Problem 208 - Elementary Algebra

Question

In the following exercises, simplify. $$ 3 \sqrt{75 y^{2}}+8 y \sqrt{48}-\sqrt{300 y^{2}} $$

Step-by-Step Solution

Answer

37 y \sqrt{3}\

Step 1 - Simplify each radical separately

Begin by simplifying each square root term separately. Start with the term \(3 \sqrt{75 y^{2}}\). Since \(75 = 25 \cdot 3\), we can rewrite the term as \(3 \sqrt{25 \cdot 3 y^{2}}\). This simplifies to \(3 \cdot 5 y \sqrt{3} = 15 y \sqrt{3}\).

Step 2 - Simplify the next term

Next, simplify \(8 y \sqrt{48}\). Since \(48 = 16 \cdot 3\), we can rewrite the term as \(8 y \sqrt{16 \cdot 3}\). This simplifies to \(8 y \cdot 4 \sqrt{3} = 32 y \sqrt{3}\).

Step 3 - Simplify the last term

Lastly, simplify \(\backslash \sqrt{300 y^{2}}\). Since \(300 = 100 \backslashcdot 3\), we can rewrite the term as \(\ \sqrt{100 \cdot 3 y^{2}}\). This simplifies to \(\backslash \cdot 10 y \sqrt{3} = 10 y \sqrt{3}\).

Step 4 - Combine like terms

Now, combine the simplified terms. We have \(15 y \sqrt{3} + 32 y \sqrt{3} - 10 y \sqrt{3}\). Combine the coefficients of like terms to get \( (15 + 32 - 10) y \sqrt{3}\). This simplifies to \(37 y \sqrt{3}\).

Key Concepts

Radical Expressions

A radical expression involves roots, such as square roots or cube roots. Working with radicals can initially seem difficult, but once you know the steps, it becomes quite simple. For instance, in the exercise, we are dealing with square roots. Radicals provide a way to express numbers that are not perfect squares, and they allow us to simplify expressions involving these roots.

Simplifying Square Roots

Simplifying square roots is a crucial skill in algebra. It involves breaking down numbers inside the square root so they can be easily managed. The goal is to express square roots in their simplest form.
In our example, we simplified each term step by step:
1. For \(3 \sqrt{75 y^{2}}\), rewriting 75 as \(25 \cdot 3\) makes it easier: \(3 \sqrt{25 \cdot 3 y^{2}} = 3 \cdot 5 y \sqrt{3} = 15 y \sqrt{3}\).
2. For \(8 y \sqrt{48}\), rewriting 48 as \(16 \cdot 3\): \(8 y \sqrt{16 \cdot 3} = 8 y \cdot 4 \sqrt{3} = 32 y \sqrt{3}\).
3. For \(\sqrt{300 y^{2}}\), rewriting 300 as \(100 \cdot 3\): \( \sqrt{100 \cdot 3 y^{2}} = 10 y \sqrt{3}\).

Combining Like Terms

Combining like terms means merging terms with the same variables and exponents. For radical expressions, this includes combining the coefficients of terms with the same radicand (number inside the square root).
In the exercise, we combined terms that all simplified to a form involving \( y \sqrt{3}\):
- \(15 y \sqrt{3}\)
- \(32 y \sqrt{3}\)
- \(10 y \sqrt{3}\)
Combining these, we add and subtract the coefficients: \( (15 + 32 - 10) y \sqrt{3} = 37 y \sqrt{3}\).

Algebraic Expressions

An algebraic expression includes numbers, variables, and arithmetic operations. In the context of this exercise, expressing terms and simplifying them are key. Understanding the structure of algebraic expressions helps in performing operations accurately.
The original expression \(3 \sqrt{75 y^{2}} + 8 y \sqrt{48} - \sqrt{300 y^{2}}\) involved radicals and variables. By carefully simplifying each radical and then combining like terms, we simplified it to a more manageable form: \(37 y \sqrt{3}\).
This shows how algebraic manipulation can transform complex expressions into simpler ones.