Problem 68 - A Graphical Approach to College Algebra

Question

Complete the following. (a) Simplify the given expression so that it does not have negative exponents. (b) Set the expression from part (a) equal to 0 and solve the resulting equation. $$\frac{x^{-2 / 3}+x^{1 / 3}}{x}$$

Step-by-Step Solution

Answer

No solution exists since the expression cannot equal zero.

Step 1: Simplify the Expression with Positive Exponents

Original expression: \( \frac{x^{-2/3} + x^{1/3}}{x} \).First, separate the terms in the numerator:\[ \frac{x^{-2/3}}{x} + \frac{x^{1/3}}{x} \]Apply the rules of exponents, which state that \( \frac{x^a}{x^b} = x^{a-b} \):- \( \frac{x^{-2/3}}{x^1} = x^{-2/3 - 1} = x^{-5/3} \)- \( \frac{x^{1/3}}{x^1} = x^{1/3 - 1} = x^{-2/3} \)So the expression becomes:\[ x^{-5/3} + x^{-2/3} \]Rewrite with positive exponents:\[ \frac{1}{x^{5/3}} + \frac{1}{x^{2/3}} \]

Step 2: Set the Expression Equal to Zero

We need to solve the equation:\[ \frac{1}{x^{5/3}} + \frac{1}{x^{2/3}} = 0 \]This is not possible since \( \frac{1}{x^{5/3}} \) and \( \frac{1}{x^{2/3}} \) are both positive for any real \( x \) except \( x = 0 \), where it is undefined. Since the sum of positive terms cannot be zero, there are no solutions to this equation.

Key Concepts

Exponents

Exponents are a fundamental concept in mathematics, particularly when dealing with algebraic expressions. An exponent indicates how many times a number, known as the base, is multiplied by itself. For example, in the expression \( x^n \), \( x \) is the base, and \( n \) is the exponent.

• If \( n \) is positive, \( x^n \) corresponds to \( x \) multiplied by itself \( n \) times.
• If \( n \) is zero, \( x^0 \) equals 1, unless \( x \) equals zero (where it is undefined).
• If \( n \) is negative, \( x^{-n} \) equals \( \frac{1}{x^n} \).

In the provided expression, we encounter both positive and negative exponents. Notice how negative exponents essentially "flip" the base into the denominator, transforming \( x^{-n} \) into a fraction. By changing these negative exponents to positive exponents, it eases the simplification process and helps maintain a standard form for further algebraic operations.

Simplification

Simplification is the process of making an expression easier to interpret and work with by removing complexities. It involves applying fundamental arithmetic operations, reducing fractions, and eliminating negative exponents.
In our exercise, the goal was to express the given rational expression with positive exponents. This involves using the property: \( \frac{x^a}{x^b} = x^{a-b} \), which helps in re-evaluating each term.First, the numerator's terms are split: \( \frac{x^{-2/3}}{x} + \frac{x^{1/3}}{x} \). By applying the rule of exponents, each term is simplified:

• \( x^{-2/3} \div x = x^{-2/3 - 1} = x^{-5/3} \)
• \( x^{1/3} \div x = x^{1/3 - 1} = x^{-2/3} \)

Next, converting these to positive exponents gives us \( \frac{1}{x^{5/3}} + \frac{1}{x^{2/3}} \). This simplification makes subsequent algebraic manipulations more straightforward and meets the requirement of the problem by eliminating negative exponents.

Algebraic Equations

An algebraic equation is a mathematical statement that expresses the equality of two expressions. Solving algebraic equations often involves finding values of variables that make the equation true.
In part (b) of the problem, we aim to set the simplified expression equal to zero and solve for \( x \). Here, the equation \( \frac{1}{x^{5/3}} + \frac{1}{x^{2/3}} = 0 \) is presented.To assess the solvability:

• Each term, \( \frac{1}{x^{5/3}} \) and \( \frac{1}{x^{2/3}} \), remains positive for any real \( x \) other than zero, where these terms become undefined.
• The equation implies adding two positive numbers and expecting the result to be zero, which is impossible.

Therefore, logically, no real values for \( x \) will satisfy this equation, indicating no solutions exist. This example reinforces the understanding that some algebraic equations cannot have solutions, especially when involving sums of positive terms equaled to zero.