Solving Systems Of Equations: A Graphical Approach

Let's dive into the fascinating world of systems of equations, where we'll explore how to find the solutions to a set of equations graphically. Our focus will be on the following system:

$\{y^2 + x^2 = 20 \\ y = -x^2$

This system presents a unique challenge, mixing a quadratic equation with a circle. To unravel this, we'll need to understand how each equation behaves geometrically and then find the points where they intersect. These intersection points represent the real solutions to the system.

Understanding the Equations

The Circle Equation: $y^2 + x^2 = 20$

First, let's look at the equation $y^2 + x^2 = 20$. This equation represents a circle. The general form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. In our case, the equation is centered at the origin $(0, 0)$. To find the radius, we take the square root of 20, which is approximately 4.47. This means the circle has a radius of about 4.47 units, extending from the origin in all directions.

Now, imagine plotting this circle on a coordinate grid. You'd have a perfect circle centered at (0, 0) and extending out to touch the points (4.47, 0), (-4.47, 0), (0, 4.47), and (0, -4.47) and all the other points in the x-y coordinates that satisfy the equation. This circle encompasses all the points whose distance from the origin is exactly 4.47 units.

The Parabola Equation: $y = -x^2$

Next, let's analyze the second equation, $y = -x^2$. This is a quadratic equation, and it represents a parabola. The negative sign in front of the $x^2$ indicates that the parabola opens downwards. The vertex of this parabola is at the origin $(0, 0)$. As $x$ increases or decreases from zero, $y$ becomes increasingly negative, forming the characteristic U-shape of a downward-opening parabola.

Think about plugging in some values for x. When x = 0, y = 0. When x = 1 or -1, y = -1. When x = 2 or -2, y = -4. These points define the shape of the parabola, starting at the origin and curving downwards.

Finding the Intersection Points

The real solutions to the system of equations are the points where the circle and the parabola intersect. To find these points, we can use the substitution method. We know that $y = -x^2$, so we can substitute this into the first equation:

$y^2 + x^2 = 20$

Substitute $y$:

$(-x^2)^2 + x^2 = 20$

Simplify:

$x^4 + x^2 = 20$

This is a quartic equation, but we can treat it as a quadratic equation in terms of $x^2$. Let $u = x^2$, then we have:

$u^2 + u - 20 = 0$

Now we can factor this quadratic equation:

$(u + 5)(u - 4) = 0$

This gives us two possible values for u: $u = -5$ or $u = 4$.

Since $u = x^2$, we can find the values of x. For $u = -5$, we have $x^2 = -5$. This gives us imaginary solutions for x, which we don't need for this problem, as the question requests the real solutions.

For $u = 4$, we have $x^2 = 4$. This gives us $x = 2$ and $x = -2$. Now we can use the equation $y = -x^2$ to find the corresponding y values.

When $x = 2$, $y = -(2)^2 = -4$. So, one solution is $(2, -4)$.

When $x = -2$, $y = -(-2)^2 = -4$. So, another solution is $(-2, -4)$.

Plotting the Solutions

Now, let's plot the real solutions we found on the coordinate grid. We have two points: $(2, -4)$ and $(-2, -4)$.

To visualize this, plot these two points on the same graph as the circle and parabola. The points will be on the downward-opening parabola, and inside the circle.

1. Plot Point 1: (2, -4)
   • Move 2 units to the right along the x-axis. From there, move 4 units down along the y-axis. Mark this point.
2. Plot Point 2: (-2, -4)
   • Move 2 units to the left along the x-axis. From there, move 4 units down along the y-axis. Mark this point.

These two points are the real solutions to the system of equations. They represent the places where the circle and the parabola intersect. In other words, these are the only points that satisfy both equations simultaneously.

Graphical Representation

To make this even clearer, imagine sketching the circle and the parabola on the same coordinate plane. The circle would be centered at the origin, with a radius of approximately 4.47. The parabola would open downwards, also with its vertex at the origin. You would observe that the parabola intersects the circle at the two points we calculated: (2, -4) and (-2, -4). The graph would visually confirm that these are the only points that satisfy both the circle and parabola equations.

Conclusion: Solutions Unveiled

We successfully found the real solutions to the system of equations $\{y^2 + x^2 = 20, y = -x^2\}$. The solutions are the points where the circle and the parabola intersect on the coordinate grid. By using substitution and solving the resulting equations, we determined that the real solutions are (2, -4) and (-2, -4). These points represent the specific locations where the circle and parabola share the same x and y coordinates, effectively 'solving' the system.

In summary, the key steps included:

• Recognizing the geometric shapes represented by each equation.
• Using substitution to solve for x and y.
• Graphically plotting the solutions to visualize the intersections.

This process highlights the power of combining algebraic techniques with graphical representations to solve complex problems in mathematics. Understanding how different equations behave and how they interact allows us to find solutions, providing a deep insight into mathematical relationships.

For a deeper dive, consider checking out this article from Khan Academy Systems of equations intro. It has a great overview of the topic and provides additional practice problems. This will further solidify your understanding of solving systems of equations. Keep practicing, and you'll become a master in no time!