Problem 69 - Algebra for College Students

Question

Use the distributive property to help simplify each of the following. All variables represent positive real numbers. \(5 \sqrt{27 n}-\sqrt{12 n}-6 \sqrt{3 n}\)

Step-by-Step Solution

Answer

The simplified expression is \( 7\sqrt{3n} \).

Step 1: Simplify Radicals

First, we'll simplify each of the radicals present in the expression. Start with each square root separately:- \( \sqrt{27n} = \sqrt{9 \times 3n} = \sqrt{9} \times \sqrt{3n} = 3\sqrt{3n} \).- \( \sqrt{12n} = \sqrt{4 \times 3n} = \sqrt{4} \times \sqrt{3n} = 2\sqrt{3n} \).- The radical \( 6\sqrt{3n} \) is already simplified.

Step 2: Apply Distributive Property

After simplifying the radicals, rewrite the entire expression using these simplifications:\( 5 \times 3\sqrt{3n} - 2\sqrt{3n} - 6\sqrt{3n} \).Now, distribute the constants outside the radicals:- \( 5 \times 3\sqrt{3n} = 15\sqrt{3n} \).

Step 3: Combine Like Terms

Now, combine the like terms:\( 15\sqrt{3n} - 2\sqrt{3n} - 6\sqrt{3n} \).Combine these terms since they are all multiples of \( \sqrt{3n} \):- \( (15 - 2 - 6) \sqrt{3n} = 7\sqrt{3n} \).

Key Concepts

Simplify Radicals

When dealing with expressions involving radicals, it's essential to simplify them as much as possible. Simplifying radicals means breaking down the expression within the square root into more manageable parts. This often involves finding perfect squares or other factors that make the radical simpler to handle. In the given exercise, we handled three radicals. For \( \sqrt{27n} \), notice that 27 can be expressed as the product of a perfect square: \( 27 = 9 \times 3 \).

• The next step is to separate the square root along this factorization: \( \sqrt{9 \times 3n} = \sqrt{9} \times \sqrt{3n} \).
• Since \( \sqrt{9} = 3 \), we simplify it to \( 3\sqrt{3n} \).

This simplification process makes it easier to manage various terms when solving the problem. Remember that not all numbers under the square root will have a perfect square, but attempting to find such factors can be very helpful. In this case, 12 was decomposed to 4 and 3, yielding \( \sqrt{12n} = 2\sqrt{3n} \) because \( 4 \) is also a perfect square. Always confirm if your radical can be simplified further, as it aids in combining like terms later.

Combining Like Terms

After simplifying radicals, the next crucial step is combining like terms. This concept is especially important in algebra to make expressions more manageable or see relationships more clearly. Like terms refer to terms in an expression that have the same variable components and powers. In our example expression:

• Once simplified, the expression becomes \( 15\sqrt{3n} - 2\sqrt{3n} - 6\sqrt{3n} \).
• All these terms are like terms as each term is a multiple of \( \sqrt{3n} \).
• To combine them, focus on the coefficients of these like terms.

Therefore, adding and subtracting the coefficients (15 - 2 - 6) results in \( 7\sqrt{3n} \). This illustrative operation shows how even seemingly complex expressions can collapse into simpler forms through strategic combination. It's important to consolidate these terms correctly to maintain the integrity of the expression throughout the task.

Positive Real Numbers

In this exercise, it's mentioned that all variables represent positive real numbers. Understanding positive real numbers is fundamental when working with radicals because it influences how we simplify and interpret expressions.Positive real numbers are numbers that are greater than zero and do not include imaginary or negative components. Crucially, the context of positive real numbers ensures:

• Results obtained from radicals are real and positive, maintaining consistency in solutions.
• This context eliminates possible negative results or undefined values.

When working with radicals, knowing that \( n \) and any numbers multiplied by \( n \) are positive simplifies the process. It ensures that we don't have to worry about complexities like imaginary numbers. This assumption also helps optimize simplifications, where potential complications related to negative values under a square root do not arise. This way, a clear, straightforward path to the right solution is possible, ensuring calculations remain accurate.