Problem 17 - Intermediate Algebra

Question

Find the following products and express answers in simplest radical form. All variables represent nonnegative real numbers. $$ 3 \sqrt{5}(2 \sqrt{2}-\sqrt{7}) $$

Step-by-Step Solution

Answer

The product is \(6\sqrt{10} - 3\sqrt{35}\) in simplest radical form.

Step 1: Distribute the Expression

To find the product, first distribute the term outside the parentheses, which is \(3\sqrt{5}\), through each term inside the parentheses: \(2\sqrt{2}\) and \(-\sqrt{7}\). This gives us: \[ 3\sqrt{5} \cdot 2\sqrt{2} - 3\sqrt{5} \cdot \sqrt{7} \] We will solve each part separately.

Step 2: Multiply the Radicals in the First Product

Let's solve the first part: \(3\sqrt{5} \cdot 2\sqrt{2}\). Multiply the coefficients (outside the radicals) and the numbers inside the radicals separately. This results in:\[ (3 \times 2) \times (\sqrt{5} \times \sqrt{2}) = 6\sqrt{10} \] Simplifying the product inside the radicals gives us \(\sqrt{10}\).

Step 3: Multiply the Radicals in the Second Product

Now solve the second part: \(-3\sqrt{5} \cdot \sqrt{7}\).Multiply the coefficients and the radicals separately:\[ -3 \times (\sqrt{5} \times \sqrt{7}) = -3\sqrt{35} \] So this simplifies to \(-3\sqrt{35}\).

Step 4: Combine Both Products

Now that we've found both products, combine them to get the final expression:\[ 6\sqrt{10} - 3\sqrt{35} \]This is the expression in its simplest radical form.

Key Concepts

Distribution in Algebra

In Algebra, distribution involves spreading out one element across another. This is often done using the Distributive Property. The Distributive Property states that for any numbers a, b, and c, this holds:

• a(b + c) = ab + ac

In our specific example, we are distributing the term \(3\sqrt{5}\) across two terms inside the parentheses, \(2\sqrt{2}\) and \(-\sqrt{7}\).
By applying the property, each term in the parentheses gets multiplied by \(3\sqrt{5}\), resulting in:

• \(3\sqrt{5} \times 2\sqrt{2}\)
• and \(-3\sqrt{5} \times \sqrt{7}\)

This step is essential because it allows us to break down more complex expressions into simpler parts, making them easier to tackle.
Always remember when you see an expression in parentheses preceded by another term, it's a signal to distribute.

Multiplying Radicals

Multiplying radicals may seem tricky, but it can be simplified. When multiplying radical expressions, focus on two parts: numbers outside the radical (coefficients) and those inside.
In our exercise, when multiplying \(3\sqrt{5}\) by \(2\sqrt{2}\), we separate the coefficients and the radicals:

• Coefficients: \(3\times2 = 6\)
• Radicals: \(\sqrt{5}\times\sqrt{2} = \sqrt{10}\)

Combine the two results to get \(6\sqrt{10}\).
For the second multiplication, \(-3\sqrt{5}\times\sqrt{7}\), you do:

• Coefficients: Multiply \(-3\) by 1 (since there's no coefficient in front of \(\sqrt{7}\))
• Radicals: Multiply \(\sqrt{5}\) by \(\sqrt{7}\), resulting in \(\sqrt{35}\)

This simplifies to \(-3\sqrt{35}\).
By multiplying radicals this way, you maintain clarity and precision in your computations.

Simplifying Radicals

Simplifying radicals involves breaking down expressions into their simplest form. The goal is to keep radicals as concise as possible. In the given exercise, we need to know the prime factorization of numbers to simplify properly.
Simplifying radicals comes into play especially when dealing with multiplication or when asked to provide an answer in simplest radical form.
First, ensure that any radicals multiplied together are simplified, if possible. For instance, if you had \(\sqrt{18}\), you would factor it as \(\sqrt{9 \times 2}\), simplifying to \(3\sqrt{2}\). In our problem, however, \(\sqrt{10}\) and \(\sqrt{35}\) are already in their simplest forms, so there's no further simplification needed.
When attempting to simplify radicals:

• Factorize the number inside the radical.
• Look for perfect squares, cubes, etc., to simplify.
• Combine like terms as necessary.

By consistently applying these techniques, you'll become adept at simplifying radical expressions efficiently.