Problem 55 - College Algebra

Question

For the following exercises, construct a system of nonlinear equations to describe the given behavior, then solve for the requested solutions. Two numbers add up to 300. One number is twice the square of the other number. What are the numbers?

Step-by-Step Solution

Answer

The numbers are 12 and 288.

Step 1: Understand the Problem

We are given two numbers that add up to 300 and one number is twice the square of the other number. We need to find these two numbers.

Step 2: Define Variables

Let \( x \) represent the first number and \( y \) represent the second number.

Step 3: Formulate Equations

Using the information given, we can write the first equation as \( x + y = 300 \). For the second piece of information where one number is twice the square of the other, we write \( y = 2x^2 \). So, our system of equations is: 1. \( x + y = 300 \) 2. \( y = 2x^2 \)

Step 4: Substitute Equation

Substitute the second equation \( y = 2x^2 \) into the first equation \( x + y = 300 \). This gives us: \[ x + 2x^2 = 300 \]

Step 5: Rearrange into a Quadratic Equation

Rearrange the equation from Step 4 to form a standard quadratic equation: \[ 2x^2 + x - 300 = 0 \]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -300 \). Calculate: \[ x = \frac{-1 \pm \sqrt{1 + 2400}}{4} \] \[ x = \frac{-1 \pm 49}{4} \] This gives \( x = 12 \) or \( x = -25.5 \).

Step 7: Verify Solutions for x

Since numbers must be positive, discard \( x = -25.5 \) as it doesn't satisfy the condition of being a positive number. Thus, \( x = 12 \) is the only valid solution.

Step 8: Find the Corresponding y

Substitute \( x = 12 \) back into the second equation \( y = 2x^2 \) to find \( y \): \[ y = 2(12)^2 = 2 \times 144 = 288 \]

Step 9: Conclusion

The two numbers that satisfy the given conditions are 12 and 288. They add up to 300 and one is twice the square of the other.

Key Concepts

Quadratic Equations

Quadratic equations are foundational in the study of nonlinear equations. They are polynomial equations of degree two, which means the highest power of the variable is squared. A standard quadratic equation has the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In our problem, we derived the quadratic equation \( 2x^2 + x - 300 = 0 \) by substituting one equation into another. The solution to a quadratic equation can be found using various methods such as:

• Factoring, if possible, where the quadratic can be expressed as a product of two binomials.
• Completing the square to make the quadratic equation into a perfect square trinomial.
• The quadratic formula, which provides a solution even if the quadratic isn't easily factorable as used in the steps above.

These tools allow us to find solutions for the variable \( x \), which are necessary to solve real-world problems like the one we're examining.

Systems of Equations

A system of equations involves solving two or more equations with the same set of unknowns. In this exercise, we have a system of two equations: one linear \( x + y = 300 \) and one nonlinear \( y = 2x^2 \). Systems of equations can be:

• Linear, where all equations are linear.
• Nonlinear, where at least one equation involves nonlinear terms, like squares or higher powers, as in this exercise.

To solve such systems, the ultimate aim is to find the values of the unknowns that satisfy all the equations simultaneously. The methods for solving these systems include substitution, elimination, and graphical representation. Finding a solution often involves manipulating one or more equations to reduce the number of variables or transform the system into a simpler form.

Solving Equations

Solving equations is at the heart of algebra. It involves finding the value of the variable(s) that make the equation true. In this exercise, solving the equation \( x + 2x^2 = 300 \) allowed us to discover one of the numbers in the problem. Here are some common steps involved:

• Identify: Clearly state what equations are being solved.
• Simplify: Use algebraic techniques like rearranging terms or factoring.
• Solve: Apply methods like substitution or elimination for systems of equations.

For the quadratic equation derived from the substitution process, the quadratic formula was used. Solving equations effectively bridges the gap between the problem posed and the solution obtained. It is essential to understand the underlying principles of each method to apply them correctly.

Substitution Method

The substitution method is a handy technique for solving systems of equations, particularly when one of the equations is easy to express in terms of one variable. This method involves these steps:

• Express one variable in terms of the other using one equation.
• Substitute this expression into the other equation(s), reducing the number of variables.
• Solve the resulting equation for one of the variables.
• Substitute back to find the remaining variable.

In our given problem, the substitution method was crucial. We expressed \( y \) in terms of \( x \) as \( y = 2x^2 \) and substituted this into the linear equation \( x + y = 300 \) to solve for \( x \). This technique is efficient and often simplifies solving systems, especially when dealing with nonlinear equations.