Problem 19 - Algebra 2 and Trigonometry

Question

In \(3-41\) , express each product in simplest form. Variables in the radicand with an even index are non-negative. $$ 7 \sqrt{a} \cdot 5 \sqrt{\frac{a}{9}} $$

Step-by-Step Solution

Answer

The simplest form is \(\frac{35a}{3}\).

Step 1: Express the Products as a Single Radical Expression

We start by multiplying the two radicals: \(7 \sqrt{a} \cdot 5 \sqrt{\frac{a}{9}}\). We can rewrite this as a single multiplication of the coefficients and the radicals: \( (7 \cdot 5) \cdot \sqrt{a \cdot \frac{a}{9}} \). This simplifies to \(35 \cdot \sqrt{\frac{a^2}{9}}\).

Step 2: Simplify the Expression Inside the Radical

Next, simplify the expression inside the radical: \(\sqrt{\frac{a^2}{9}}\). Since the square of a variable divided by a square of a number is the square of the quotient, this becomes \(\sqrt{\left(\frac{a}{3}\right)^2}\).

Step 3: Simplify the Square Root

For \(\sqrt{\left(\frac{a}{3}\right)^2}\), the square root and the square cancel each other. Hence, it simplifies to \(\frac{a}{3}\).

Step 4: Multiply Simplified Radical with Coefficient

Now multiply this simplified term back with the coefficient outside: \(35 \cdot \frac{a}{3}\). Multiply the numbers: \(35 \div 3\) gives you \(\frac{35}{3} \cdot a\).

Step 5: Write the Simplified Answer

The expression \(\frac{35}{3} \cdot a\) is the simplest form of the original product. Thus, the final simplified form is \(\frac{35a}{3}\).

Key Concepts

Radicals

Radicals are expressions that involve roots, most commonly the square root which is denoted by the symbol \(\sqrt{}\). In our given problem, we deal with square roots, indicated by \(\sqrt{a}\) and \(\sqrt{\frac{a}{9}}\). A radical expression might include numbers, variables, or a combination of both under the root symbol. Understanding radicals is essential because they allow us to express complex numbers and quantities in a simpler form.

• The part inside the radical symbol is called the radicand. In \(\sqrt{a}\), \(a\) is the radicand.
• It is important to note the symbol \(\sqrt{}\) usually implies a positive root.
• Even when dealing with variables, if the index of the root is even, we assume the value under the root is non-negative, as stated in the exercise.

Learning to manipulate and simplify radical expressions prepares you for more intricate algebraic operations that might not always result in an integer or a simple fraction.

Simplifying Expressions

Simplifying expressions, particularly those involving radicals, requires careful manipulation to arrive at the most reduced form. Our task was to simplify the product \(7 \sqrt{a} \cdot 5 \sqrt{\frac{a}{9}}\). To simplify, we:

• First combined the coefficients (7 and 5) and the terms under the radicals into a single radical: \(7 \cdot 5\) and \(\sqrt{a \cdot \frac{a}{9}}\).
• This was simplified by computing \((7 \cdot 5) \cdot \sqrt{\frac{a^2}{9}}\), giving the expression \(35 \cdot \sqrt{\frac{a^2}{9}}\).
• Recognize and apply the fact that \(\sqrt{\frac{a^2}{9}} = \sqrt{(\frac{a}{3})^2}\).

With this identification, the square root \(\sqrt{(\frac{a}{3})^2}\) simplifies directly to the term itself, \(\frac{a}{3}\). This reduction is due to the basic property of square roots where the square and the square root cancel each other.

Algebraic Manipulation

Algebraic manipulation involves the strategic arrangement and transformation of algebraic expressions to simplify or solve them. In simplifying radical expressions like the one in our exercise, understanding how to manipulate algebraic components becomes crucial. Let's focus on how this works:

• Combining like terms: In each step, we combined terms with identical roots to simplify our expression into a single radical form.
• Canceling terms: The expression inside the radical \(\sqrt{\frac{a^2}{9}}\) was transformed through a series of manipulations into \(\sqrt{(\frac{a}{3})^2}\), allowing us to cancel the square root with the square.
• Coefficients and multiplication: Once simplified, we multiply the remaining elements to complete the algebraic manipulation: \(35 \cdot \frac{a}{3}\) simplifies to \(\frac{35 a}{3}\).

Applying these skills efficiently helps in solving algebraic expressions quickly and accurately. Always aim to make expressions as simple as possible, as this greatly eases the process of problem-solving and understanding complex algebraic concepts.