Problem 64 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to find the point(s) of intersection of the graphs. Then confirm your solution algebraically. $$\left\\{\begin{array}{l}x^{2}+y^{2}=8 \\ y=x^{2}+4\end{array}\right.$$

Step-by-Step Solution

Answer

Based on these steps, there are no real points of intersection between the given two graphs.

Step 1: Graph the Equations

Using a graphing utility, plot the equations \(x^2 + y^2 = 8\) and \(y = x^2 + 4\). Observe where they intersect each other. The coordinates of these intersection points are the solutions.

Step 2: Set Equations Equal to Each Other

In order to confirm the graphical solution algebraically, set the two equations equal to each other and solve for x. Thus we have: \(x^2 + y^2 = y\). Since \(y = x^2 + 4\), substitute it into the equation to get \(x^2 + (x^2 + 4)^2 = x^2 + 4\).

Step 3: Solve for x

Expand and simplify the equations to get a quadratic equation in terms of x. The equation becomes \(x^4 + 8x^2 + 16 = x^2 + 4\). This simplifies to \(x^4 + 7x^2 + 12 = 0\).

Step 4: Factor the Quadratic

Factor the quadratic equation as follows: \(x^4 + 4x^2 + 3x^2 + 12 = 0 \rightarrow x^2(x^2 + 4) + 3(x^2 + 4) = 0 \rightarrow (x^2 + 3)(x^2 + 4) = 0\).

Step 5: Solve for x

After factoring, set each factor equal to zero and solve for x: \((x^2 + 3) = 0\) and \((x^2 + 4) = 0\). This gives \(x = \pm \sqrt{-3}\), and \(x = \pm \sqrt{-4}\). However, in the real number system, square roots of negative numbers do not exist. So, there are no real solutions for x.

Step 6: Find y values

By substituting \(x\) values in \(y = x^2 + 4\), it is found that there are no real solutions for y either as the x solutions do not exist.

Key Concepts

Graphing Utility

A graphing utility is a tool that can help visualize mathematical functions, making it easier to understand the relationship between different equations. For the given exercise, a graphing utility can quickly show where the graphs of the circle equation \(x^2 + y^2 = 8\) and the parabola equation \(y = x^2 + 4\) intersect.
By entering these equations into a graphing calculator or software, students can observe how the shapes relate to each other on a coordinate plane. The intersection points, if any, represent solutions to the system of equations, indicating where both conditions described by the equations are simultaneously satisfied.
Using technology in these contexts allows students to experiment and quickly try different scenarios, giving them a hands-on understanding of graphical solutions.

• Plot both equations in the graphing utility.
• Identify where the parabola and circle meet on the graph.
• These points potentially represent solutions to the system of equations.

Quadratic Equations

Quadratic equations are a fundamental part of algebra and form the base for various complex mathematical problems. They generally take the form \(ax^2 + bx + c = 0\) and describe parabolas when graphed on a coordinate plane.
In the context of the exercise, the equation \(y = x^2 + 4\) is quadratic. It represents a parabola that has been shifted upwards by 4 units along the y-axis. The standard solving approach for quadratic equations involves factoring, completing the square, or using the quadratic formula to find the values of \(x\) that satisfy the equation.

• Recognize different forms of quadratic equations.
• Graph them as parabolas to find visual solutions.
• Use algebraic methods to confirm graphical findings.

Algebraic Solution

An algebraic solution involves manipulating equations to find exact answers rather than approximations available through graphs. This method often confirms or helps further explore the graphical findings.
For these equations, setting \(y = x^2 + 4\) from the parabola into the circle's equation \(x^2 + y^2 = 8\), we derive the equation \(x^4 + 7x^2 + 12 = 0\). This results in two quadratic expressions that must be factored to solve for \(x\).
Factoring methods like grouping or using identities simplify the equation into factors whose solutions are solvable in real numbers. However, in this scenario, the solutions involve square roots of negative numbers, indicating no real intersection points. Therefore,

• Perform algebraic operations to substitute and combine equations.
• Factor the resulting quadratics.
• Identify whether solutions exist in the real number system.

By confirming graphical solutions or finding potential discrepancies, algebraic solutions enhance understanding and ensure concept clarity.