Problem 22 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation by hand. $$\frac{2 x}{x-3}+\frac{4}{x}-6=0$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Common Denominator

The given equation is \( \frac{2x}{x-3} + \frac{4}{x} - 6 = 0 \). First, identify a common denominator for the fractions. The common denominator of \( x(x-3) \) is derived from the denominator \( x-3 \) and \( x \).

Step 2: Clear the Fractions

Multiply both sides of the equation by the common denominator, \( x(x-3) \), to eliminate the fractions. This gives:\[ x(x-3) \left( \frac{2x}{x-3} + \frac{4}{x} - 6 \right) = 0 \]Simplifying each term, we get:\[ 2x^2 + 4(x-3) - 6x(x-3) = 0 \]

Step 3: Simplify the Equation

Expand and simplify the terms:- \( 2x^2 \) is already simplified.- \( 4(x-3) = 4x - 12 \)- \( 6x(x-3) = 6x^2 - 18x \)Combine these results: \[ 2x^2 + 4x - 12 - 6x^2 + 18x = 0 \]

Step 4: Combine Like Terms

Combine the like terms to simplify the equation:\[ (2x^2 - 6x^2) + (4x + 18x) - 12 = 0 \]This simplifies to:\[ -4x^2 + 22x - 12 = 0 \]

Step 5: Solve the Quadratic Equation

Divide the entire equation by -1 for simplicity:\[ 4x^2 - 22x + 12 = 0 \]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 4 \), \( b = -22 \), and \( c = 12 \). First, calculate the discriminant:\[ b^2 - 4ac = (-22)^2 - 4 \times 4 \times 12 = 484 - 192 = 292 \]Then, solve:\[ x = \frac{22 \pm \sqrt{292}}{8} \]

Step 6: Simplify the Roots

Calculate the exact solutions:First, find \( \sqrt{292} = \sqrt{4 \times 73} = 2\sqrt{73} \).So, the solutions are:\[ x = \frac{22 + 2\sqrt{73}}{8} \] and \[ x = \frac{22 - 2\sqrt{73}}{8} \].Simplify further:\[ x_1 = \frac{11 + \sqrt{73}}{4} \] and \[ x_2 = \frac{11 - \sqrt{73}}{4} \].

Key Concepts

Quadratic Equation

A quadratic equation is a mathematical statement set in the form of a polynomial where the highest exponent of the variable is two. It can be expressed as

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. The solutions to a quadratic equation can be found using various methods such as factoring, completing the square, or using the quadratic formula. Solving a quadratic equation often involves finding complex numbers, especially when working with negative discriminants or more complicated equations. In our exercise, we worked through transforming a fractional equation into a quadratic equation that is easier to handle.

Discriminant

The discriminant is a key component in figuring out the nature of the roots in a quadratic equation. It appears in the quadratic formula as the part under the square root:

• \( b^2 - 4ac \)

This part tells us whether the roots are real or complex:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there are exactly two real roots, which are the same (a double root).
• If \( b^2 - 4ac < 0 \), the roots are complex and cannot be expressed as real numbers.

In our solution, the discriminant was calculated to be 292, signifying that the quadratic equation has two distinct real solutions.

Fractions

Fractions represent numbers through division, creating an expression with a numerator and a denominator. In solving equations, especially ones involving variables, dealing with fractions can sometimes be challenging. The key technique often involves finding a common denominator to eliminate the fractional form. In our problem,

• We found a common denominator of \( x(x-3) \)

This step transformed the fractional equation into a polynomial form, which is more straightforward to solve computationally. Tackling fractions in this way helps to simplify and manage equations, paving the way for further simplification processes.

Simplification

Simplification in solving equations means reducing them into their simplest or most compact forms without changing their solutions. This makes the equation more manageable and the solution process clearer.
In the exercise:

• We began by multiplying through by the common denominator to eliminate fractions.
• Next, we expanded and combined like terms to streamline the equation further.
• We eventually simplified it into a standard quadratic form.

Simplification is a critical process in algebra as it helps to avoid complex computations and reduces the risk of errors. By continually combining like terms and factoring, you can make the solving process more efficient and pave the way for easily applying solving techniques such as using the quadratic formula.