Problem 9 - Algebra for College Students

Question

Solve each equation. $$ 1+\frac{1}{n-1}=\frac{1}{n^{2}-n} $$

Step-by-Step Solution

Answer

Solutions: \(n = \frac{1 + \sqrt{5}}{2}\) and \(n = \frac{1 - \sqrt{5}}{2}\).

Step 1: Simplify Left Side

First, simplify the left side of the equation. You have: \[ 1 + \frac{1}{n-1} = \frac{n-1}{n-1} + \frac{1}{n-1} = \frac{n}{n-1} \]This simplifies to: \[ \frac{n}{n-1} \]

Step 2: Rewrite Right Side

Now simplify the right side of the equation. You have: \[ \frac{1}{n^2-n} \]Notice that \(n^2-n\) can be factored:\[ n^2-n = n(n-1) \]Thus, the equation becomes: \[ \frac{1}{n(n-1)} \]

Step 3: Set Equal and Cross Multiply

Now your equation looks like this:\[ \frac{n}{n-1} = \frac{1}{n(n-1)} \]Cross multiply to eliminate the fractions:\[ n \cdot n(n-1) = 1 \cdot (n-1) \]

Step 4: Simplify and Solve for n

Expand the terms:\[ n^2(n-1) = n \]Distribute the \(n^2\):\[ n^3 - n^2 = n \]Rearrange the equation:\[ n^3 - n^2 - n = 0 \]Factor the equation:\[ n(n^2 - n - 1) = 0 \]This gives a trivial solution \(n = 0\) and a quadratic equation \(n^2 - n - 1 = 0\) to solve.

Step 5: Solve the Quadratic

Use the quadratic formula for \(n^2 - n - 1 = 0\):\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a=1\), \(b=-1\), \(c=-1\):\[ n = \frac{1 \pm \sqrt{1 + 4}}{2} \]\[ n = \frac{1 \pm \sqrt{5}}{2} \]This gives solutions \(n = \frac{1 + \sqrt{5}}{2}\) and \(n = \frac{1 - \sqrt{5}}{2}\).

Step 6: Check the Solutions

Evaluate whether each potential solution makes both sides of the original equation undefined, particularly note that for \(n = 0\), the equation is undefined.Thus, the valid solutions are the two quadratic solutions: \(n = \frac{1 + \sqrt{5}}{2}\) and \(n = \frac{1 - \sqrt{5}}{2}\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation with a degree of two. They take the standard form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The solutions to quadratic equations can be found by several methods, including factoring, completing the square, or using the quadratic formula. Quadratic equations often appear in algebraic problems and play a significant role in solving rational equations like the one in the exercise. The roots of a quadratic equation provide the points where a parabola represented by the equation intersects the x-axis.

Factoring Expressions

Factoring is the process of breaking down a complex expression into simpler ones that can be multiplied to obtain the original expression. This strategy is common in simplifying polynomial expressions, and it greatly helps in solving equations.
In our original exercise, we needed to factor the expression \( n^2 - n \). We did this by recognizing that it could be expressed as \( n(n - 1) \), using the common factor \( n \). Factoring is crucial because it allows for the simplification of equations so that methods like cross multiplication can be used more effectively.

Cross Multiplication

Cross multiplication is a technique used to simplify equations involving fractions. It involves multiplying the numerator of one fraction by the denominator of the other and equating the two products obtained. This method helps to eliminate the fractions, making the equation easier to solve.
In the exercise, we had the equation \( \frac{n}{n-1} = \frac{1}{n(n-1)} \). Cross-multiplying helped us transition from a rational form to a polynomial one, leading to the form \( n^2(n-1) = n \). By removing the denominators through cross multiplication, we streamline the process required to solve for \( n \).

Quadratic Formula

The quadratic formula is a tool for finding the solutions of quadratic equations \( ax^2 + bx + c = 0 \). It's expressed as \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula is essential because it provides a systematic way to find the roots of any quadratic equation, even those that cannot be easily factored.
For the exercise, once we simplified and found the quadratic \( n^2 - n - 1 = 0 \), we used the quadratic formula with \( a = 1 \), \( b = -1 \), and \( c = -1 \). Plugging these into the formula gave us the solutions \( n = \frac{1 \pm \sqrt{5}}{2} \), demonstrating the powerful utility of the quadratic formula in finding exact solutions to equations without needing other cumbersome techniques.