Problem 8 - 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. $$ \frac{\sqrt{150}}{\sqrt{3}} $$

Step-by-Step Solution

Answer

The simplest form of the expression is \( 5\sqrt{2} \).

Step 1: Simplify the quotient of radicals

When you have a fraction of square roots, you can simplify this by writing it as a single square root: \( \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \). So, in this case, we do: \( \frac{\sqrt{150}}{\sqrt{3}} = \sqrt{\frac{150}{3}} \).

Step 2: Divide inside the radical

Now, divide 150 by 3 inside the radical: \( \frac{150}{3} = 50 \), which simplifies the expression to \( \sqrt{50} \).

Step 3: Factor the radicand

Next, we factor 50 into its prime factors to simplify it further: \( 50 = 2 \times 5 \times 5 \).

Step 4: Simplify using square root properties

Apply the property that allows simplification of squares inside the radical: \( \sqrt{2 \times 5 \times 5} = \sqrt{5^2 \times 2} = 5\sqrt{2} \). Now the expression is in its simplest form.

Key Concepts

Quotients of Radicals

When dealing with quotients of radicals, we want to combine or simplify them, making our expression easier to work with. A quotient of radicals looks like a fraction involving square roots or other types of roots. The rule to simplify is to combine them into a single radical by taking the square root of the entire fraction rather than each part separately.

Here's how it works:

• Take the square root of the numerator and the denominator separately, as in \( \frac{\sqrt{a}}{\sqrt{b}} \).
• Combine them under a single radical: \( \sqrt{\frac{a}{b}} \).

Why is this allowed? It's because the square root of a division can be viewed as the division of two separate square roots. By compressing these separate elements into one, you simplify your calculations. It's essential in bone up by double-checking that both square roots, numerator, and denominator, make sense of their equivalences. Reducing the elements within the radical (as we did by dividing 150 by 3 to get 50) makes this task much more manageable.

Prime Factorization

Prime factorization is a technique used to break down a number into its basic building blocks, the prime numbers. We use this method especially when simplifying square roots, because it helps identify pairs of identical factors which can be "taken out" of the square root as a whole number.

The process:

• Continue dividing the number by the smallest prime numbers (like 2, 3, 5, 7) until you can't divide anymore, except by 1.
• Write the number as a product of prime factors. For instance, the prime factors of 50 are \(2 \times 5 \times 5\).

This breakdown is helpful when simplifying square roots because any pair of identical prime numbers can be taken out of the square root as a single whole number. In our example, \(5 \times 5 = 25\), for which the square root is 5. Therefore, \(\sqrt{5^2 \times 2} = 5\sqrt{2}\). This simplification process is foundational for making sense of complex radical expressions.

Simplifying Square Roots

Simplifying square roots is about making them as simple as possible. This includes reducing what's inside the square root to its lowest form or expressing it in a cleaner format.

Steps for simplification:

• Begin with prime factorization, as we discussed previously.
• Identify pairs of the same prime factor. Every pair can "exit" the square root as a whole number. This is due to the definition of a square root: the product of a number by itself equals its square root.
• The leftover primes stay inside the square root.

Returning to our example of \(\sqrt{50}\), prime factorization gave us \(\sqrt{5^2 \times 2}\). Since \(5^2\) equals 25, which is a perfect square, you "take out" a 5 from the square root, and 2 remains under it. Thus \(\sqrt{50}\) simplifies down to \(5\sqrt{2}\). Simplification helps us deal with numbers more comfortably and can be crucial for solving algebraic equations involving radicals efficiently. By mastering this, students can tackle more complex mathematical problems with greater ease.