Problem 26
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
Verified Answer
Solutions are (2, 0) and (-2, 0).
1Step 1: Set Equations Equal to Each Other
Since both equations equal \(y\), we can set the right-hand sides of the equations equal to each other:\[ 4-x^2 = x^2-4 \] This simplifies the system to a single equation involving \(x\).
2Step 2: Solve for x
Rearrange the equation \(4-x^2 = x^2-4\) to combine like terms:\[ 4 + 4 = x^2 + x^2 \]\[ 8 = 2x^2 \]Divide both sides by 2 to solve for \(x^2\):\[ 4 = x^2 \]Take the square root of both sides:\[ x = 2 \quad \text{or} \quad x = -2 \]
3Step 3: Find Corresponding y Values
Substitute the values of \(x\) into one of the original equations to find \(y\). We'll use \(y = 4 - x^2\):For \(x = 2\):\[ y = 4 - (2)^2 = 4 - 4 = 0 \]For \(x = -2\):\[ y = 4 - (-2)^2 = 4 - 4 = 0 \]
4Step 4: Write the Solution Set
The solutions for the system of equations are the points where both equations are satisfied. These points are the ordered pairs \[ (2, 0) \text{ and } (-2, 0) \].
Key Concepts
Quadratic FunctionsIntersection PointsAlgebraic Manipulation
Quadratic Functions
Quadratic functions are a vital part of algebra and are often visualized as parabolas on a graph. A standard quadratic equation has the form \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In a quadratic function, the coefficient \( a \) determines the direction of the parabola (upward if positive, downward if negative). The vertex of the parabola provides the peak or the trough point depending on its orientation.
Understanding quadratic functions is key when solving systems of equations that contain them. Such functions have distinctive properties, including:
Understanding quadratic functions is key when solving systems of equations that contain them. Such functions have distinctive properties, including:
- Their graphs are symmetric around the vertical line through the vertex (the axis of symmetry).
- They can have 0, 1, or 2 real roots, which are the points where the parabola intersects the x-axis.
Intersection Points
Intersection points of two functions on a graph are where the graphs of the equations cross or meet. For systems of equations, these points represent the solution(s) to the system. In terms of a graph, these are the coordinates where both equations hold true simultaneously.
In our particular example, the equations \( y = 4 - x^2 \) and \( y = x^2 - 4 \) represent parabolas. The intersection points are where the y-values of both equations are equal for the same x-values. To find these intersection points algebraically, we set the equations equal to each other since they both equal \( y \). Doing so, we derived the equation \( 4 - x^2 = x^2 - 4 \).
Solving this equation, we identified the \( x \) values where the parabolas intersect. These \( x \) values, when substituted back into either function, yield the corresponding \( y \) values, finalizing the intersection points as coordinates on the graph.
In our particular example, the equations \( y = 4 - x^2 \) and \( y = x^2 - 4 \) represent parabolas. The intersection points are where the y-values of both equations are equal for the same x-values. To find these intersection points algebraically, we set the equations equal to each other since they both equal \( y \). Doing so, we derived the equation \( 4 - x^2 = x^2 - 4 \).
Solving this equation, we identified the \( x \) values where the parabolas intersect. These \( x \) values, when substituted back into either function, yield the corresponding \( y \) values, finalizing the intersection points as coordinates on the graph.
Algebraic Manipulation
When solving systems of equations, algebraic manipulation is a critical skill that involves rearranging and simplifying equations to find solutions. It typically includes operations such as addition, subtraction, multiplication, division, and applying functions like squaring or taking square roots.
In our solved example, algebraic manipulation helped isolate and solve for \( x \). We started by setting the two quadratic equations equal to each other, creating a single equation from the system. We then rearranged:
In our solved example, algebraic manipulation helped isolate and solve for \( x \). We started by setting the two quadratic equations equal to each other, creating a single equation from the system. We then rearranged:
- Combined like terms by moving terms across the equals sign and reducing the equation to \( 8 = 2x^2 \).
- Divided both sides by 2, simplifying to \( 4 = x^2 \).
- Finally, took the square root of both sides to find \( x = 2 \) or \( x = -2 \).
Other exercises in this chapter
Problem 26
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