Problem 18 - Intermediate Algebra

Question

Find the following products and express answers in simplest radical form. All variables represent nonnegative real numbers. $$ 5 \sqrt{6}(2 \sqrt{5}-3 \sqrt{11}) $$

Step-by-Step Solution

Answer

The product is \(10 \sqrt{30} - 15 \sqrt{66}\).

Step 1: Apply the Distributive Property

Multiply the term outside the parenthesis, \(5\sqrt{6}\), by each term inside the parenthesis. This means we need to perform the multiplication \(5\sqrt{6} \times 2\sqrt{5}\) and \(5\sqrt{6} \times (-3\sqrt{11})\).

Step 2: Multiply Radicals and Coefficients Separately

For the first term \(5\sqrt{6} \times 2\sqrt{5}\), multiply coefficients and radicals separately: \ \(5 \times 2 = 10\) \ For the radicals, \(\sqrt{6} \times \sqrt{5} = \sqrt{30}\).\Therefore, the first term is \(10 \sqrt{30}\). For the second term \(5\sqrt{6} \times (-3\sqrt{11})\), calculate: \ \(5 \times -3 = -15\) \ For the radicals, \(\sqrt{6} \times \sqrt{11} = \sqrt{66}\).\Thus, the second term is \(-15 \sqrt{66}\).

Step 3: Combine the Products

Combine the products from Step 2 into a single expression: \[10 \sqrt{30} - 15 \sqrt{66}\].

Key Concepts

Distributive Property

The distributive property is a fundamental concept in mathematics, particularly useful when dealing with expressions involving parentheses. This property states that multiplying a single term by terms inside parentheses involves multiplying that single term by each term inside the parentheses, and then summing the results.

To clarify, in the equation given:

• The term outside the parenthesis is: \(5\sqrt{6}\)
• The terms inside the parenthesis are: \(2\sqrt{5}\) and \(-3\sqrt{11}\)

According to the distributive property, you multiply \(5\sqrt{6}\) by each of the terms inside the parentheses individually. This results in two separate products that you'll later combine. Understanding this property makes handling complex expressions simpler, as it allows you to distribute a multiplication over an addition or a subtraction operation within parentheses. This helps break down difficult expressions into manageable pieces.

Multiply Radicals

To multiply radical expressions effectively, you need to handle numbers outside the square roots (coefficients) and the numbers under the square roots (radicals) separately. This requires you to follow a simple process for each term.When we encountered the term \(5\sqrt{6} \times 2\sqrt{5}\), we first multiplied the coefficients: \(5\) and \(2\), resulting in \(10\). For the radicals, \(\sqrt{6} \times \sqrt{5}\) produces \(\sqrt{30}\). Thus, the product was \(10\sqrt{30}\).

Similarly, in \(5\sqrt{6} \times (-3\sqrt{11})\), multiplying the coefficients \(5\) and \(-3\) gives \(-15\), and for the radicals, \(\sqrt{6} \times \sqrt{11}\) results in \(\sqrt{66}\). Therefore, the product became \(-15\sqrt{66}\).

This approach benefits from being structured, helping to maintain accuracy in calculations by treating coefficients and radicals independently before combining results.

Radical Expressions

Radical expressions are expressions that contain roots, such as square roots. Simplifying such expressions to their simplest radical form is a key skill in algebra. The simplest radical form means expressing the root in its most reduced form where no perfect squares remain under the radical.In the given exercise, we dealt with radical expressions like \(\sqrt{30}\) and \(\sqrt{66}\), which were outcomes of our multiplication process. These radical expressions do not have perfect square factors, so they stay in their current form. By always trying to reduce the radicals, we aim for solutions that are as simplified as possible while still maintaining their essential properties.

Simplifying radicals is crucial because it allows us to see the fundamental components of the expression without any excess. This makes further calculations and comparisons easier and cleaner in more complex mathematical problems.