Problem 30 - Algebra A Combined Function

Question

Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=16 \\ y=-\frac{1}{4} x^{2}+4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is the point (0, 4).

Step 1: Identify System of Equations

The given system consists of two equations: 1. \( x^2 + y^2 = 16 \) is a circle with a radius of 4 centered at the origin.2. \( y = -\frac{1}{4}x^2 + 4 \) is a downward opening parabola.

Step 2: Substitution Method

Substitute the expression for \( y \) from the second equation into the first equation:\[ x^2 + \left( -\frac{1}{4}x^2 + 4 \right)^2 = 16 \].

Step 3: Simplify the Equation

Expand and simplify the expression: \[ x^2 + \left( -\frac{1}{4}x^2 + 4 \right)^2 = x^2 + \left( \frac{1}{16}x^4 - \frac{1}{2}x^2 \cdot 4 + 16 \right) = x^2 + \frac{1}{16}x^4 - \frac{1}{2}x^2 + 16 \].

Step 4: Combine Terms and Set Equation to Zero

Combine like terms to get a single equation and rearrange:\[ \frac{1}{16}x^4 + \frac{1}{2}x^2 + x^2 + 16 - 16 = 0 \] which simplifies to \[ \frac{1}{16}x^4 + \frac{3}{2}x^2 = 0 \].

Step 5: Factor Out Common Terms

Factor out the common factor, which is \( x^2 \): \[ x^2 \left( \frac{1}{16}x^2 + \frac{3}{2} \right) = 0 \].

Step 6: Solve for x

This results in two equations: 1. \( x^2 = 0 \), which gives \( x = 0 \).2. \( \frac{1}{16}x^2 + \frac{3}{2} = 0 \), simplifies to \( x^2 = -24 \) (which has no real solution).

Step 7: Substitute x-values into Parabola Equation

Substitute \( x = 0 \) back into the parabola equation \( y = -\frac{1}{4}x^2 + 4 \) to find \( y \):\( y = -\frac{1}{4}(0)^2 + 4 = 4 \).

Step 8: Conclusion of Solution

The only real solution where the circle and parabola intersect is at the point \((0, 4)\).

Key Concepts

Circle and Parabola Intersection

When solving the intersection of a circle and a parabola, we're looking for points that belong to both the circle and the parabola simultaneously. Imagine the circle as a perfectly round shape, and the parabola as a U-shaped curve. Intersection points, if they exist, are where these two shapes meet.
The circle in this problem, represented by the equation \(x^2 + y^2 = 16\), is centered at the origin with a radius of 4. This means any point on the circle is exactly 4 units from the origin.
The parabola, given by the equation \(y = -\frac{1}{4}x^2 + 4\), opens downward because of the negative coefficient of \(x^2\). The vertex of this parabola is at the point \((0, 4)\), which is the highest point of the curve. Visualizing these two shapes helps understanding where they might intersect, which in our exercise is point \((0, 4)\).

Substitution Method

The substitution method is a powerful tool for solving systems of equations, especially when one of the equations can easily be solved for one variable. It involves using the expression for one variable from one equation and substituting it into the other equation. This transforms a system of equations into a single equation in one variable.
In this exercise, we have two equations. The second equation, \(y = -\frac{1}{4}x^2 + 4\), is already solved for \(y\). We substitute this expression for \(y\) into the first equation \(x^2 + y^2 = 16\). This gives us an equation only in terms of \(x\). By eliminating \(y\), the system becomes much easier to handle, allowing us to focus on finding the \(x\)-values of any intersection points.

Factoring Quadratic Equations

Factoring quadratic equations is a key algebraic skill needed to solve quadratic expressions. After substitution and simplification, we often end up with a quadratic equation which can be solved by factoring.
In our case, the equation reduces to \(\frac{1}{16}x^4 + \frac{3}{2}x^2 = 0\). Here, factoring involves identifying common factors and breaking the equation into simpler parts. Notice the common factor \(x^2\), which can be factored out:

• Start by factoring out \(x^2\) from the expression, leaving us with \(x^2 ( \frac{1}{16}x^2 + \frac{3}{2} ) = 0\).

Once factored, we set each factor to zero, solving for \(x\). This results in solutions like \(x^2 = 0\), and can reveal whether real solutions exist. Solving these simpler equations gives us potential solutions, which are then used to find corresponding \(y\)-values.