Problem 510 - Elementary Algebra

Question

In the following exercises, simplify. \(\sqrt[5]{-486}\)

Step-by-Step Solution

Answer

-3

Step 1: Identify the type of root

Determine what the index of the radical is. In this case, the root is a fifth root, as indicated by the 5 in the radical symbol: \(\forall x \ \text{to simplify} \ \sqrt[5]{x}\text{, we need to find the number which when multiplied by itself five times yields } x\).

Step 2: Recognize the properties of odd roots

Notice that the root in question is an odd root (fifth root) and odd roots of negative numbers are possible. This means the simplified form will be a real number.

Step 3: Factor the radicand

Factor \(-486 \text{\textbf{ into its prime factors: } -1, 2, 3, \text{\textbf{ and 81}}} \ : -486 = -1 * 2 * 3^5\)

Step 4: Apply the root to each factor

Take the fifth root of each factor: our main component(which our aim simplifies) involves the avoiding – (negative sign) as \(-\) can be taken as –; and \( \sqrt[5]{-486} => \sqrt[5]{-1 * 3^5 => -\ 3} \Rightarrow -3\).

Key Concepts

fifth roots

Fifth roots are about finding a number that, when multiplied by itself five times, equals the original number under the radical. For example, if we have \(\root[5]{32}\), we are looking for a value that satisfies \(n^5 = 32\).
Think about it in simple terms:

• If \(n = 2\), then \(2^5 = 32\).

This means \(\(\root[5]{32} = 2\)\).
This process can seem challenging, but understanding this core concept is essential in solving these types of exercises.

odd roots

Odd roots, such as the fifth root, have interesting properties. Unlike even roots like the square root, odd roots can be taken from negative numbers. For example, \(\root[5]{-32}\) is possible and valid!
Why does this work? When you multiply an odd number of negative numbers together (e.g., \(\(-2) * (-2) * (-2) * (-2) * (-2) = -32\)\), the result is a negative number. This means odd roots of negative numbers will be negative, unlike even roots which don’t apply to negative numbers. Mastering this principle helps in simplifying roots more efficiently.

prime factorization

Prime factorization involves breaking a number into its prime factors. Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves (like 2, 3, 5, etc.).
Let's take \(-486\) for example:

• Break it down: \(-486 = -1 * 486\).
• Next, factorize 486 itself: \(486 = 2 * 243\).
• Then factorize further: \(243 = 3^5\).

Combining all factors: \(-486 = -1 * 2 * 3^5\).
This breakdown helps in simplifying the given radical.

negative numbers

Negative numbers are numbers less than zero, and they follow distinct rules in arithmetic operations and functions. When dealing with radicals, knowing whether the root is odd or even is crucial.
For instance, odd roots allow for negative radicands. \(\root[5]{-486}\) takes advantage of this property. By breaking \(-486\) into its factors: \(-1 * 2 * 3^5\), the key is recognizing \(\root[5]{-1 * 3^5}\)

• Using the property of odd roots and prime factors, we simplify it down to \(-3\).

This understanding ensures you correctly manage negative numbers in radical expressions.