Problem 56 - College Algebra with Corequisite Support

Question

For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. The squares of two numbers add to 360 . The second number is half the value of the first number squared. What are the numbers?

Step-by-Step Solution

Answer

The numbers are 6 and 18 or -6 and 18.

Step 1: Define the Variables

Let's define the variables. Let \( x \) be the first number and \( y \) be the second number.

Step 2: Formulate the Equations

According to the problem, the squares of the two numbers add to 360. This gives us the first equation: \( x^2 + y^2 = 360 \).

Step 3: Relationship Between the Numbers

The problem states that the second number \( y \) is half the value of the first number squared, which gives us the second equation: \( y = \frac{1}{2}x^2 \).

Step 4: Substitute to Solve for x

Substitute \( y = \frac{1}{2}x^2 \) into the first equation: \[ x^2 + \left(\frac{1}{2}x^2\right)^2 = 360 \] Simplify to get: \[ x^2 + \frac{1}{4}x^4 = 360 \] Multiply through by 4 to clear the fraction: \[ 4x^2 + x^4 = 1440 \].

Step 5: Solve the Quartic Equation

Rearrange the equation to standard form: \[ x^4 + 4x^2 - 1440 = 0 \]To simplify solving, perform substitution: let \( z = x^2 \). Then, \( z^2 + 4z - 1440 = 0 \). Solve this quadratic equation using the quadratic formula:\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 5760}}{2} \]\[ z = \frac{-4 \pm 76}{2} \].

Step 6: Solve for Possible x Values

From simplifying, we find \( z = 36 \) or \( z = -40 \). Since \( z = x^2 \), it must be non-negative, so \( z = 36 \). Thus, \( x^2 = 36 \) leading to \( x = 6 \) or \( x = -6 \).

Step 7: Determine Corresponding y Values

Using the equation \( y = \frac{1}{2}x^2 \), for \( x = 6 \), \( y = \frac{1}{2}(6^2) = 18 \). For \( x = -6 \), \( y = \frac{1}{2}((-6)^2) = 18 \).

Step 8: Solution Validation

Check both solutions: For \( x = 6 \) and \( y = 18 \), \( 6^2 + 18^2 = 36 + 324 = 360 \). For \( x = -6 \) and \( y = 18 \), \( (-6)^2 + 18^2 = 36 + 324 = 360 \). Both combinations satisfy the original equation.

Key Concepts

System of Equations

A system of equations comprises multiple equations that need to be solved together, as they're interconnected. In many real-world problems, the relationships between variables are not isolated, and this is where systems come into play.
In our example exercise, the system consists of two nonlinear equations. By declaring two variables, we can write equations that express the relationships mentioned in the problem:

• The sum of the squares of the numbers
• A specific relationship between the numbers

The task is to find values for these variables where both equations are true simultaneously. This forms the backbone of solving a system of equations, as one equation alone often doesn't capture all necessary constraints.

Quadratic Equations

At the heart of solving the system we encounter a quadratic equation. A quadratic equation is an equation that can be written in the standard form \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants. Quadratic equations are second-degree polynomial equations and have a characteristic U-shaped plot called a parabola.
In solving the given nonlinear system, the substitution of variables reduces the problem to solving a quadratic equation. This approach leverages the quadratic formula, which is a powerful method for finding the roots of any quadratic equation:

• \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula provides solutions in terms of \( z \), representing our transformed variable \( x^2 \), essential for finding the actual numbers in the problem.

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves expressing one variable in terms of another and then substituting this expression into another equation.
In our exercise, we utilized the substitution method by expressing \( y \) in terms of \( x \) (\( y = \frac{1}{2}x^2 \)) and replacing \( y \) in the first equation.
This substitution transforms a complicated system into a single equation in terms of one variable, making it easier to solve. The goal is to simplify the system to ultimately find specific values for the variables:

• First, reduce the number of equations
• Then, manage the complexity of the system more effectively

This method is particularly useful when dealing with nonlinear systems, as it can transform a cumbersome problem into more manageable parts.

Quartic Equation

In our problem, after substitution, we derive an equation that is quartic in nature. A quartic equation involves fourth-degree terms, and it can generally be written as \[ ax^4 + bx^3 + cx^2 + dx + e = 0 \] where at least the coefficient \( a \) is nonzero. These kinds of equations are more complex than quadratic or cubic equations and often require special techniques to solve.
In this exercise, we

• First represented the quartic equation in terms of \( z = x^2 \), simplifying it to a quadratic form \( z^2 + 4z - 1440 = 0 \).
• This technique reduced the complexity by finding solutions for \( z \), which subsequently gave us the values for \( x \).

Understanding how to manipulate quartic equations is crucial in solving more complex polynomial expressions, where the fourth-degree component is significant in the problem.