Problem 14 - Intermediate Algebra

Question

For Problems \(1-20\), 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

Answer

The simplified expression is \(-\frac{1}{2}\sqrt{6}\).

Step 1: Simplify Square Roots

First, simplify the square root terms. The number 96 can be factored into 16 and 6, where 16 is a perfect square. Thus, \(\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16}\sqrt{6} = 4\sqrt{6}\). Similarly, 54 can be factored into 9 and 6, where 9 is a perfect square. Thus, \(\sqrt{54} = \sqrt{9 \times 6} = \sqrt{9}\sqrt{6} = 3\sqrt{6}\).

Step 2: Substitute Simplified Square Roots

Substitute the simplified square roots back into the expression:\[ \frac{3}{8} \times 4\sqrt{6} - \frac{2}{3} \times 3\sqrt{6} \].

Step 3: Apply Distributive Property

Apply the distributive property to simplify the expression by multiplying the constants:- \(\frac{3}{8} \times 4\sqrt{6} = \frac{12}{8}\sqrt{6} = \frac{3}{2}\sqrt{6}\),- \(\frac{2}{3} \times 3\sqrt{6} = 2\sqrt{6}\).

Step 4: Combine Like Terms

Combine the terms as they both include \(\sqrt{6}\):\[ \frac{3}{2}\sqrt{6} - 2\sqrt{6} \].Convert \(2\sqrt{6}\) to \(\frac{4}{2}\sqrt{6}\) to have a common denominator: \( \frac{3}{2}\sqrt{6} - \frac{4}{2}\sqrt{6} = -\frac{1}{2}\sqrt{6} \).

Key Concepts

Simplifying Square Roots

Simplifying square roots is a key concept in algebra, especially when dealing with non-perfect squares. To simplify square roots efficiently, factor the number under the square root into its prime factors and extract any perfect square factors. For instance, in our exercise, the square root term \( \sqrt{96} \) can be broken down into the factors 16 and 6, where 16 is a perfect square. This allows us to separate and simplify: \( \sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6} \). Similarly, the square root of 54 can be simplified by recognizing that 9 is a perfect square, giving us \( \sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6} \). To simplify square roots, remember these steps:

• Identify any perfect square factors of the number under the root.
• Separate these perfect squares from non-perfect factors.
• Take the square root of the perfect square factor outside the square root.

This process helps simplify expressions and is critical in solving algebraic problems. Look for patterns and practice this technique to enhance your skills in algebra.

Combining Like Terms

After simplifying square roots, the next step is often to combine like terms to simplify the expression further. Combining like terms involves adding or subtracting terms with the same variable or radical part, treating them like single algebraic entities. In the exercise, we ultimately arrive at terms like \( \frac{3}{2}\sqrt{6} \) and \( 2\sqrt{6} \). Since both terms contain \( \sqrt{6} \), we can consolidate them. First, convert \( 2\sqrt{6} \) to \( \frac{4}{2}\sqrt{6} \) to match the denominator of the first term. Now, subtract: - \( \frac{3}{2}\sqrt{6} - \frac{4}{2}\sqrt{6} = -\frac{1}{2}\sqrt{6} \). Key steps involve:

• Identifying terms with the same radical or variable.
• Aligning fractions to similar denominators if needed.
• Performing addition or subtraction as required.

By combining like terms, expressions become simpler and more manageable to solve, representing an essential skill in algebra.

Intermediate Algebra

Intermediate algebra introduces a range of concepts that bridge basic arithmetic and more complicated forms of mathematics, essential for students advancing in math. This level not only covers operations like simplifying square roots and combining like terms but also introduces the distributive property, functions, and systems of equations.In the exercise provided, you begin by leveraging the distributive property, which states that \(a(b + c) = ab + ac\). This concept simplifies complex expressions, which is evident in how we multiplied constants with square roots to apply operations separately. Here's a glimpse of how it’s used:

• Using the distributive property when simplifying expressions like \( \frac{3}{8} \times 4\sqrt{6} \).
• Applying property keeps the operations manageable and concise.

Mastering these elements in intermediate algebra builds a strong foundation for tackling advanced algebra topics and different areas of math like calculus or statistics. It encourages analytical thinking and problem-solving skills, equipping you for all mathematical challenges ahead.