Problem 22 - Precalculus Mathematics 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 are (2, 0) and (-2, 0).

Step 1: Set Equations Equal

We know that both equations represent expressions for the same variable, y. Therefore, set them equal to each other. That is, set \( 4 - x^2 = x^2 - 4 \).

Step 2: Solve for x

Now we solve the equation from Step 1. Rearrange the terms: \( 4 - x^2 = x^2 - 4 \) gives us \( 4 + 4 = x^2 + x^2 \), simplifying to \( 8 = 2x^2 \). Divide both sides by 2 to obtain \( x^2 = 4 \).

Step 3: Find Values of x

Take the square root of both sides to solve for x. We get \( x = \pm 2 \). So, the possible values for x are 2 and -2.

Step 4: Substitute x-values into Original Equations

Substitute \( x = 2 \) into both original equations to find corresponding y-values. In both cases, \( y = 4 - 2^2 = 0 \) and \( y = 2^2 - 4 = 0 \). Substitute \( x = -2 \) into both equations and similarly find \( y = 4 - (-2)^2 = 0 \) and \( y = (-2)^2 - 4 = 0 \). Both equations confirm that \( y = 0 \) for both x-values.

Step 5: Conclusion

There are two solutions to the system of equations. The solutions are \( (2, 0) \) and \( (-2, 0) \).

Key Concepts

Solving Quadratic Equations

Quadratic equations are equations where the highest exponent of the variable (usually x) is 2. Such equations often take the form \(ax^2 + bx + c = 0\). In the exercise, we rearranged and solved the quadratic-like equation \(4 - x^2 = x^2 - 4\). Here, we collected like terms to create a true quadratic equation: \(2x^2 = 8\). Solving this involves:

• Rearranging: Combine like terms to simplify the equation.
• Dividing both sides by the same number: Divide by 2 to lessen complexity, leading to \(x^2 = 4\).
• Finding x: Take the square root of both sides to solve for x, yielding \(x = \pm 2\).

Quadratic equations can have two solutions, as shown by the \(\pm\) sign in our solutions. It is essential to remember that the square root of a number can be both positive and negative.

Simultaneous Equations

Simultaneous equations are sets of equations that share common variables. The challenge is finding variable values that satisfy all equations. In this case, the two equations given were both in terms of \(y\).

To solve simultaneous equations:

• Set equations equal: Since they represent the same quantity (y), we set \(4 - x^2 = x^2 - 4\). This lets us focus on solving for x first.
• Solve for one variable: Solve this combined equation to find values of one variable—in this case \(x = \pm 2\).
• Substitute back: Use found values to calculate the other variable, y.

By solving simultaneously, we account for all the conditions presented by each equation, ensuring no solutions are missed.

Roots of Equations

The roots of an equation are the solutions, or values of x, that satisfy that equation. When solving quadratic equations like in the exercise, finding the roots means identifying where the given function equals zero.

For our quadratic equation, the process involved:

• Rewriting: Combining equations yielded \(2x^2 = 8\), simplified to \(x^2 = 4\).
• Determining roots: Solving \(x^2 = 4\) gives roots \(x = 2\) and \(x = -2\), as both these values satisfy the equation.

These roots represent the x-values where the graphs of the equations intersect or touch the x-axis, which denotes the values that make the original system of equations true.
Roots are crucial in understanding the behavior and intersection points of graphs relating to these equations.