = x^2 + 4

Solving this equation will give us the x-coordinates of the intersection points. Subtracting x^2 from both sides, we get:

5x^2 + 1 = 4

Subtracting 1 from both sides gives:

5x^2 = 3

Dividing by 5, we have:

x^2 = 3/5

Taking the square root of both sides, we get two possible values for x:

x = ±√(3/5)

This result shows that there are two x-values where the parabolas intersect: one positive and one negative. These x-values correspond to the two distinct solutions of the system. To find the corresponding y-values, we can substitute these x-values back into either of the original equations. Using y = x^2 + 4, we have:

For x = √(3/5):

y = (√(3/5))^2 + 4 = 3/5 + 4 = 23/5

For x = -√(3/5):

y = (-√(3/5))^2 + 4 = 3/5 + 4 = 23/5

So, the two points of intersection are (√(3/5), 23/5) and (-√(3/5), 23/5). These points are the graphical representations of the solutions to the system. The fact that there are two intersection points confirms that the system has two solutions. This graphical analysis provides a clear and intuitive understanding of why the system has two solutions, reinforcing the concept that solutions to a system of equations are the points where the graphs of the equations intersect. The visual representation is a powerful tool for understanding the nature of the solutions and the behavior of the equations.

Refuting the Misconception About Vertices

One common misconception about systems of equations involving parabolas is that the vertices of the parabolas are the solutions. This is not always the case, and it's crucial to understand why. In this specific system of equations:

y = 6x^2 + 1
y = x^2 + 4

the vertices of the parabolas are (0, 1) and (0, 4), respectively. However, these points are not the solutions to the system. The solutions are the points where the parabolas intersect, which, as we calculated earlier, are (√(3/5), 23/5) and (-√(3/5), 23/5).

The reason the vertices are not the solutions is that the vertices represent the minimum (or maximum) points of each individual parabola, but they do not necessarily satisfy both equations simultaneously. A solution to a system of equations must satisfy all equations in the system. In this case, the y-values of the vertices are different (1 and 4), meaning they cannot be solutions to both equations at the same x-value (x=0). The vertices are critical points for understanding the shape and position of the parabolas, but they do not directly indicate the solutions to the system.

The misconception arises from a misunderstanding of what a solution to a system of equations represents. A solution is a point that lies on the graph of every equation in the system. Graphically, this means the solution is a point of intersection. The vertices, on the other hand, are specific points on each individual parabola that indicate its lowest or highest point. They are not necessarily points of intersection unless the parabolas are specifically configured to intersect at their vertices, which is not the case here.

To further illustrate this point, consider a simple example where two lines intersect. The intersection point is the solution to the system of linear equations, but the intercepts (where the lines cross the axes) are not solutions unless they happen to coincide with the intersection point. Similarly, for parabolas, the vertices are important features, but they are not inherently solutions to the system. It is the intersection points that define the solutions, as they are the points where both equations hold true simultaneously. Therefore, it is essential to differentiate between the characteristics of individual equations (like vertices) and the solutions of a system, which are the points of intersection of the graphs of the equations. By focusing on the graphical representation and the points of intersection, we can avoid the misconception about vertices and accurately identify the solutions to the system.

Alternative Methods for Solving the System

While the graphical approach provides an intuitive understanding of why the system of equations has two solutions, it's also important to explore alternative algebraic methods for solving the system. These methods provide a more rigorous way to find the solutions and reinforce the concepts discussed earlier.

One common method for solving systems of equations is substitution. In this case, we have:

y = 6x^2 + 1
y = x^2 + 4

Since both equations are already solved for y, we can set them equal to each other:

6x^2 + 1 = x^2 + 4

This is the same equation we derived when discussing the intersection points graphically. Simplifying this equation, we get:

5x^2 = 3
x^2 = 3/5
x = ±√(3/5)

As we found earlier, there are two possible values for x: √(3/5) and -√(3/5). To find the corresponding y-values, we substitute these x-values into either of the original equations. Using y = x^2 + 4:

For x = √(3/5):

y = (√(3/5))^2 + 4 = 3/5 + 4 = 23/5

For x = -√(3/5):

y = (-√(3/5))^2 + 4 = 3/5 + 4 = 23/5

Thus, the solutions are (√(3/5), 23/5) and (-√(3/5), 23/5), which confirms the graphical analysis.

Another method is elimination, although it is less straightforward in this case since both equations are already solved for y. However, we can still apply it by rearranging the equations if needed. The key principle of elimination is to manipulate the equations so that when they are added or subtracted, one of the variables is eliminated. In this case, since the y terms are already isolated, substitution is the more efficient method.

The algebraic solutions confirm the graphical interpretation that the system has two solutions. The substitution method provides a precise way to find these solutions by equating the two expressions for y and solving for x. The two distinct values of x indicate the two points where the parabolas intersect. By substituting these x-values back into one of the original equations, we find the corresponding y-values, completing the solutions. This algebraic approach reinforces the understanding that the solutions to a system of equations are the points that satisfy all equations simultaneously. Both the graphical and algebraic methods are essential tools for analyzing and solving systems of equations, each providing unique insights into the nature of the solutions.

Conclusion

In conclusion, the system of equations:

y = 6x^2 + 1
y = x^2 + 4

has two solutions because the graphs of the two equations, which are parabolas, intersect at two distinct points. These points of intersection represent the (x, y) coordinates that satisfy both equations simultaneously. The first equation, y = 6x^2 + 1, is a narrower parabola with its vertex at (0, 1), while the second equation, y = x^2 + 4, is a wider parabola with its vertex at (0, 4). The difference in their shapes and vertical positions leads to two intersection points.

We demonstrated this graphically by plotting the two parabolas and identifying their points of intersection. These points, (√(3/5), 23/5) and (-√(3/5), 23/5), are the solutions to the system. It is crucial to recognize that the vertices of the parabolas are not the solutions; the solutions are the intersection points. This understanding is essential for avoiding common misconceptions and accurately solving systems of equations.

We also explored alternative algebraic methods, such as substitution, to solve the system. The substitution method involves setting the two equations equal to each other and solving for x. This approach yielded the same two x-values as the graphical analysis, confirming the solutions. By substituting these x-values back into one of the original equations, we found the corresponding y-values, thus obtaining the solutions algebraically.

The combination of graphical and algebraic methods provides a comprehensive understanding of the system. The graphical method offers an intuitive visualization of the solutions as intersection points, while the algebraic methods provide a rigorous way to calculate these solutions. Both methods reinforce the concept that the solutions to a system of equations are the points that satisfy all equations simultaneously. Understanding the properties of parabolas and their interactions is crucial for analyzing and solving systems of quadratic equations. By recognizing that solutions are points of intersection and avoiding the misconception about vertices, we can effectively solve such systems and interpret their solutions accurately.

This analysis highlights the importance of both graphical and algebraic approaches in understanding and solving systems of equations. The ability to visualize the equations and their solutions graphically, combined with the precision of algebraic methods, provides a robust toolkit for tackling mathematical problems. Understanding the underlying principles allows us to confidently address similar problems and deepen our grasp of mathematical concepts.