Problem 14
Question
Use the distributive property to help simplify each of the following. \(\frac{3}{8} \sqrt{96}-\frac{2}{3} \sqrt{54}\)
Step-by-Step Solution
Verified Answer
\( -\frac{\sqrt{6}}{2} \).
1Step 1: Simplify radicals
\( \sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6} \) and \( \sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6} \).
2Step 2: Substitute and simplify
\( \frac{3}{8}(4\sqrt{6}) - \frac{2}{3}(3\sqrt{6}) = \frac{3}{2}\sqrt{6} - 2\sqrt{6} = -\frac{1}{2}\sqrt{6} \).
Key Concepts
Simplifying RadicalsPrime FactorizationPerfect Squares
Simplifying Radicals
Simplifying radicals can make complex expressions more manageable. The first step involves writing the number under the square root as a product of its factors. Think of the whole number under the root as a combination of smaller numbers. For example, to simplify \( \sqrt{96} \) and \( \sqrt{54} \), we find a pair of factors where at least one factor is a perfect square.
Once you have these factors, pull out the square root of the perfect square factor, placing it in front of the radical. For \( \sqrt{96} \), you can use \( \sqrt{16 \times 6} \) since 16 is a perfect square. Simplifying gives \( 4\sqrt{6} \). Similarly, \( \sqrt{54} \) becomes \( \sqrt{9 \times 6} \) and simplifies to \( 3\sqrt{6} \).
This technique reduces radicals to their simplest form, which makes subsequent calculations easier.
Once you have these factors, pull out the square root of the perfect square factor, placing it in front of the radical. For \( \sqrt{96} \), you can use \( \sqrt{16 \times 6} \) since 16 is a perfect square. Simplifying gives \( 4\sqrt{6} \). Similarly, \( \sqrt{54} \) becomes \( \sqrt{9 \times 6} \) and simplifies to \( 3\sqrt{6} \).
This technique reduces radicals to their simplest form, which makes subsequent calculations easier.
Prime Factorization
Understanding prime factorization is crucial in simplifying radicals. Every whole number can be expressed as a product of prime numbers. These are numbers that have exactly two distinct positive divisors: 1 and the number itself.
To start with prime factorization, break down the given number into its smallest "prime" multipliers repeatedly, until all factors are themselves prime numbers. For example, number 96 can be factorized into prime numbers like this:
The prime factorization of 96 is therefore \(2^5 \times 3^1\). Once you have the prime factors, identifying perfect squares becomes much easier.
To start with prime factorization, break down the given number into its smallest "prime" multipliers repeatedly, until all factors are themselves prime numbers. For example, number 96 can be factorized into prime numbers like this:
- 96 ÷ 2 = 48
- 48 ÷ 2 = 24
- 24 ÷ 2 = 12
- 12 ÷ 2 = 6
- 6 ÷ 2 = 3
- 3 ÷ 3 = 1
The prime factorization of 96 is therefore \(2^5 \times 3^1\). Once you have the prime factors, identifying perfect squares becomes much easier.
Perfect Squares
A perfect square is a number that can be expressed as the product of an integer with itself. Recognizing perfect squares can greatly aid in simplifying radicals.
When simplifying, look for pairs in the prime factorization. Each pair represents a perfect square. For instance, while factorizing \( 96 \), you extract \( 2^4 = 16 \), which is a perfect square.
If you have \( n \times n = n^2 \), then the square root of \( n^2 \) is \( n \). Recognizing these perfect squares allows you to simplify equations like \( \sqrt{16 \times 6} \) to \( 4\sqrt{6} \), because \( \sqrt{16} = 4 \).
Spotting perfect squares within an expression not only simplifies, but it also aids in solving complex mathematical problems with greater efficiency.
When simplifying, look for pairs in the prime factorization. Each pair represents a perfect square. For instance, while factorizing \( 96 \), you extract \( 2^4 = 16 \), which is a perfect square.
If you have \( n \times n = n^2 \), then the square root of \( n^2 \) is \( n \). Recognizing these perfect squares allows you to simplify equations like \( \sqrt{16 \times 6} \) to \( 4\sqrt{6} \), because \( \sqrt{16} = 4 \).
Spotting perfect squares within an expression not only simplifies, but it also aids in solving complex mathematical problems with greater efficiency.
Other exercises in this chapter
Problem 14
Solve each equation. Don't forget to check each of your potential solutions. \(\sqrt{4 x-1}-3=2\)
View solution Problem 14
Multiply and simplify where possible. \((9 \sqrt[3]{6})(2 \sqrt[3]{9})\)
View solution Problem 14
Evaluate each of the following. For example, \(\sqrt{25}=5\). \(\sqrt{\frac{16}{64}}\)
View solution Problem 14
Simplify each numerical expression. \(-\left(\frac{5}{6}\right)^{0}\)
View solution