Problem 91 - Algebra and Trigonometry

Question

\(89-96\) m State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$ \frac{2}{4+x}=\frac{1}{2}+\frac{2}{x} $$

Step-by-Step Solution

Answer

The equation is not true for all values of \(x\), except \(x = -4\) and \(x = 0\).

Step 1: Identify the Domain of the Equation

First, identify the values of the variable \(x\) that make the denominators zero. The given equation is \(\frac{2}{4+x} = \frac{1}{2} + \frac{2}{x}\). For the first fraction, the denominator \(4+x = 0\) gives \(x = -4\). For the second fraction, the denominator \(x = 0\). Excluding these values, the domain is all real numbers except \(x = -4\) and \(x = 0\).

Step 2: Multiply Both Sides by the Least Common Denominator

The least common denominator of \(4+x\) and \(x\) is \(x(4+x)\). Multiply every term by this common denominator to clear the fractions:\[ x(4+x) \times \frac{2}{4+x} = x(4+x) \times \frac{1}{2} + x(4+x) \times \frac{2}{x} \] which simplifies to: \[ 2x = \frac{x(4+x)}{2} + 2(4+x) \].

Step 3: Distribute and Simplify the Equation

First, simplify the expression \(\frac{x(4+x)}{2}\):\[ \frac{x(4+x)}{2} = \frac{x \cdot 4 + x^2}{2} = 2x + \frac{x^2}{2} \].Now substitute back into the equation: \[ 2x = 2x + \frac{x^2}{2} + 8 + 2x \].

Step 4: Combine Like Terms and Simplify

Combine like terms on the right side: \[ 2x = \frac{x^2}{2} + 4x + 8 \]. Since \(2x\) is present on both sides, we subtract \(2x\) from each side to simplify:\[ 0 = \frac{x^2}{2} + 2x + 8 \].

Step 5: Analyze the Resulting Equation

The equation \(0 = \frac{x^2}{2} + 2x + 8\) is a quadratic equation, which doesn't hold true for every possible value of \(x\). Instead, it may have specific solutions based on its factors or roots.

Key Concepts

Domain of a Function

In mathematics, a function's domain is the set of all possible input values that won't lead to undefined conditions, such as division by zero. Let's take a deeper look at the provided algebraic equation, \( \frac{2}{4+x} = \frac{1}{2} + \frac{2}{x} \).
Here, the denominators, \( 4+x \) and \( x \), must not be zero, as dividing by zero is undefined. To identify these values, solve:

• \( 4+x = 0 \rightarrow x = -4 \)
• \( x = 0 \)

Thus, the domain of this equation includes all real numbers except \( x = -4 \) and \( x = 0 \). This is crucial before solving the equation, as it helps determine the values for which the function is valid.
Understanding the domain ensures accuracy and avoids misleading conclusions.

Simplifying Expressions

Simplifying algebraic expressions is a vital skill when tackling equations, particularly those with multiple components like fractions. In our step-by-step solution, we started by eliminating the fractions in \( \frac{2}{4+x} = \frac{1}{2} + \frac{2}{x} \) using the least common denominator (LCD), \( x(4+x) \).
By multiplying every term by the LCD, the equation becomes easier to manage as it removes the fractions:

• \( x(4+x) \times \frac{2}{4+x} = 2x \)
• \( x(4+x) \times \frac{1}{2} = \frac{x(4+x)}{2} \)
• \( x(4+x) \times \frac{2}{x} = 2(4+x) \)

This process transforms the original equation into a polynomial equation that is simpler to manipulate.
After substitution, we combine and simplify terms, making the expression easier to analyze and solve. Simplification is essential to make complex problems more manageable and to focus on the core aspects of an equation.

Quadratic Equations

A quadratic equation is a second-degree polynomial of the form \( ax^2 + bx + c = 0 \). In our exercise, after simplifying the expression, we encountered \( 0 = \frac{x^2}{2} + 2x + 8 \). This represents a quadratic equation where:

• \( a = \frac{1}{2} \)
• \( b = 2 \)
• \( c = 8 \)

Quadratic equations may have two solutions (real or complex), one solution, or no real solutions, determined using the discriminant \( b^2 - 4ac \).
To solve this equation, you'd typically isolate \( x \) by factoring, completing the square, or using the quadratic formula.
These solutions provide the specific values for \( x \) where the equation equates to zero.
Understanding and solving quadratic equations is key in algebra, since they frequently appear in various mathematical problems and real-world phenomena.