Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, simplify each expression. \(9 \sqrt{3 m^{2}}+\sqrt{27}\)

Step-by-Step Solution

Answer

The expression simplifies to \(3\sqrt{3}(3m + 1)\).

Step 1: Simplify the radical inside each square root

First, let's address each square root separately. We start with the term under the first square root, \(\sqrt{3m^2}\). The term \(m^2\) inside the square root can be simplified as \(m\) because \(\sqrt{m^2} = m\). Therefore, \(\sqrt{3m^2} = m\sqrt{3}\). Next, for \(\sqrt{27}\), we know that \(27\) can be written as \(9 \times 3\). Since \(9\) is a perfect square, we get \(\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}\).

Step 2: Substitute simplified terms back into the expression

Now that we have simplified the radicals, we substitute them back into the original expression. This gives us: \(9(m\sqrt{3}) + 3\sqrt{3}\). This simplifies to \(9m\sqrt{3} + 3\sqrt{3}\).

Step 3: Factor out the common term

We see that both terms in the expression \(9m\sqrt{3} + 3\sqrt{3}\) have a common factor of \(3\sqrt{3}\). Factor \(3\sqrt{3}\) out of the terms: \(3\sqrt{3}(3m + 1)\).

Step 4: Conclusion: Simplified Expression

The expression \(9\sqrt{3m^2} + \sqrt{27}\) simplifies to \(3\sqrt{3}(3m + 1)\).

Key Concepts

Radicals

Radicals are expressions that include a root symbol, most commonly square roots. These symbols indicate that we are looking to find the original number which, when multiplied by itself a certain number of times, gives the number inside the radical. For example, \( \sqrt{9} \) represents a number which, when squared, equals 9. The answer, of course, is 3 because \( 3 \times 3 = 9 \).
Radicals often appear in algebraic expressions, including those with variables, like \( \sqrt{3m^2} \). In this, \( \sqrt{3m^2} \) simplifies by breaking down the product under the square root. \( m^2 \) is a perfect square, allowing it to be taken out of the radical as a simple \( m \). You are left with \( m\sqrt{3} \) in a simplified form.
Simplifying radicals makes it easier to handle additional algebraic operations, like addition or multiplication. Knowing how to separate numbers or expressions within a radical into their simplest form is key to mastering these tasks.

Factoring

Factoring is the process of breaking down an expression into its simplest parts, or factors, that can be multiplied together to give the original expression. It is a crucial skill in algebra, particularly when simplifying expressions or solving equations.
Consider the expression \( 9m\sqrt{3} + 3\sqrt{3} \). Both terms contained a common factor of \( 3\sqrt{3} \). To factor this expression, we extract \( 3\sqrt{3} \) from each term, simplifying the expression to \( 3\sqrt{3}(3m + 1) \).

• Look for common variables and constants between terms.
• Simplify the expression by pulling out the greatest common factor.
• Rewrite the expression to reflect this simplification.

Factoring not only makes expressions simpler to work with, but also lays the groundwork for solving more complex algebraic equations.

Perfect Squares

Perfect squares are numbers or expressions that are the square of an integer. They are important in algebra because they simplify expressions significantly when they appear under a square root.

When simplifying radicals, recognizing perfect squares is essential. For example, when we encounter \( \sqrt{27} \), it can be rewritten as \( \sqrt{9 \times 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3} \). Here, 9 is a perfect square, making calculations straightforward. Since \( \sqrt{9} = 3 \), simplifying the radical becomes a trivial matter.

• Identify perfect squares, which include numbers like 1, 4, 9, 16, and so on.
• Use perfect squares to simplify expressions containing radicals.
• This approach reduces complexity in calculations and furthers understanding of algebraic manipulation.

Understanding perfect squares helps you recognize opportunities for simplification in an expression, streamlining the process of working through algebraic problems.