Problem 29 - Algebra for College Students

Question

Solve each equation. $$\frac{6}{x}+\frac{40}{x+5}=7$$

Step-by-Step Solution

Answer

The equation has no real solutions because the discriminant is negative.

Step 1: Determine Common Denominator

The denominators in the equation are \( x \) and \( x+5 \). The least common denominator (LCD) for these two fractions is \( x(x+5) \).

Step 2: Clear the Fractions

Multiply every term in the equation by the LCD \( x(x+5) \) to eliminate the fractions: \[ (x(x+5))\left(\frac{6}{x}\right) + (x(x+5))\left(\frac{40}{x+5}\right) = 7(x(x+5))\]Simplifying gives:\[ 6(x+5) + 40x = 7x(x+5)\]

Step 3: Expand and Simplify Both Sides

Start by expanding the terms: \[ 6x + 30 + 40x = 7x^2 + 35x\]Now combine like terms on the left side of the equation:\[ 46x + 30 = 7x^2 + 35x\]

Step 4: Rearrange to Form a Quadratic Equation

Move all terms to one side of the equation to set it to zero:\[ 0 = 7x^2 + 35x - 46x + 30\]Simplify by combining like terms:\[ 0 = 7x^2 - 11x + 30\]

Step 5: Solve the Quadratic Equation Using the Quadratic Formula

The quadratic equation is \( 7x^2 - 11x + 30 = 0 \). This can be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( a = 7 \), \( b = -11 \), and \( c = 30 \).Calculate the discriminant:\[ b^2 - 4ac = (-11)^2 - 4 \times 7 \times 30 = 121 - 840 = -719\]Since the discriminant is negative, there are no real solutions.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It's super handy when the equation can't be easily factored. The formula helps to find the roots of a quadratic equation in the form \( ax^2 + bx + c = 0 \). The roots or solutions can be found using:

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

Here, \( a \), \( b \), and \( c \) are coefficients. This formula can give both real and complex solutions. The part under the square root sign, called the discriminant (\( b^2 - 4ac \)), tells us about the types of solutions:

• If it's positive, expect real and distinct solutions.
• If it's zero, the solutions are real and identical.
• If negative, expect complex solutions.

This formula is derived from completing the square method and always provides a solution, making it a sure bet!

Common Denominators

When solving rational equations, finding a common denominator is key. It's similar to what we do with adding fractions. We look for the lowest common denominator (LCD) to combine the fractions into a single equation. For instance, in the equation \[ \frac{6}{x} + \frac{40}{x+5} = 7 \]we have two denominators: \( x \) and \( x+5 \). The LCD here is \( x(x+5) \), combining features of both denominators.
By multiplying through by the LCD, we "clear" the fractions, simplifying the equation to something that's easier to handle and solve. This step is crucial for eliminating complex fraction calculations, leaving us with a more straightforward polynomial equation.

Quadratic Equations

A quadratic equation is one of the most common forms of polynomial equations. It takes the form \[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations graph as parabolas and are central to many algebra problems.
To solve these equations, methods like factoring, completing the square, or using the quadratic formula exist. The goal is to find the value of \( x \) where the equation equals zero (the roots/solutions). Depending on the discriminant's value (\( b^2 - 4ac \)), quadratic equations can have two real solutions, one real solution, or two complex solutions.

Real and Complex Solutions

Solutions to quadratic equations come in two types: real and complex. Real solutions exist when the discriminant \( b^2 - 4ac \) is zero or positive. They can be distinct or identical depending on whether the discriminant is greater than or equal to zero.

• If greater than zero, two distinct real solutions emerge.
• If equals zero, only one real solution exists, repeated.

Complex solutions arise when the discriminant is less than zero. In this scenario, the solutions are not real numbers, but complex numbers, involving the imaginary unit \( i \), where \( i = \sqrt{-1} \).
Quadratic equations can always be solved, real or complex, providing a complete set of solutions. This concept connects algebra and complex analysis, showing the depth and breadth of quadratic equations.