Problem 56 - College Algebra

Question

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 fi st number squared. What are the numbers?

Step-by-Step Solution

Answer

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

Step 1: Define Variables

Let's denote the first number as \( x \), and the second number as \( y \).

Step 2: Translate Relationships into Equations

We are given two main relationships in the problem. First, the squares of the two numbers add up to 360: \( x^2 + y^2 = 360 \). Second, the second number, \( y \), is half of the first number squared: \( y = \frac{x^2}{2} \).

Step 3: Substitute and Simplify

Substitute the expression for \( y \) from the second equation into the first equation to eliminate \( y \):\[ x^2 + \left(\frac{x^2}{2}\right)^2 = 360 \]This simplifies to:\[ x^2 + \frac{x^4}{4} = 360 \]

Step 4: Clear Fractions

To simplify solving the equation, multiply every term by 4 to clear the fraction:\[ 4x^2 + x^4 = 1440 \]

Step 5: Rearrange into a Polynomial Equation

Rearrange the equation to form a quadratic in terms of \( x^2 \): \[ x^4 + 4x^2 - 1440 = 0 \]Let \( z = x^2 \), then we have a quadratic \( z^2 + 4z - 1440 = 0 \).

Step 6: Solve the Quadratic Equation

Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = 4, c = -1440 \):\[ z = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-1440)}}{2 \times 1} \]\[ z = \frac{-4 \pm \sqrt{16 + 5760}}{2} \]\[ z = \frac{-4 \pm 76}{2} \]This yields \( z = 36 \) and \( z = -40 \). Since \( z = x^2 \) and must be positive, we take \( z = 36 \).

Step 7: Find Values of x from z

Since \( x^2 = 36 \), solving gives \( x = 6 \) or \( x = -6 \).

Step 8: Find Corresponding y Values

Use the equation \( y = \frac{x^2}{2} \) to find \( y \):- If \( x = 6 \), then \( y = \frac{36}{2} = 18 \).- If \( x = -6 \), then \( y = \frac{36}{2} = 18 \).Thus, the pairs \((x, y)\) are \((6, 18)\) and \((-6, 18)\).

Key Concepts

System of Equations

A system of equations involves more than one equation considered simultaneously. In our exercise, we have a **nonlinear system of equations**. This means that at least one of the equations in the system is not linear. For the given problem, the system is constructed from the two statements:

• The squares of two numbers add to 360: \( x^2 + y^2 = 360 \).
• The second number is half the value of the first number squared: \( y = \frac{x^2}{2} \).

To solve such systems, we often use substitution or elimination methods. The idea is to express one variable in terms of others using one equation, and then substitute it into the other, simplifying the problem to a single-variable equation.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of quadratic equations, expressed as \( ax^2 + bx + c = 0 \). In this instance, we rephrase the problem to replace the role of \( x \) with \( z = x^2 \). This rearranges our equation into a more familiar quadratic form:

• \( z^2 + 4z - 1440 = 0 \)

The quadratic formula is given as:\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substituting the appropriate values, \( a = 1 \), \( b = 4 \), and \( c = -1440 \) into the formula helps us find the values of \( z \) that satisfy the equation. These values represent the potential squares of our original variables.

Polynomial Equations

Polynomial equations are equations involving sums of powers of variables. In our exercise, we've ultimately simplified a complex situation into a simple polynomial equation involving \( x^2 \). Polynomial equations can have various degrees, leading to different numbers of possible solutions.

• The non-linear origin of our problem involves both quadratic (\( x^2 \)) and quartic (\( x^4 \)) terms.
• By substituting \( z = x^2 \), we translate the quartic equation into the quadratic form \( z^2 + 4z - 1440 = 0 \).

Solving polynomial equations efficiently often requires reducing them to known forms—like quadratics—allowing application of specific methods such as factoring or using formulas.

Solution of Equations

Finding the solution of an equation means determining the variable values that satisfy the equation. After calculating the potential values of \( z = x^2 \) using the quadratic formula, we find that:

• \( z = 36 \): This leads to \( x^2 = 36 \), thus \( x = 6 \) or \( x = -6 \).
• The negative value of \( z = -40 \) is discarded since a square cannot be negative.

With these \( x \) values, we use the relationship \( y = \frac{x^2}{2} \) to find corresponding \( y \) values. Thus, the solutions are \((6, 18)\) and \((-6, 18)\). In a system context, each pair \((x, y)\) represents a valid solution where both equations are satisfied simultaneously.