Problem 28 - Intermediate Algebra

Question

Simplify each radical expression. All variables represent positive real numbers. $$ \sqrt[5]{64 x^{10} y^{5}} $$

Step-by-Step Solution

Answer

The simplified expression is \( 2x^2y \).

Step 1: Understand the Problem

We need to simplify the radical expression \( \sqrt[5]{64 x^{10} y^5} \). This means we extract or simplify any parts that can be taken out of the fifth root.

Step 2: Prime Factorization of Constants

The number 64 can be expressed as \( 2^6 \). This means \( 64 = 2^5 \times 2^1 \). We can take \( 2^5 \) out of the fifth root.

Step 3: Simplify Variables inside the Radical

For the variable \( x^{10} \), since its exponent (10) is a multiple of 5, we can take it completely out of the radical: \( (x^{10})^{1/5} = x^2 \). The variable \( y^5 \) can also be taken out completely as \( (y^5)^{1/5} = y \).

Step 4: Extract from the Radical

Using the results from Steps 2 and 3, we extract all the terms out of the fifth root: the fifth root of \( 64 = 2^5 \times 2^1 \) is \( 2 \), therefore, \( \sqrt[5]{2^5} = 2 \). Combining with variables, we get \( 2 \times x^2 \times y \).

Step 5: Simplified Expression

Combine all extracted terms: our simplified expression is \( 2x^2y \).

Key Concepts

Radicals

A radical is a mathematical symbol that represents the root of a number or an expression. It is denoted by the radical sign, √, or its cubic and higher order counterparts.
In this exercise, we are dealing with a fifth root, represented as \( \sqrt[5]{\cdot} \).

• A square root (like √) is the most common radical, used to find a number which, when multiplied by itself, yields a given number.
• Higher order roots, such as the fifth root, ask for what number raised to the fifth power equals the number inside the radical.
• For example, the fifth root \( \sqrt[5]{a} \) asks for the number that when raised to the 5th power results in \( a \).

Remember, when simplifying radicals, you're finding numbers or expressions that can be pulled out of the radical completely to simplify the expression.

Prime Factorization

Prime factorization involves expressing a number as the product of its prime numbers. This is essential for simplifying radicals because it helps identify how many times a factor appears within a root bracket. For instance, the number 64 in our exercise can be broken down into prime factors.

• Beginning with the smallest prime, which is 2, divide 64 repeatedly by 2 until you reach 1.
• \( 64 \div 2 = 32 \)
• \( 32 \div 2 = 16 \)
• \( 16 \div 2 = 8 \)
• \( 8 \div 2 = 4 \)
• \( 4 \div 2 = 2 \)
• \( 2 \div 2 = 1 \)

This gives us \( 64 = 2^6 \). Recognizing these factors helps us simplify \( \sqrt[5]{64} \) to \( 2^5 \times 2^1 \), where we can extract \( 2 \) out of the root.

Exponentiation

Exponentiation is a mathematical operation that involves raising a number, known as the base, to a power. It helps simplify expressions, including radicals, by rewriting them in terms of their powers. For variables inside radical expressions, exponentiation allows us to determine what can come out of the radical:

• In the case of \( x^{10} \), since \( 10 \div 5 = 2 \), it can be fully extracted from the fifth root as \( x^2 \).
• Similarly, for \( y^5 \), since \( 5 \div 5 = 1 \), it comes out as \( y \).

Exponentiation helps us see how many complete sets of the power fit the root, ensuring accurate simplification of the radical expression.

Variables in Algebra

Variables in algebra represent unknown or generalized numbers in an expression. These can be manipulated and simplified, especially when working with radical expressions. Understand how to handle variables such as \( x \) and \( y \) when they are embedded within radicals:

• Variables often possess exponents themselves, which tell us how many times a variable is used as a factor.
• In our inverse problem example, the variable expression \( x^{10} \) has been raised to a power, making it necessary to extract it from the radical through exponentiation.
• Positive real numbers as variables allow straightforward simplification, as we assume they are greater than zero, bypassing considerations of negative values or imaginary numbers.

By understanding variables and using exponent rules, you can simplify expressions as shown in the exercise, achieving the expression \( 2x^2y \).