Problem 30 - Intermediate Algebra

Question

Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=16 \\ y &=-\frac{1}{4} x^{2}+4 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The real solution is \((0, 4)\).

Step 1: Understand the System

We have two equations here. The first equation \(x^2 + y^2 = 16\) is a circle with radius 4 centered at the origin. The second equation \(y = -\frac{1}{4}x^2 + 4\) is a downward opening parabola that intersects the y-axis at 4.

Step 2: Substitute Expression for y

Substitute \( y = -\frac{1}{4}x^2 + 4 \) from the second equation into the first equation \( x^2 + y^2 = 16 \). This gives \( x^2 + \left(-\frac{1}{4}x^2 + 4\right)^2 = 16 \).

Step 3: Simplify Substituted Equation

Simplify the resulting equation: \[ x^2 + \left(-\frac{1}{4}x^2 + 4\right)^2 = 16 \] First, expand the squared term: \[ \left(-\frac{1}{4}x^2 + 4 \right)^2 = \frac{1}{16}x^4 - \frac{1}{2}x^2 \cdot 4 + 16 = \frac{1}{16}x^4 - \frac{1}{2}x^2 + 16 \].

Step 4: Combine Terms

Combine terms from Simplify Substituted Equation:\[ x^2 + \frac{1}{16}x^4 - \frac{1}{2} x^2 + 16 = 16 \] Combine the \(x^2\) terms:\[ \frac{1}{16}x^4 + \frac{1}{2}x^2 = 0 \].

Step 5: Factor the Polynomials

Factor out \(\frac{1}{16}x^2\) from \(\frac{1}{16}x^4 + \frac{1}{2}x^2 = 0\):\[ \frac{1}{16}x^2(x^2 + 8) = 0 \].

Step 6: Solve for x

Set each factor to zero:1. \(\frac{1}{16}x^2 = 0\) implies \(x = 0\).2. \(x^2 + 8 = 0\) has no real solutions since \(x^2 = -8\) is not possible for real numbers.

Step 7: Solve for Corresponding y

Use \(x = 0\) in the linear equation for \(y\):\[ y = -\frac{1}{4}(0)^2 + 4 = 4 \].

Step 8: Conclude Real Solutions

The only real solution to the system is \((x, y) = (0, 4)\).

Key Concepts

Circle Equation

A circle equation is a mathematical representation of a circle in a coordinate plane. It expresses the relationship between the coordinates of the points on the circle and its center and radius. In its simplest form, the equation of a circle centered at the origin is given by:

• \(x^2 + y^2 = r^2\)

Here, \(r\) is the radius of the circle. For example, in the given exercise, the circle equation is \(x^2 + y^2 = 16\). This indicates a circle centered at the origin \((0, 0)\) with a radius of 4, since \(\sqrt{16} = 4\). Each point \((x, y)\) satisfying this equation lies on the circumference of the circle. Understanding how to interpret and manipulate circle equations is essential in solving systems that involve circles.

Parabola Equation

A parabola is a U-shaped curve that can open up, down, left, or right. The equation of a parabola in the standard form is usually written as:

• \(y = ax^2 + bx + c\)

This form represents a parabola that opens up or down depending on the sign of \(a\). In the exercise, the equation \(y = -\frac{1}{4}x^2 + 4\) defines a parabola that opens downward. The vertex of this parabola is at the point \((0, 4)\), meaning it reaches its highest point at \(y = 4\) when \(x = 0\). Recognizing the shape and position of a parabola on a graph helps visualize how it might intersect with other curves such as circles, providing potential solutions or points of intersection for those systems.

Real Solutions

Real solutions are solutions to equations that result in real numbers. When dealing with nonlinear systems of equations involving circles and parabolas, we often look for intersection points that have real coordinates as solutions. In this exercise, we seek real solutions for the system:

• \(x^2 + y^2 = 16\)
• \(y = -\frac{1}{4}x^2 + 4\)

Real solutions are practical and can be plotted or visualized on a graph. Here, we found that \((x, y) = (0, 4)\) is the only real solution since all other potential intersection points resulted in non-real values. Recognizing real solutions in different contexts helps in solving and understanding mathematical problems.

Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of equations. It involves solving one of the equations for one variable and then substituting this expression into another equation. This method enables us to reduce the number of variables and simplify the problem. In the given problem, we started with two equations:

• \(x^2 + y^2 = 16\)
• \(y = -\frac{1}{4}x^2 + 4\)

We used the expression obtained from the second equation to replace \(y\) in the circle equation. The substitution resulted in a single equation with one variable, \(x\). This approach allowed us to easily determine the x-values that could potentially solve the system. Later, these values of \(x\) are used to find corresponding \(y\) values. Using substitution is effective in resolving complex systems systematically.