Problem 24 - College Algebra

Question

Use any method to solve the nonlinear system. $$ \begin{aligned} x^{2}+y^{2} &=9 \\ y &=3-x^{2} \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((0, 3)\), \((\sqrt{5}, -2)\), and \((-\sqrt{5}, -2)\).

Step 1: Analyze the Equations

The system consists of two equations: \(x^2 + y^2 = 9\) is a circle with radius 3 centered at the origin, and \(y = 3 - x^2\) is a downward-facing parabola. We need to find the points where these equations intersect.

Step 2: Substitute for y

From the second equation, \(y = 3 - x^2\). Substitute this expression into the first equation to eliminate \(y\): \(x^2 + (3-x^2)^2 = 9\).

Step 3: Simplify the Equation

Expand \((3-x^2)^2\) to get \(9 - 6x^2 + x^4\). Substitute it back into the equation: \(x^2 + 9 - 6x^2 + x^4 = 9\). Simplify to obtain \(x^4 - 5x^2 = 0\).

Step 4: Factor the Quadratic in Disguise

Rewrite \(x^4 - 5x^2 = 0\) as \(x^2(x^2 - 5) = 0\). This gives us two possible equations to solve: \(x^2 = 0\) and \(x^2 = 5\).

Step 5: Solve for x

Solving \(x^2 = 0\) gives \(x = 0\). Solving \(x^2 = 5\) gives \(x = \sqrt{5}\) and \(x = -\sqrt{5}\).

Step 6: Solve for y

Use the expressions for \(x\) to find \(y\): - When \(x = 0\), \(y = 3 - 0^2 = 3\).- When \(x = \sqrt{5}\), \(y = 3 - (\sqrt{5})^2 = 3 - 5 = -2\).- When \(x = -\sqrt{5}\), \(y = 3 - (-\sqrt{5})^2 = 3 - 5 = -2\).

Step 7: Compile the Solutions

The system's solutions, or intersection points of the circle and the parabola, are: \((0, 3)\), \((\sqrt{5}, -2)\), and \((-\sqrt{5}, -2)\).

Key Concepts

Circle Equation

A circle is a two-dimensional shape where every point on its boundary is equidistant from a fixed center point. This distance is called the radius. The standard equation of a circle in the Cartesian coordinate plane is given by

• \( x^2 + y^2 = r^2 \)

where \( r \) is the radius of the circle. In the given problem, the circle's equation is \( x^2 + y^2 = 9 \), which means our circle has a radius of 3 (since \( r^2 = 9 \) and \( r = \sqrt{9} = 3 \)) and is centered at the origin, \((0,0)\).
Circles play an important role in various mathematical contexts and applications, often used to model situations involving constant distances. Understanding their equations helps solve geometric problems, such as finding intersection points with other shapes like lines and parabolas.

Parabola Equation

A parabola is a symmetric curve that can model various real-world phenomena, such as projectile motion or optimization problems. In the Cartesian plane, the standard form of a parabola is typically given by

• \( y = ax^2 + bx + c \)

where \( a \), \( b \), and \( c \) are constants. The given parabola equation \( y = 3 - x^2 \) is a specific case of the standard form with \( a = -1 \), \( b = 0 \), and \( c = 3 \). This means our parabola opens downwards because the leading coefficient \( a \) is negative. The vertex of this parabola, which is its highest point due to the negative \( a \), is located at \((0, 3)\).
Parabolas have distinct properties, such as symmetry and a single vertex point, making them easy to recognize and graph. They are not only important in algebra but also in fields like physics, where they describe the trajectory of objects under the force of gravity without air resistance.

Intersection Points

Finding intersection points between two curves involves determining where they meet or cross each other on the graph. The intersection points represent solutions that satisfy both equations in a system simultaneously. In this particular problem, we look for points where the circle and the parabola equations are equal.
To find these intersection points:

• Substitute the expression for \( y \) from the parabola equation into the circle equation to eliminate one variable.
• Solve the resulting single-variable equation to find values of \( x \).
• Plug these \( x \) values back into one of the original equations to solve for corresponding \( y \) values.

The solutions of the problem gave us the points \((0, 3)\), \((\sqrt{5}, -2)\), and \((-\sqrt{5}, -2)\), meaning that the circle and parabola intersect at these coordinates. Understanding how to find intersection points is critical in mathematics since it involves solving systems of equations, a common task in various scientific fields.

Algebraic Substitution

Algebraic substitution is a fundamental technique in solving systems of equations. It involves replacing a variable in one equation with an expression derived from another equation, simplifying the process into one equation with a single variable.
For this problem, we started with the system:

• \( x^2 + y^2 = 9 \)
• \( y = 3 - x^2 \)

By substituting \( y = 3 - x^2 \) into the circle equation, we eliminated \( y \), leaving an equation with only \( x \). This turned the problem into solving the one-variable equation \( x^4 - 5x^2 = 0 \). Next, factor the equation and solve for \( x \) to discover possible intersection points, further substituting back to find matching \( y \) values. Thus, substituting helps reduce complexity in equations, making it easier to isolate and solve for unknowns.