Problem 26 - Precalculus: Mathematical for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{l}y=4-x^{2} \\\y=x^{2}-4\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions to the system are (2,0) and (-2,0).

Step 1: Set the Equations Equal to Each Other

The given system of equations is \( y = 4 - x^2 \) and \( y = x^2 - 4 \). Since they both equal \( y \), set them equal to each other: \[ 4 - x^2 = x^2 - 4. \]

Step 2: Simplify the Equation

Rearrange the equation \( 4 - x^2 = x^2 - 4 \) to get all terms involving \( x \) on one side. This gives us \[ 4 + 4 = x^2 + x^2. \] Thus, the equation simplifies to \[ 8 = 2x^2. \]

Step 3: Solve for \( x^2 \)

Divide each side of the equation by 2 to solve for \( x^2 \): \[ x^2 = 4. \]

Step 4: Solve for \( x \)

Take the square root of both sides of the equation \( x^2 = 4 \) to find \( x \): \[ x = \pm 2. \]

Step 5: Solve for \( y \) When \( x = 2 \) and \( x = -2 \)

Substitute \( x = 2 \) and \( x = -2 \) into one of the original equations to find \( y \). Using \( y = 4 - x^2 \):- For \( x = 2 \): \( y = 4 - (2)^2 = 4 - 4 = 0. \)- For \( x = -2 \): \( y = 4 - (-2)^2 = 4 - 4 = 0. \)

Step 6: Write the Solutions

The intersection points where both equations are satisfied are the solutions. Since both cases resulted in \( y = 0 \), the solutions are \((2, 0)\) and \((-2, 0)\).

Key Concepts

Solution of Equations

To find the solution of a system of equations, such as the one given in the exercise, it means identifying the points where both equations share a common solution in their graph. In our case, we have two equations both equal to \( y \):

• \( y = 4 - x^2 \)
• \( y = x^2 - 4 \)

By setting these two expressions equal to each other, we establish a new equation: \( 4 - x^2 = x^2 - 4 \). This simplifies the problem to finding the correlation between \( x \) values where these two curves intersect. By solving this equation, we can determine the particular values for \( x \) and, subsequently, \( y \).
The solution of equations involves finding values that satisfy both equations at the same time, resulting in specific points that are solutions for the system. In simpler terms, a solution to a system represents the point(s) of intersection on a graph of these equations.

Quadratic Equations

Quadratic equations often appear in systems, as shown in the exercise. These equations are polynomial equations in the form of \( ax^2 + bx + c = 0 \). Quadratic equations have distinct characteristics:

• Their highest degree is 2, which is why they are called quadratic.
• They graph as parabolas, which have a specific shape that is symmetric.

In our exercise, each function is shaped like a parabola. The first function \( 4 - x^2 \) opens downwards (since the coefficient of \( x^2 \) is negative), while the second function \( x^2 - 4 \) opens upwards. Recognizing that both are quadratic helps in understanding why setting them equal allows their intersection points to be found efficiently.
Solving a quadratic often involves steps like rearranging terms, factoring, or using the quadratic formula. But in systems like ours, recognizing symmetry often simplifies finding the exact solutions, because it directly involves finding where these parabolas cross each other.

Algebraic Simplification

Solving the system of equations often requires algebraic simplification, an essential tool in transforming complex expressions into more manageable ones. Simplification can involve:

• Reorganizing equations by moving terms to either side.
• Combining like terms.
• Dividing or multiplying through by constants to isolate variables.

In the exercise solution, the original equation \( 4 - x^2 = x^2 - 4 \) was simplified to \( 8 = 2x^2 \) by moving all terms involving \( x \) to one side and constants to the other. Simplification is not merely an arithmetic process; it involves logical decisions about how best to approach a solution.
This process is vital because it can make apparently complicated equations much more straightforward. Through simplifying, solving the mathematics behind these curves becomes feasible, and clearer, culminating in finding particular solutions like \( x = 2 \) and \( x = -2 \), leading us to the final solutions \((2,0)\) and \((-2,0)\).