Problem 10 - Algebra 2 and Trigonometry

Question

In \(3-29\) write each quotient in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero. $$ \frac{\sqrt{300}}{\sqrt{25}} $$

Step-by-Step Solution

Answer

The quotient in simplest form is \( 2 \sqrt{3} \).

Step 1: Simplify Each Radical

Start by simplifying the radicals in both the numerator and the denominator separately. The numerator is \( \sqrt{300} \), and the denominator is \( \sqrt{25} \).

Step 2: Simplify \( \sqrt{25} \)

The square root of 25 is a simple expression: \( \sqrt{25} = 5 \). This gives us a simplified denominator.

Step 3: Factor and Simplify \( \sqrt{300} \)

To simplify \( \sqrt{300} \), first factor 300 into its prime factors: 300 = 2^2 \times 3 \times 5^2. Thus, \( \sqrt{300} = \sqrt{(2^2)(3)(5^2)} = \sqrt{2^2} \times \sqrt{3} \times \sqrt{5^2} = 2 \times \sqrt{3} \times 5 = 10 \sqrt{3} \).

Step 4: Divide Simplified Radicals

Now, divide the simplified radicals: \( \frac{10 \sqrt{3}}{5} \).

Step 5: Simplify the Quotient

Divide the coefficients outside the radicals: \( \frac{10}{5} = 2 \). Therefore, the quotient becomes \( 2 \sqrt{3} \).

Key Concepts

Prime Factorization

Understanding prime factorization is the key to simplifying many mathematical expressions, especially those involving radicals. It involves breaking down a number into its simplest building blocks—prime numbers. Prime numbers are numbers greater than 1 that only have two divisors, 1 and themselves. For example, consider the number 300. By repeatedly dividing by the smallest prime numbers like 2, 3, and 5, we find that:

• 300 is divisible by 2, giving us 150.
• 150 is divisible by 2 again, giving us 75.
• 75 is divisible by 3, providing 25.
• Finally, 25 is divisible by 5 twice.

Thus, the prime factorization of 300 is expressed as \( 2^2 \times 3 \times 5^2 \). This step is crucial before moving on to simplifying radicals, as it helps identify pairs of numbers that can be extracted out of the radical.

Square Roots

The square root function is a way to find a number that, when multiplied by itself, gives the original number. It is denoted by the radical sign, \( \sqrt{} \). In practice, simplifying square roots involves expressing the number under the radical as a product of squares. When taking the square root of a product, any squares can be taken out of the radical. For instance, \( \sqrt{300} \) simplifies as follows:First, use the prime factorization: \( \sqrt{(2^2)(3)(5^2)} \). Noting the pairs, we can separate them:

• \( \sqrt{2^2} = 2 \)
• \( \sqrt{5^2} = 5 \)
• The \( \sqrt{3} \) remains as it is not a perfect square.

Therefore, \( \sqrt{300} = 2 \times 5 \times \sqrt{3} = 10 \sqrt{3} \). Understanding this concept is pivotal for manipulating and simplifying expressions involving radicals.

Fraction Simplification

Simplifying fractions is about reducing them to their simplest form, where the numerator and denominator share no common divisors other than 1. When it comes to fractions involving radicals, such as \( \frac{10 \sqrt{3}}{5} \), the process remains the same: focus on simplifying any coefficients outside the radical first.For this example, divide the coefficients:

• The numbers outside the radicals are 10 and 5.
• \( \frac{10}{5} = 2 \), leaving us with the simplified fraction.

Thus, the entire expression simplifies to \( 2 \sqrt{3} \) due to 3 having no corresponding pair to simplify further. The key is always to look for common factors between the numerator and denominator to reduce any fractions involving radicals efficiently.