Problem 24 - College Algebra

Question

For the following exercises, 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: Identify the Equations

We're given two equations. Equation 1 is \( x^2 + y^2 = 9 \) and equation 2 is \( y = 3 - x^2 \). Our task is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Substitute y in Equation 1

Using equation 2, substitute \( y = 3 - x^2 \) into equation 1. This gives us \( x^2 + (3 - x^2)^2 = 9 \).

Step 3: Expand the Substituted Equation

Expand \( (3 - x^2)^2 \) to get \( 9 - 6x^2 + x^4 \). Substitute in the expanded form into the first equation to get: \( x^2 + 9 - 6x^2 + x^4 = 9 \).

Step 4: Simplify the Equation

Simplify the equation: \( x^4 - 5x^2 = 0 \). Factor the equation as \( x^2(x^2 - 5) = 0 \).

Step 5: Solve for x

Set each factor to zero: \( x^2 = 0 \) or \( x^2 - 5 = 0 \). From \( x^2 = 0 \), we get \( x = 0 \). From \( x^2 = 5 \), we get \( x = \pm \sqrt{5} \).

Step 6: Find Corresponding Values of y

Substitute \( x = 0 \) into \( y = 3 - x^2 \), yielding \( y = 3 \). Substitute \( x = \sqrt{5} \) and \( x = -\sqrt{5} \) into the same equation, yielding \( y = -2 \).

Step 7: Verify the Solutions

For \( (0, 3) \), substitute into \( x^2 + y^2 \) to check: \( 0^2 + 3^2 = 9 \), which satisfies equation 1. For \( (\sqrt{5}, -2) \) and \((-\sqrt{5}, -2)\), both satisfy: \( (sqrt{5})^2 + (-2)^2 = 5 + 4 = 9 \), meeting equation 1's requirement.

Key Concepts

Substitution Method

The substitution method is a technique for solving systems of equations. It involves expressing one variable from one equation in terms of the other variable, and then substituting this expression into the other equations.
This helps to reduce the number of variables, making the system easier to solve.In the context of our nonlinear system:

• We had two equations: \( x^2 + y^2 = 9 \) and \( y = 3 - x^2 \).
• By using \( y = 3 - x^2 \) (from equation 2), we substituted \( y \) in the first equation.
• This gave us \( x^2 + (3 - x^2)^2 = 9 \), which simplifies the system into one equation with one variable.

Substituting in this way simplifies complex systems, transforming it into simpler quadratic equations, which are easier to handle.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2 and are typically in the form \( ax^2 + bx + c = 0 \). Solving these involves finding the roots, or values of \( x \), which might be real or complex numbers.
Quadratic equations can have up to two distinct real solutions.For the exercise problem:

• After substitution and simplification, the equation we obtained was \( x^4 - 5x^2 = 0 \).
• This equation can be treated as a quadratic in terms of \( x^2 \): specifically, as \( (x^2)^2 - 5(x^2) = 0 \).

By recognizing this format, the complex-looking equation transforms into a standard quadratic type, which can be solved using well-known methods such as factoring or the quadratic formula.

Factoring

Factoring is a technique used to simplify and solve polynomial equations. It involves breaking down an equation into a product of simpler terms, making the solutions more apparent.
It's particularly effective for solving quadratic equations.In this exercise:

• We had \( x^4 - 5x^2 = 0 \).
• This expression was factored as \( x^2(x^2 - 5) = 0 \), which separates the polynomial into two components \( x^2 \) and \( x^2 - 5 \).
• Setting each factor equal to zero gives us \( x^2 = 0 \) and \( x^2 = 5 \).

Solving these simpler equations is straightforward, and in this case, they yield the solutions \( x = 0 \), \( x = \pm \sqrt{5} \). This process of factoring not only simplifies finding the solutions but also confirms whether any multiple solutions exist.