Problem 97 - Algebra for College Students

Question

Find all real or imaginary solutions to each equation. Use the method of your choice. $$\frac{1}{x}+\frac{1}{x-1}=\frac{1}{4}$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{9 + \sqrt{65}}{2}\) and \(x = \frac{9 - \sqrt{65}}{2}\).

Step 1: Combine the Fractions

Combine the fractions on the left side of the equation. Use the common denominator, which is \(x(x-1)\): \[\frac{1}{x} + \frac{1}{x-1} = \frac{(x-1) + x}{x(x-1)} = \frac{2x-1}{x(x-1)}\]

Step 2: Set Up the Equation

Now set the left side equal to the right side of the equation: \[\frac{2x-1}{x(x-1)} = \frac{1}{4}\]

Step 3: Cross Multiply

Cross multiply to eliminate the fractions: \[4(2x-1) = x(x-1)\]

Step 4: Expand and Simplify

Expand both sides of the equation: \[8x - 4 = x^2 - x\]

Step 5: Rearrange and Form a Quadratic Equation

Rearrange terms to form a standard quadratic equation: \[x^2 - x - 8x + 4 = 0\] \[x^2 - 9x + 4 = 0\]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -9\), and \(c = 4\): \[x = \frac{9 \pm \sqrt{81 - 16}}{2} = \frac{9 \pm \sqrt{65}}{2}\]

Step 7: Calculate the Solutions

Calculate the two solutions: \[x = \frac{9 + \sqrt{65}}{2}\] and \[x = \frac{9 - \sqrt{65}}{2}\]

Key Concepts

quadratic equations

A quadratic equation is a second-degree polynomial equation in a single variable. It has the form: \[ ax^2 + bx + c = 0 \] where \(a\), \(b\), and \(c\) are constants. The quadratic equation always has at least one solution in the set of complex numbers.We commonly use the quadratic formula to solve these equations: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] here, the term under the square root sign, \(b^2 - 4ac\), is known as the discriminant. It determines the nature of the solutions:

• If the discriminant is positive, there are two real and distinct solutions.
• If the discriminant is zero, there is exactly one real solution.
• If the discriminant is negative, there are two complex solutions.

Understanding quadratic equations helps in solving rational equations that transform into a quadratic, which is exactly what happens in this exercise. After combined and simplified, our initial rational equation took the form of a quadratic equation that can be solved with the quadratic formula.

cross multiplication

Cross multiplication is a method used to solve equations involving fractions. This technique involves multiplying the numerator of each fraction by the denominator of the other fraction. For example, if you have an equation of the form: \[ \frac{a}{b} = \frac{c}{d} \] cross multiply to get: \[ a \cdot d = b \cdot c \] This can eliminate the denominators, making it easier to solve.In our exercise above, after finding a common denominator and combining the fractions, we got the simplified form: \[ \frac{2x-1}{x(x-1)} = \frac{1}{4} \] By cross multiplying, we eliminate the fractions, giving us a straightforward equation: \[ 4(2x-1) = x(x-1) \] Cross multiplication is crucial for converting complex fraction equations into simpler polynomial equations.

common denominator

Finding a common denominator is essential when adding or subtracting fractions. The common denominator is a shared multiple of the denominators of the fractions involved. It allows you to combine multiple fractions into a single fraction easily.In our exercise, we started with the equation: \[ \frac{1}{x} + \frac{1}{x-1} = \frac{1}{4} \] To combine the fractions on the left side, we needed to find a common denominator, which in this case is: \[ x(x-1) \] After rewriting each fraction with this common denominator, we combine them: \[ \frac{(x-1) + x}{x(x-1)} = \frac{2x-1}{x(x-1)} \] It's important to recognize common denominators because it simplifies the process of dealing with multiple fractions, making the equation easier to solve.Understanding how to find the common denominator is key to simplifying and solving rational equations.