Problem 28 - Algebra 2 and Trigonometry

Question

In \(11-38,\) evaluate each expression in the set of real numbers. $$ \sqrt[5]{-\frac{1}{32}} $$

Step-by-Step Solution

Answer

\(\sqrt[5]{-\frac{1}{32}} = -\frac{1}{2}\).

Step 1: Understand the Problem

We need to evaluate the expression \( \sqrt[5]{-\frac{1}{32}} \) within the set of real numbers. This means we are looking to find a real number that, when raised to the fifth power, results in \(-\frac{1}{32}\).

Step 2: Express as a Power

The fifth root of a number can be expressed as raising that number to the power of \( \frac{1}{5} \). So, we rewrite \( \sqrt[5]{-\frac{1}{32}} \) as \( \left( -\frac{1}{32} \right)^{\frac{1}{5}} \).

Step 3: Apply Negative Exponent Rule

We recognize that \( -\frac{1}{32} = -1 \times \frac{1}{32} \). Therefore, \( \left( -\frac{1}{32} \right)^{\frac{1}{5}} = (-1)^{\frac{1}{5}} \times \left(\frac{1}{32}\right)^{\frac{1}{5}} \).

Step 4: Evaluate \((-1)^{\frac{1}{5}}\)

The fifth root of \(-1\) is \(-1\) because \((-1) \times (-1) \times (-1) \times (-1) \times (-1) = -1\). So, \((-1)^{\frac{1}{5}} = -1\).

Step 5: Evaluate \(\left(\frac{1}{32}\right)^{\frac{1}{5}}\)

Since \(\frac{1}{32} = 2^{-5}\), it follows that \(\left(2^{-5}\right)^{\frac{1}{5}} = 2^{-1} = \frac{1}{2}\).

Step 6: Multiply the Results

Combine the results from Steps 4 and 5: \((-1) \times \frac{1}{2} = -\frac{1}{2}\).

Step 7: Conclude the Solution

Thus, \( \sqrt[5]{-\frac{1}{32}} = -\frac{1}{2} \). This is the evaluated real number of the expression.

Key Concepts

Fifth Root

In mathematics, the fifth root of a number refers to a value that, when multiplied by itself five times, becomes the original number. We denote the fifth root of a number, say \( x \), as \( \sqrt[5]{x} \). This operation can also be interpreted as raising the number to the power \( \frac{1}{5} \). For example, \( \sqrt[5]{32} \) means what number multiplied five times equals 32.

• This can be puzzling when dealing with negative numbers because traditional, even roots of negative numbers are not real. However, odd roots, like the fifth root, are applicable to negative numbers.
• For example, since \(-1\) raised to the power of five is \(-1\), the fifth root of \(-1\) is \(-1\). This characteristic is what makes odd roots special and applicable to negative values within the real numbers.

Exponentiation

Exponentiation is a mathematical operation involving two numbers, often termed the base and the exponent. In expressions like \( a^b \), \( a \) is the base, and \( b \) is the exponent, indicating the number of times the base is multiplied by itself.

• When the exponent is a fraction, like \( \frac{1}{5} \), this represents a root. Thus, \( a^{\frac{1}{5}} \) is the fifth root of \( a \).
• Exponentiation is crucial for simplifying expressions. It helps in expressing roots as power operations, and usually clarifies calculations, especially when combining rules for manipulating powers and roots.

In our example, transforming the fifth root into a power of \( \frac{1}{5} \) clarified the process, allowing further simplification by applying rules about negative and fractional exponents.

Negative Numbers

Handling negative numbers can complicate calculations, especially within roots and powers. Yet, understanding their properties simplifies processes in expressions.

• In roots: Odd roots (such as the cube root or fifth root) of negative numbers yield real numbers. That's because the odd multiplicative process inherently accounts for the negative sign.
• In our scenario: \((-1)^{\frac{1}{5}} = -1\) since multiplication by \(-1\) five times (odd times) still results in a negative result.
• Multiplier effect: Recognize \(-\frac{1}{32}\) as \(-1 \times \frac{1}{32}\) assists in isolating the negative component and dealing with the fractional part separately when evaluating expressions.

Understanding these distinctions helps in breaking down complex expressions involving roots and powers effectively.

Evaluating Expressions

Evaluating expressions involves systematically working through steps to simplify and compute values accurately. This requires understanding of operations such as roots, powers, and rules of arithmetic, especially when dealing with complex components.

• Decompose the expression into manageable components: For instance, break down \( \sqrt[5]{-\frac{1}{32}} \) into \((-1)^{\frac{1}{5}}\) and \(\left(\frac{1}{32}\right)^{\frac{1}{5}}\) before recalculating.
• Apply known results: Use special results like powers of negative numbers and simple fraction manipulations to express complex roots as simpler powers, which are easier to evaluate.
• Multiplication of results: Distinct parts of the expression, once evaluated separately, are multiplied together to reach the final value.

All these strategies, when applied, lead to a logically simplified process to evaluate complex mathematical expressions effectively and accurately.