Problem 13 - Intermediate Algebra

Question

For Problems \(1-20\), use the distributive property to help simplify each of the following. For example, $$ \begin{aligned} 3 \sqrt{8}-\sqrt{32} &=3 \sqrt{4} \sqrt{2}-\sqrt{16} \sqrt{2} \\ &=3(2) \sqrt{2}-4 \sqrt{2} \\ &=6 \sqrt{2}-4 \sqrt{2} \\ &=(6-4) \sqrt{2}=2 \sqrt{2} \end{aligned} $$ $$ \frac{3}{5} \sqrt{40}+\frac{5}{6} \sqrt{90} $$

Step-by-Step Solution

Answer

\(\frac{37}{10} \sqrt{10}\)

Step 1: Break Down the Radicals

To simplify \(\frac{3}{5} \sqrt{40}+\frac{5}{6} \sqrt{90}\), first break down the radicals into their prime factorizations.- \( \sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2 \sqrt{10} \) - \( \sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \times \sqrt{10} = 3 \sqrt{10} \)

Step 2: Apply the Distributive Property

Substitute the simplified radicals back into the expression:\[ \frac{3}{5} \cdot 2 \sqrt{10} + \frac{5}{6} \cdot 3 \sqrt{10} \]

Step 3: Simplify the Expression

Multiply the coefficients with the radicals:- \( \frac{3}{5} \cdot 2 \sqrt{10} = \frac{6}{5} \sqrt{10} \) - \( \frac{5}{6} \cdot 3 \sqrt{10} = \frac{15}{6} \sqrt{10} \) = \( \frac{5}{2} \sqrt{10} \).

Step 4: Add the Terms

Add the two terms using a common denominator. Convert \( \frac{6}{5} \) and \( \frac{5}{2} \) to a common denominator of 10.- \( \frac{6}{5} = \frac{12}{10} \) - \( \frac{5}{2} = \frac{25}{10} \) Now, add them: \( \frac{12}{10} \sqrt{10} + \frac{25}{10} \sqrt{10} = \frac{37}{10} \sqrt{10} \)

Key Concepts

Simplifying Radicals

The idea of simplifying radicals is to make expressions involving roots as simple as possible. For instance, when dealing with square roots, you can break down numbers inside a radical into their factors to simplify things. Let's take an example:

• Consider the number 40. The square root of 40 can be broken down using its factors. You identify that 40 equals 4 times 10, i.e., 40 = 4 \( \times \) 10.
• The square root of 4 is easy because it's a perfect square, resulting in 2. So, \( \sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2 \sqrt{10} \).

An important detail is understanding perfect squares like 4, 9, 16, etc., as these make the simplification process much more straightforward. Knowing these, you can efficiently breakdown radicals and simplify the problem.

Prime Factorization

Prime factorization is the process of expressing any number as a product of prime numbers. This is a valuable tool in simplifying expressions. Here's how it connects to simplifying radicals:

• For example, take the number 90. To find its prime factors, divide it by the smallest prime number, which is 2, but 90 is not divisible by 2. The next smallest is 3. Divide 90 by 3 to get 30.
• Continue with 30, which is also divisible by 3, giving you 10. Finally, 10 is broken down into 2 and 5, both prime numbers.
• Thus, 90 = 3 \( \times \) 3 \( \times \) 2 \( \times \) 5. When simplifying \( \sqrt{90} \), this becomes \( \sqrt{9 \times 10} \), where \( \sqrt{9} = 3 \).

Knowing prime factorization allows you to simplify radicals by identifying perfect squares within them, aiding in reducing complex expressions to simpler forms.

Common Denominator

A common denominator is essential when adding and subtracting fractions within an expression. Let's see why it's crucial:

• Consider fractions \( \frac{6}{5} \) and \( \frac{5}{2} \) . When adding these, their denominators must be the same to directly combine them.
• The least common denominator (LCD) here is 10. So, you convert \( \frac{6}{5} \) into \( \frac{12}{10} \), and \( \frac{5}{2} \) into \( \frac{25}{10} \).
• Now, with common denominators, you can easily add them to get \( \frac{12}{10} + \frac{25}{10} = \frac{37}{10} \).

This adjustment is necessary as it ensures that addition or subtraction operates on terms with equivalent bases, making calculations simpler and more intuitive. Understanding this process aids in correctly simplifying expressions that involve fractions.