Problem 60 - Intermediate Algebra

Question

Simplify by combining like radicals. $$ \sqrt{20}+\sqrt{125}-\sqrt{80} $$

Step-by-Step Solution

Answer

\( 3\sqrt{5} \)

Step 1: Simplify Each Radical

Start by simplifying each square root term individually. For \( \sqrt{20} \), recognize that 20 can be factored into 4 and 5, then \( \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5} \).For \( \sqrt{125} \), recognize that 125 can be factored into 25 and 5, then \( \sqrt{125} = \sqrt{25 \times 5} = \sqrt{25} \times \sqrt{5} = 5\sqrt{5} \).For \( \sqrt{80} \), recognize that 80 can be factored into 16 and 5, then \( \sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5} \).

Step 2: Combine Like Radicals

Now that each term is simplified, combine the like radicals. All terms have \( \sqrt{5} \) as the common radical.The expression is now \( 2\sqrt{5} + 5\sqrt{5} - 4\sqrt{5} \).Combine the coefficients of \( \sqrt{5} \): \( 2 + 5 - 4 = 3 \).The simplified expression becomes: \( 3\sqrt{5} \).

Key Concepts

Combining Like Terms

When simplifying mathematical expressions, one crucial step is combining like terms. This process helps in reducing the complexity of an expression. Like terms are terms that share the same variable and exponent. In the context of radical expressions, "like terms" are those that have the same radical part.
In our expression \( \sqrt{20} + \sqrt{125} - \sqrt{80} \), simplifying each term reveals that all terms share the common radical \( \sqrt{5} \).

• \( 2\sqrt{5} \) from \( \sqrt{20} \)
• \( 5\sqrt{5} \) from \( \sqrt{125} \)
• \( -4\sqrt{5} \) from \( \sqrt{80} \)

To combine them, simply add or subtract the coefficients (numbers in front) of these like terms. Here, add the coefficients: 2, 5, and -4, which results in 3. Thus, the expression simplifies to \( 3\sqrt{5} \). Breaking the problem into steps like these and ensuring all terms are simplified individually makes combining a quicker and easier task.

Factoring Under a Radical

Factoring numbers under a radical is an essential skill when simplifying radicals. It involves breaking down the number into factors, where one of the factors is a perfect square. Perfect squares are numbers like 4, 9, 16, 25, etc., whose square roots are whole numbers.
To simplify \( \sqrt{20} \), \( 20 \) can be broken into 4 and 5, because 4 is a perfect square. Therefore, \( \sqrt{20} = \sqrt{4\times 5} = \sqrt{4}\times\sqrt{5} \). This results in \( 2\sqrt{5} \), as the square root of 4 is 2.

• For \( \sqrt{125} \), factor it as \( 25 \times 5 \), because the square root of 25 is 5.
• Similarly, \( \sqrt{80} \) is factored as \( 16 \times 5 \), because \( \sqrt{16} \) is 4.

By always factoring using perfect squares, you ensure that each radical is simplified completely, which is a crucial step before combining like terms in expressions.

Simplifying Square Roots

Simplifying square roots involves expressing a square root in its simplest form. This means extracting whole numbers from radicals, which are rooted in perfect square factors.
For example, to simplify \( \sqrt{20} \), split 20 into 4 and 5 (since 4 is a perfect square). This results in \( \sqrt{4} \times \sqrt{5} \), which simplifies to \( 2\sqrt{5} \) because \( \sqrt{4} = 2 \).

• In \( \sqrt{125} \), the perfect square factor is 25. Therefore, \( \sqrt{125} = \sqrt{25\times 5} = 5\sqrt{5} \).
• For \( \sqrt{80} \), extract the \( \sqrt{16} \), giving \( 4\sqrt{5} \).

Simplifying each radical before further operations helps in maintaining clarity in calculation. It's a foundational step that paves the way for successfully combining radicals or performing additional operations.