Problem 20 - Elementary and Intermediate Algebra

Question

Solve each of these number problems. See Example \(1 .\) The sum of the reciprocals of two consecutive even integers is \(\frac{7}{24} \cdot\) Find each integer.

Step-by-Step Solution

Answer

The integers are 6 and 8.

Step 1: Define the Variables

Let's define the two consecutive even integers. Let the first integer be \( x \) and the second be the next even integer, \( x + 2 \).

Step 2: Express the Reciprocals

The reciprocals of these integers are \( \frac{1}{x} \) and \( \frac{1}{x+2} \).

Step 3: Set Up the Equation

According to the problem, the sum of the reciprocals is \( \frac{7}{24} \). We can write the equation as: \[\frac{1}{x} + \frac{1}{x+2} = \frac{7}{24}\]

Step 4: Find a Common Denominator

To combine the fractions on the left, find a common denominator: \[\frac{x+2}{x(x+2)} + \frac{x}{x(x+2)} = \frac{7}{24}\]which simplifies to: \[\frac{x + 2 + x}{x(x+2)} = \frac{7}{24}\] or \[\frac{2x + 2}{x(x+2)} = \frac{7}{24}\]

Step 5: Cross Multiply to Solve for x

Cross-multiply to eliminate the fractions:\[24(2x + 2) = 7x(x + 2)\]which simplifies to:\[ 48x + 48 = 7x^2 + 14x\] Rearrange to form a quadratic equation:\[ 7x^2 - 34x - 48 = 0\]

Step 6: Solve the Quadratic Equation

Utilize the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 7 \), \( b = -34 \), and \( c = -48 \):\[ x = \frac{-(-34) \pm \sqrt{(-34)^2 - 4(7)(-48)}}{2(7)}\]Calculating inside the square root:\[ x = \frac{34 \pm \sqrt{1156 + 1344}}{14}\]\[ x = \frac{34 \pm \sqrt{2500}}{14}\]\[ x = \frac{34 \pm 50}{14}\] This gives us two potential solutions: \( x = 6 \) or \( x = -1.142857 \). Since we are looking for even integers, we choose \( x = 6 \).

Step 7: Identify Both Integers

With \( x = 6 \), the two consecutive even integers are \( 6 \) and \( 6 + 2 = 8 \).

Key Concepts

Quadratic Equation

A quadratic equation is an equation that can be represented in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In our problem-solving process, we arrived at the quadratic equation \( 7x^2 - 34x - 48 = 0 \). This equation was derived from the setup where we had formed the equation to solve for consecutive even integers based on their reciprocals.

An important method to solve quadratic equations is using the quadratic formula, which is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Quadratic equations often have two solutions, corresponding to the \( + \) and \( - \) in the formula. In this exercise, after substituting values \( a = 7 \), \( b = -34 \), and \( c = -48 \), and simplifying, we found:

• \( x = \frac{34 \pm 50}{14} \)

yielding the solutions \( x = 6 \) or \( x = -1.14 \). However, since we need even integers, only \( x = 6 \) is appropriate.

Understanding quadratic equations is vital as they appear frequently in real-world applications and various mathematical contexts.

Reciprocals

The reciprocal of a number is simply one divided by that number. Take any number, say \( n \), its reciprocal is \( \frac{1}{n} \). When dealing with consecutive even integers, it's important to pay attention to their numerical order and properties.

Here, the first consecutive even integer was defined as \( x \), and thus its reciprocal is \( \frac{1}{x} \). The next consecutive even integer is \( x + 2 \), making its reciprocal \( \frac{1}{x+2} \).

Reciprocals are crucial in algebra because they help us with equations involving fractions and can also play a role in analyzing ratios and proportions. In our exercise, understanding the reciprocals helped in setting up the equation to find the sum of the two reciprocals as \( \frac{7}{24} \). This setup paved the way for forming a solvable equation for the problem.

Common Denominator

Finding a common denominator is a foundational skill when working with fractions. It allows us to add or subtract fractions more efficiently by converting them into equivalent fractions that have the same denominator.

In our exercise, we had the equation: \( \frac{1}{x} + \frac{1}{x+2} = \frac{7}{24} \). To add the fractions on the left side, identifying a common denominator is necessary. Both \( x \) and \( x+2 \) multiply to form \( x(x+2) \) which serves as a common denominator.

By rewriting the fractions with this common denominator, we obtained:

• \( \frac{x+2}{x(x+2)} + \frac{x}{x(x+2)} = \frac{7}{24} \)

Thus, combining the fractions results in \( \frac{2x + 2}{x(x+2)} \).

This compatibility allows the setup to be expressed as a single fraction, making it easier to manage and solve through cross-multiplication later. Using common denominators is a key technique in many mathematical problems involving fractions and rational expressions.