Problem 28 - Elementary Algebra

Question

Simplify each expression by performing the indicated operation. $$ 5 \sqrt{20}+3 \sqrt{45}-3 \sqrt{40} $$

Step-by-Step Solution

Answer

Question: Simplify the expression: $5\sqrt{20}+3\sqrt{45}-3\sqrt{40}$. Answer: $7\sqrt{5}$

Step 1: Break down each term into its prime factors

To simplify each square root, find the prime factors of each number inside the square roots. $$ \sqrt{20}= \sqrt{2^2 \cdot 5} $$ $$ \sqrt{45}= \sqrt{3^2 \cdot 5} $$ $$ \sqrt{40}= \sqrt{2^3 \cdot 5} $$

Step 2: Simplify each square root term

Now that we have the prime factors, we can simplify each square root term. Remember that a square root of a number squared is just the number itself. $$ 5\sqrt{20}= 5\cdot 2\sqrt{5}= 10\sqrt{5} $$ $$ 3\sqrt{45}= 3\cdot 3\sqrt{5}= 9\sqrt{5} $$ $$ 3\sqrt{40}= 3\cdot 2^2\sqrt{5}=12\sqrt{5} $$

Step 3: Substitute the simplified terms into the expression

Now that each term is simplified, substitute these values back into the original expression: $$ 10\sqrt{5}+9\sqrt{5}-12\sqrt{5} $$

Step 4: Add or subtract the like terms

Finally, combine the like terms to simplify the expression: $$ (10\sqrt{5}+9\sqrt{5})-12\sqrt{5} = (10+9-12) \sqrt{5} = 7\sqrt{5} $$ So, the simplified expression is: $$ 7\sqrt{5} $$

Key Concepts

Prime Factorization

Prime factorization involves breaking down a number into a product of its prime numbers. These are the building blocks of the number. For example, 20 can be expressed as the product of its prime factors, which are 2 and 5. To do this, you start by dividing the number by the smallest prime number (2, 3, 5, etc.) and continue dividing the quotient by prime numbers until you reach 1.

• For 20, you divide by 2, twice, until you are left with 5. So, 20 = 2 × 2 × 5 or 2² × 5.
• For 45, divide by 3, twice to get 5, thus 45 = 3 × 3 × 5 or 3² × 5.
• For 40, divide by 2 three times to get 5, leading to 40 = 2 × 2 × 2 × 5 or 2³ × 5.

This method will help in simplifying square roots later on by identifying pairs of factors.

Square Root Simplification

When simplifying square roots, prime factorization makes it easier. The idea is to pair prime numbers since \( \sqrt{a^2} = a \). Simplification means taking these pairs out of the square root.

• For example, in \(\sqrt{20} = \sqrt{2^2 \cdot 5} \), the pair of 2s can be pulled out, leaving \(2 \cdot \sqrt{5} \). Multiply this result by any coefficient outside of the square root for the term. Thus \(5 \sqrt{20} = 10\sqrt{5} \).
• Similarly, \(\sqrt{45} = \sqrt{3^2 \cdot 5} \) simplifies to \( 3\cdot\sqrt{5} \) since the pair of 3s can be taken out. Therefore, \(3\sqrt{45} = 9\sqrt{5} \).
• And \(\sqrt{40} = \sqrt{2^2 \cdot 10} \) which simplifies to \(2\cdot \sqrt{10} \) but finding simplified factors gives \(2^3 \cdot 5 \) = \(2^2 \cdot 2 \cdot 5 \), enabling \(\sqrt{2^2} = 2 \sqrt{5} \), so \(3 \sqrt{40} = 12 \sqrt{5} \).

This helps make the entire expression easier to work with by reducing the complexity of the terms.

Like Terms

In algebra, like terms refer to terms that have identical variable parts or radicands in the context of radicals. They can be combined by performing arithmetic operations on their coefficients.

• In the expression \(10\sqrt{5}+9\sqrt{5}-12\sqrt{5} \), the radicals are like terms because they each have the same square root, \(\sqrt{5} \). This means you can combine them by adding or subtracting their coefficients.
• First, add \(10\sqrt{5} \) and \(9\sqrt{5} \) to get \(19\sqrt{5} \).


• Next, subtract \(12\sqrt{5} \) from \(19\sqrt{5} \) to end up with \(7\sqrt{5} \).

Handling like terms this way simplifies the expression significantly, making it manageable and easy to interpret.