Problem 21 - Elementary Algebra

Question

$$ \text { For Problems 1-32, solve each equation. (Objective 1) } $$ $$ n+\frac{1}{n}=\frac{17}{4} $$

Step-by-Step Solution

Answer

The solutions are \( n = 4 \) and \( n = \frac{1}{4} \).

Step 1: Eliminate the Fraction

To eliminate the fraction, multiply every term by \( n \) to get rid of the denominator. Doing this, the equation becomes \( n^2 + 1 = \frac{17}{4}n \).

Step 2: Rearrange to Form a Quadratic Equation

Rearrange the equation by moving all terms to one side to form a quadratic equation: \( n^2 - \frac{17}{4}n + 1 = 0 \).

Step 3: Clear the Decimal Coefficient

Multiply every term of \( n^2 - \frac{17}{4}n + 1 = 0 \) by 4 to eliminate the fraction \( \frac{17}{4} \). This gives \( 4n^2 - 17n + 4 = 0 \).

Step 4: Solve the Quadratic Equation Using the Quadratic Formula

Use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4 \), \( b = -17 \), and \( c = 4 \) to solve the equation. Calculating inside the formula: \[ n = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \]Simplify: \[ n = \frac{17 \pm \sqrt{289 - 64}}{8} \]\[ n = \frac{17 \pm \sqrt{225}}{8} \]\[ n = \frac{17 \pm 15}{8} \].

Step 5: Find the Solutions

Now solve \( n = \frac{17 + 15}{8} \) and \( n = \frac{17 - 15}{8} \):- For \( n = \frac{17 + 15}{8} \), calculate \( n = \frac{32}{8} = 4 \).- For \( n = \frac{17 - 15}{8} \), calculate \( n = \frac{2}{8} = \frac{1}{4} \).

Key Concepts

Solving Quadratics

Solving quadratic equations is an essential skill in algebra. These equations include a variable raised to the power of two, such as \( ax^2 + bx + c = 0 \). When solving, our goal is to find the value of the variable that makes this equation true.
In general, there are several methods to solve these equations:

• Factoring: This involves expressing the quadratic as a product of two binomials. But this is only possible if the equation can be factored easily.
• Completing the Square: A method of rewriting the equation in a perfect square form to easily solve for the variable.
• Quadratic Formula: A powerful tool for finding solutions, even when the other methods won't work. This is what we'll focus on later.

Understanding when and how to apply these methods will help in finding solutions quickly and accurately. In our specific exercise, we've used the quadratic formula for a precise and clear solution.

Fractions in Equations

Dealing with fractions in equations can sometimes be tricky. A common goal is to simplify the equation by eliminating fractions, making it easier to solve. In the exercise, the initial equation given was \( n + \frac{1}{n} = \frac{17}{4} \).
To eliminate the fraction involving the variable, we multiply through by \( n \), which resulted in \( n^2 + 1 = \frac{17}{4}n \). This step effectively removes the fractions, transforming it into a manageable equation.
It’s important to handle fractions carefully to avoid errors, particularly when they involve variables. This process is useful when converting an equation to a standard quadratic form, as seen in our example.

Quadratic Formula

The quadratic formula is a reliable method for solving any quadratic equation. It derives from the standard form of quadratics, \( ax^2 + bx + c = 0 \). The formula is given as:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our problem, we identified \( a = 4 \), \( b = -17 \), and \( c = 4 \).
This formula allows us to find the solutions by plugging these coefficients in. We calculate the value under the square root, known as the discriminant \( b^2 - 4ac \). In our case, it is \( 225 \), which is a perfect square leading to real and distinct solutions.
This method works consistently, especially when the quadratic cannot be easily factored.

Mathematical Solutions

Mathematical solutions usually involve clear, defined steps to find the correct answer. For our quadratic equation \( 4n^2 - 17n + 4 = 0 \), following the quadratic formula method, we calculated:

• Determining the Discriminant: Here, the discriminant was \( 225 \), allowing us to proceed under the square root sign.
• Solving for \( n \): We found two solutions: \( n = \frac{17 + 15}{8} = 4 \) and \( n = \frac{17 - 15}{8} = \frac{1}{4} \).

Finding the solutions involves careful computation and understanding of each step. Ensure each arithmetic operation is correct, especially when working with square roots and fractions. Practicing these steps will help build familiarity and confidence in solving such equations.