Problem 17 - Beginning and Intermediate Algebra

Question

Write an equation and solve. The product of two consecutive odd integers is 1 less than three times their sum. Find the integers.

Step-by-Step Solution

Answer

The two pairs of consecutive odd integers that satisfy the given condition are (5, 7) and (-1, 1). This solution is derived from the quadratic equation \(x^2 - 4x - 5 = 0\), which has two possible solutions for x (5 and -1). Using the definition of y as \(y = x + 2\), the consecutive odd integers are obtained.

Step 1: Define Variables

Let x be the first odd integer and y be the second consecutive odd integer. Since they are consecutive, we can write y = x + 2.

Step 2: Write the Equation using the given information

The product of the two integers is 1 less than three times their sum. We can write this as: xy = 3(x + y) - 1 We can replace y with x + 2, so our equation becomes: x(x + 2) = 3(x + (x + 2)) - 1

Step 3: Solve the Equation for x

Now we solve the equation for x: x(x + 2) = 3(2x + 2) - 1 Expanding the equation, we get: x^2 + 2x = 6x + 6 - 1 Combining like terms, we get: x^2 + 2x - 6x - 5 = 0 x^2 - 4x - 5 = 0

Step 4: Factor the Quadratic Equation

Factor the quadratic equation to solve for x: (x - 5)(x + 1) = 0

Step 5: Solve for x

Setting each factor to zero, we get two possible solutions for x: x - 5 = 0 => x = 5 x + 1 = 0 => x = -1

Step 6: Find the Consecutive Odd Integers

Now we find the corresponding two consecutive odd integers using x = 5 and x = -1 and our previous definition of y = x + 2. For x = 5: y = x + 2 = 5 + 2 = 7 So, the first pair of consecutive odd integers are 5 and 7. For x = -1: y = x + 2 = -1 + 2 = 1 So, the second pair of consecutive odd integers are -1 and 1. Hence, the two pairs of consecutive odd integers that satisfy the given condition are (5, 7) and (-1, 1).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. They come in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Solving these equations typically yields two solutions. Quadratics can represent many real-world problems, from projectile motion in physics to economics models.In our original exercise, the problem transformed into a quadratic equation: \( x^2 - 4x - 5 = 0 \). The unknowns represent odd integers, making quadratics a useful tool to find these specific numbers. Our goal is to find the values of \( x \) that make this equation true. With quadratic equations, we often proceed to factor them to simplify finding the solutions.

Factoring Equations

Factoring is a crucial step in solving quadratic equations. It involves rewriting the quadratic equation as a product of two binomials. Let's take our problem's quadratic equation, \( x^2 - 4x - 5 = 0 \).

• First, we look for two numbers that multiply to \( -5 \) (the constant term) and add up to \( -4 \) (the coefficient of the linear term).
• Those numbers are \( -5 \) and \( 1 \).
• Thus, we factor the equation as \( (x - 5)(x + 1) = 0 \).

Factoring breaks the equation into simpler parts. It clearly shows where the solution spots are found by setting each factor to zero. This step is often where algebra students triumph over confusion.

Integer Equations

Integer equations involve finding whole number solutions for the variables. Our exercise required us to work with consecutive odd integers.Consecutive odd integers differ by two. So, if one integer is \( x \), the next is \( x + 2 \). With our solution:

• Solve \( x - 5 = 0 \) gives \( x = 5 \).
• The consecutive odd integer is \( y = 5 + 2 = 7 \).
• Alternatively, solve \( x + 1 = 0 \) gives \( x = -1 \).
• Then, the other integer is \( y = -1 + 2 = 1 \).

Thus, the integer pairs are (5, 7) and (-1, 1). These reflect whole-number solutions that make sense in the context of the problem.

Algebra Problem Solving

Algebra problem solving is an essential skill that involves various methods or strategies. It enables students to break down problems into manageable steps.In our exercise:

• Start by defining variables based on the problem context. We used \( x \) and \( y = x + 2 \) for the integers.
• Write an equation that translates the problem's word statement into mathematical language.
• Transform this equation into a recognizable algebraic form, like a quadratic equation.
• Apply techniques such as factoring and setting equations to zero to find potential solutions.

This step-by-step approach is typical in algebra and provides a structured method to solve equations of various types, including those with integers and polynomials.