Problem 343 - Elementary Algebra

Question

The product of two consecutive integers is 56. Find the integers.

Step-by-Step Solution

Answer

The consecutive integers are either -8 and -7, or 7 and 8.

Step 1: Set up the equation

Let the two consecutive integers be represented by and +1. The product of these integers is given by the equation: (n) (n + 1) = 56.

Step 2: Expand the equation

Expand the left side of the equation to form a quadratic equation: n^2 + n = 56.

Step 3: Rearrange the equation

Rearrange the quadratic equation to set it to zero: n^2 + n - 56 = 0.

Step 4: Factor the quadratic equation

Factor the quadratic equation to find the possible values of : (n + 8)(n - 7) = 0.

Step 5: Solve for

Set each factor equal to zero and solve for : + 8 = 0 or - 7 = 0, which gives = -8 or = 7.

Step 6: Determine the consecutive integers

Substitute back into and +1 to find the pairs of consecutive integers. For = -8, the integers are -8 and -7. For = 7, the integers are 7 and 8.

Key Concepts

quadratic equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents the variable. In this exercise, we encounter a quadratic equation when looking for the product of two consecutive integers that equals 56. In our case, after expanding the initial equation \(n(n + 1) = 56\) and rearranging it, we get \(n^2 + n - 56 = 0\).
This is a standard quadratic equation where the coefficients are:

• \(a = 1\)
• \(b = 1\)
• \(c = -56\)

Solving quadratic equations usually involves methods like factoring, using the quadratic formula, or completing the square. In this problem, we use factoring as the primary method to find the integer solutions.

factoring

Factoring is a method used to break down a complex expression into simpler components, called factors, that when multiplied together give the original expression. In solving the quadratic equation \(n^2 + n - 56 = 0\), our goal is to find two numbers that multiply to \(c\) (in this case, -56) and add to \(b\) (in this case, 1). By inspection or trial and error, we find that these numbers are 7 and -8 because:

• \(7 \times -8 = -56\)
• \(7 + (-8) = -1\)

Thus, we can write the quadratic equation as \((n + 8)(n - 7) = 0\). This factored form reveals the roots or solutions of the equation.

integer solutions

Finding integer solutions involves determining the specific integer values that satisfy the equation. From our factored quadratic equation \((n + 8)(n - 7) = 0\), we set each factor equal to zero and solve for \(n\):

• \(n + 8 = 0\)
• \(n - 7 = 0\)

Solving these, we get:

• \(n = -8\)
• \(n = 7\)

These values of \(n\) are the integer solutions to the equation. By substituting \(n\) back into our expressions for the consecutive integers, we get the pairs of consecutive integers that multiply to 56:

• For \(n = -8\), the integers are -8 and -7.
• For \(n = 7\), the integers are 7 and 8.

product of numbers

The product of numbers is simply the result of multiplying them together. In this problem, we are given that the product of two consecutive integers is 56. Consecutive integers are numbers that follow each other in order, differing by 1.
For example:

• The consecutive integers following -8: -8, -7
• The consecutive integers following 7: 7, 8

By setting up an equation \((n)(n + 1) = 56\), we represent the product of two unknown consecutive integers. Expanding and solving this equation eventually points to pairs of integers that multiply to give the desired product of 56. Through solving, we find these pairs:

• The negative pair: -8 and -7
• The positive pair: 7 and 8

Understanding how to work with the product of numbers and consecutive integers is crucial in solving such algebraic problems.