Problem 103 - Algebra and Trigonometry

Question

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 false for all allowed values of \(x\).

Step 1: Find a common denominator

The equation given is \( \frac{2}{4+x} = \frac{1}{2} + \frac{2}{x} \). First, identify the denominators: \(4+x\), 2, and \(x\). To combine the fractions on the right-hand side, find a common denominator. The least common denominator (LCD) is \((4+x) \cdot 2 \cdot x\).

Step 2: Express each term with the common denominator

Rewrite each term to have the common denominator \((4+x) \cdot 2 \cdot x\):\1. \( \frac{2}{4+x} = \frac{2 \cdot 2 \cdot x}{(4+x) \cdot 2 \cdot x} \).\2. \( \frac{1}{2} = \frac{(4+x) \cdot x}{(4+x) \cdot 2 \cdot x} \).\3. \( \frac{2}{x} = \frac{2 \cdot (4+x)}{(4+x) \cdot 2 \cdot x} \).

Step 3: Combine the right-hand side

Now, add the fractions on the right-hand side of the equation: \( \frac{(4+x) \cdot x + 2(4+x)}{(4+x) \cdot 2 \cdot x} \). This becomes \( \frac{4x + x^2 + 8 + 2x}{(4+x) \cdot 2 \cdot x} = \frac{x^2 + 6x + 8}{(4+x) \cdot 2 \cdot x} \).

Step 4: Compare numerators

The equation is \( \frac{4x}{(4+x) \cdot 2 \cdot x} = \frac{x^2 + 6x + 8}{(4+x) \cdot 2 \cdot x} \). For the fractions to be equal, \( 4x = x^2 + 6x + 8 \) must hold. Simplifying yields \( 0 = x^2 + 2x + 8 \). This quadratic equation has no real solutions since the discriminant, \(2^2 - 4\cdot1\cdot8 = 4 - 32 = -28\), is negative.

Step 5: Conclusion

Since the quadratic equation has no real solutions, the original equation cannot be true for any value of \(x\) (except where the original denominators are nonzero). Therefore, the equation is false for all values of \(x\) allowed in the domain.

Key Concepts

Understanding the Common Denominator

The term 'common denominator' is crucial when dealing with fractions, especially in equations that compare or combine fractions with different denominators. When we have an equation like \( \frac{2}{4+x} = \frac{1}{2} + \frac{2}{x} \), we need a common denominator to combine the fractions on the right side effectively.

To find a common denominator, identify all the denominators involved: \(4+x\), \(2\), and \(x\). The least common denominator (LCD) is found by multiplying these denominators together, which in this case is \((4+x) \cdot 2 \cdot x\).

This LCD is the smallest expression that can evenly accommodate each of the individual denominators, allowing us to rewrite each fraction so that they all share this common base. Once they are rewritten with the common denominator, you can perform operations like addition or subtraction on the numerators directly.

• Find denominators: \(4+x\), \(2\), and \(x\).
• Calculate the LCD: \((4+x) \cdot 2 \cdot x\).
• Rewrite each fraction with the LCD to simplify operations.

Introducing Quadratic Equations

Quadratic equations are a central concept in algebra, represented in the standard form as \(ax^2 + bx + c = 0\). These equations involve an unknown, usually \(x\), raised to the second power, and they can have up to two real solutions. In the context of our example from the exercise, once we equated the original fractions and simplified, we arrived at a quadratic equation:

\[x^2 + 2x + 8 = 0\]

To determine if this equation has any real solutions, you solve for \(x\). If the solutions exist, they indicate critical points where the original equation can be true. The nature of these solutions is evaluated using a determinant called the 'discriminant.'

• A quadratic equation typically has the form: \(ax^2 + bx + c = 0\).
• Solutions can tell you where or if the original equation can be true.
• Real solutions relate directly to the characteristics of the discriminant.

Understanding the Discriminant

The discriminant is a handy tool for determining the number of real solutions a quadratic equation might have. For a quadratic equation in the form \(ax^2 + bx + c = 0\), the discriminant is calculated as \( b^2 - 4ac \). This value can tell us if the equation has real solutions:

1. **Positive Discriminant**: Two distinct real solutions exist.
2. **Zero Discriminant**: The equation has exactly one real solution. It's a precise point where the function touches the x-axis, known as a 'repeated root.'
3. **Negative Discriminant**: No real solutions exist because the square root of a negative number is not a real number in standard algebraic terms.

In our example, we computed the discriminant for the quadratic \(x^2 + 2x + 8 = 0\) as:

\[(2)^2 - 4 \cdot 1 \cdot 8 = 4 - 32 = -28\]

With a negative discriminant, the equation doesn't intersect the x-axis, signifying no real solutions and thus no value of \(x\) makes the original equation valid.

• The discriminant formula: \(b^2 - 4ac\).
• Positive: Two real solutions.
• Zero: One repeated real solution.
• Negative: No real solutions.