Problem 22 - A Graphical Approach to College Algebra

Question

Find all complex solutions for each equation by hand. Do not use a calculator. $$\frac{2 x}{x-3}+\frac{4}{x}-6=0$$

Step-by-Step Solution

Answer

The complex solutions are \(x = \frac{11 \pm \sqrt{73}}{4}\).

Step 1: Identify a Common Denominator

The equation given is \( \frac{2x}{x-3} + \frac{4}{x} - 6 = 0 \). The common denominator for the fractions \( \frac{2x}{x-3} \) and \( \frac{4}{x} \) is \( x(x-3) \). We will use this to combine the fractions.

Step 2: Combine Fractions

Rewrite each term with the common denominator \( x(x-3) \): \(\frac{2x \cdot x}{x(x-3)} + \frac{4 \cdot (x-3)}{x(x-3)} - \frac{6x(x-3)}{x(x-3)} = 0\). Combine the fractions: \(\frac{2x^2 + 4x - 12 - 6x^2 + 18x}{x(x-3)} = 0\). The expression simplifies to \(\frac{-4x^2 + 22x - 12}{x(x-3)} = 0\).

Step 3: Cancel the Denominator and Simplify

Since the fraction equals zero, the numerator must also be zero. Solve the quadratic equation: \(-4x^2 + 22x - 12 = 0\). Divide through by \(-2\) to simplify: \(2x^2 - 11x + 6 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -11\), \(c = 6\). Calculate the discriminant: \(b^2 - 4ac = (-11)^2 - 4 \times 2 \times 6 = 121 - 48 = 73\). The solutions are: \(x = \frac{11 \pm \sqrt{73}}{4}\).

Step 5: Verify Validity of Solutions

Ensure that neither solution makes any denominator zero in the original equation. For the expressions \(x(x-3)\), the values \(x = 0\) and \(x = 3\) are undefined. Neither \(x = \frac{11 + \sqrt{73}}{4}\) nor \(x = \frac{11 - \sqrt{73}}{4}\) is 0 or 3.

Key Concepts

Understanding the Common Denominator

When you work with fractions, finding a common denominator is crucial because it allows you to combine and compare the fractions more easily. The common denominator is essentially the lowest common multiple of the denominators in the expressions you are examining.
To find a common denominator for fractions like \( \frac{2x}{x-3} \) and \( \frac{4}{x} \), you first determine the least common multiple of the denominators \( (x-3) \) and \( x \).

• The denominator for \( \frac{2x}{x-3} \) is \( x-3 \).
• The denominator for \( \frac{4}{x} \) is \( x \).
• Thus, the common denominator is the product \( x(x-3) \), because both terms need to be present in the denominator to allow us to add the fractions together.

Using this common denominator, each fraction can be rewritten with terms that allow them to be added or subtracted. This is a fundamental step to solve equations involving fractions.

Exploring the Quadratic Equation

Quadratic equations form the backbone of many algebraic problems and are written in the general form \( ax^2 + bx + c = 0 \). These equations typically have two solutions, though the solutions can sometimes be complex numbers.
In our problem, when you formed the equation \(-4x^2 + 22x - 12 = 0\) and simplified it to \(2x^2 - 11x + 6 = 0\), you were dealing with a typical quadratic equation. Here:

• is the coefficient of \(x^2\).
• is the coefficient of \(x\).
• is the constant term.

The solutions to this equation give you the values of \(x\) that satisfy the equation. Solving quadratic equations often involves methods like factoring, completing the square, or using the quadratic formula.

Applying the Quadratic Formula

To solve a quadratic equation like \(2x^2 - 11x + 6 = 0\), you can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

• The quadratic formula is effective and can solve any quadratic equation, whether the solutions are real or complex.
• The term \(b^2 - 4ac\) is known as the discriminant.

If the discriminant is positive, you'll get two real solutions; if zero, one real solution;if negative, the solutions will be complex.
In our example, the discriminant is 73, which is positive, so the solutions are real. After plugging the values \(a = 2\), \(b = -11\), and \(c = 6\) into the formula, you get two solutions for \(x\): \(x = \frac{11 \pm \sqrt{73}}{4}\). These solutions are then checked to ensure they do not violate any conditions set by the original denominators being zero.