Problem 22 - Algebra and Trigonometry

Question

\(19-32\) . 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 to each other

Since both equations equal \( y \), we can set them equal to each other: \( 4 - x^2 = x^2 - 4 \).

Step 2: Simplify the equation

Rearrange the equation to bring all terms to one side: \( 4 - x^2 = x^2 - 4 \) becomes \( 4 + 4 = x^2 + x^2 \), which simplifies to \( 8 = 2x^2 \).

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

Divide both sides of the equation \( 8 = 2x^2 \) by 2 to get \( x^2 = 4 \).

Step 4: Solve for \( x \)

Take the square root of both sides: \( x = \pm \sqrt{4} \), so \( x = 2 \) or \( x = -2 \).

Step 5: Substitute \( x \) back into one of the original equations

Use the original equation \( y = 4 - x^2 \). For \( x = 2 \), we have \( y = 4 - 2^2 = 0 \). For \( x = -2 \), we have \( y = 4 - (-2)^2 = 0 \).

Step 6: Present the solutions

The solutions to the system of equations are the points where both equations intersect, which are \((2, 0)\) and \((-2, 0)\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically in the form \( ax^2 + bx + c = 0 \). They are crucial in algebra because they model various phenomena, such as the trajectory of a ball under gravity or the shape of a parabolic dish. Quadratics can have:

• Two distinct real roots.
• One repeated real root.
• No real roots but two complex roots.

The general properties of a quadratic equation help us identify the number of solutions without solving the equation completely. In our exercise, both equations are expressed as quadratics in a system. They take the form \( y = 4 - x^2 \) and \( y = x^2 - 4 \). The goal is to find their points of intersection, which requires setting them equal to each other and solving the resulting equation.

Solution Sets

A solution set is a collection of all values that make an equation or system of equations true. In a system of equations like the one we solved, the solution set consists of the points at which the graphs of the equations intersect. Each point in the solution set satisfies all equations in the system simultaneously.

For the given system of equations:\[\begin{align*}y &= 4 - x^2 \y &= x^2 - 4\end{align*}\]the solution set can be found by determining where these two equations are equal. Setting the equations equal to each other involves combining and simplifying them, allowing us to find the \( x \)-values that satisfy both equations. The final step is to find the corresponding \( y \)-values by substituting these \( x \)-values back into either of the original equations.

For our problem, the solution set is \((2, 0)\) and \((-2, 0)\), meaning the two parabolas intersect at these points.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions and equations to solve for unknown values. In solving systems of equations, algebraic manipulation is essential to isolate variables and uncover solutions. The steps typically include:

• Substituting equivalent expressions.
• Rearranging terms.
• Combining like terms.
• Factorizing where possible.
• Using operations such as addition, subtraction, multiplication, and division.

In this exercise, after defining the equations as equal \( 4 - x^2 = x^2 - 4 \), algebraic manipulation allows us to rearrange the equation to \( 8 = 2x^2 \). Dividing both sides by 2, we find \( x^2 = 4 \). This leads to taking square roots and considering both positive and negative solutions: \( x = 2 \) and \( x = -2 \).

Without algebraic manipulation, isolating the variables in a system of equations would be much more challenging. It not only helps simplify the equations but also aids in accurately identifying the solution set that solves the entire system.