Problem 29 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{l} y=-x^{2}-2 \\ x^{2}+y^{2}=4 \end{array}$$

Step-by-Step Solution

Answer

The point of intersection between the parabola \(y = -x^2 - 2\) and the circle \(x^2 + y^2 = 4\) is \((x, y) = (0, -2)\).

Step 1: (Step 1: Understand the equations)

The first equation represents a parabola, with the equation: $$ y = -x^2 - 2 $$ The second equation represents a circle, with the equation: $$ x^2 + y^2 = 4 $$ We need to find the points where these two curves intersect.

Step 2: (Step 2: Substitute the first equation into the second equation)

To find the points of intersection, we need to solve the system of equations. We can do this by substituting the first equation into the second equation: $$ x^2 + (-x^2 - 2)^2 = 4 $$

Step 3: (Step 3: Simplify the equation)

Now simplify the resulting equation by expanding the second term: $$ x^2 + (x^4 + 4x^2 + 4) = 4 $$

Step 4: (Step 4: Rearrange the equation to find the x-values)

Rearrange the equation and combine like terms to find the x-values: $$ x^4 + 5x^2 + 4 - 4 = 0 $$ The resulting equation is a quadratic equation in terms of \(x^2\): $$ x^4 + 5x^2 = 0 $$ Factor out \(x^2\): $$ x^2(x^2 + 5) = 0 $$ Solving for x, we find two possible values: $$ x = 0, x = \pm\sqrt{-5} $$ However, the second value for x is not a real number, so the only valid x-value is: $$ x = 0 $$

Step 5: (Step 5: Find the y-value)

Now that we have the x-value, we can substitute it back into the first equation to find the y-value: $$ y = -(0)^2 - 2 $$ So the y-value is: $$ y = -2 $$

Step 6: (Step 6: Write the solution)

The point of intersection between the parabola and the circle is: $$ (x, y) = (0, -2) $$

Key Concepts

Parabola

A parabola is one of the most common curves encountered in algebra. It is a symmetrical curve which can open either upwards or downwards depending on its equation. This curve is described by a quadratic equation in the form of \( y = ax^2 + bx + c \). In our exercise, the given equation is \( y = -x^2 - 2 \). Here, the parabola opens downwards because the coefficient of \( x^2 \) is negative.

• The vertex of this parabola is its highest point, given by the equation's constant term, \( -2 \), because it's opening downwards.
• The axis of symmetry is the line \( x = 0 \), which passes through the vertex.

Recognizing these features helps in understanding where this curve will intersect with other shapes, such as a circle, when solving algebraic systems.

Circle

A circle is a set of points equidistant from a central point, called the center. In algebraic terms, a circle in a plane can be defined by the equation \( x^2 + y^2 = r^2 \) where \( r \) is the radius of the circle.

• In the given exercise, \( x^2 + y^2 = 4 \), which means the circle has a center at the origin \((0, 0)\) and a radius of \( 2 \), as \( r^2 = 4 \) implies \( r = 2 \).
• This circle stays fixed in the coordinate plane and is fully round without any peaks or edges.

Identifying the circle's equation allows us to pinpoint how it might intersect with other curves, such as the parabola in this system.

Quadratic Equation

Quadratic equations play a pivotal role in solving systems involving parabolas and circles, because they help us find intersections. A typical quadratic equation has the form \( ax^2 + bx + c = 0 \). In our exercise, after substituting and simplifying, we get a quadratic equation in terms of \( x^2 \):

• \( x^4 + 5x^2 = 0 \)

Solving this equation involves factoring: \( x^2(x^2 + 5) = 0 \). This gives the possibilities \( x^2 = 0 \) or \( x^2 = -5 \), but only \( x^2 = 0 \) holds real solutions. Thus, \( x = 0 \).
Once the value of \( x \) is found, substitute back to find \( y \).

• In our case, substituting into \( y = -x^2 - 2 \), gives \( y = -2 \).

Understanding how to manipulate and solve quadratic equations is key to solving and finding the points where these two types of algebraic curves intersect.