Problem 41 - 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. $$ (2-\sqrt{5})(2+\sqrt{5})^{2} $$

Step-by-Step Solution

Answer

The product simplifies to \(-2 - \sqrt{5}\).

Step 1: Expand the squared term

Start by expanding the expression \((2 + \sqrt{5})^{2} = (2 + \sqrt{5})(2 + \sqrt{5})\).Apply the distributive property: \[(2 + \sqrt{5})(2 + \sqrt{5}) = 2 \times 2 + 2 \times \sqrt{5} + \sqrt{5} \times 2 + \sqrt{5} \times \sqrt{5}\]Which simplifies to:\[= 4 + 2\sqrt{5} + 2\sqrt{5} + 5 = 9 + 4\sqrt{5} \]

Step 2: Express the main expression in terms of the previous result

Substitute the expanded term back into the main expression: \((2 - \sqrt{5})(9 + 4\sqrt{5})\).

Step 3: Apply the distributive property

Distribute the first term \(2 - \sqrt{5}\) across \(9 + 4\sqrt{5}\):\[(2)(9) + (2)(4\sqrt{5}) - (\sqrt{5})(9) - (\sqrt{5})(4\sqrt{5})\]This becomes:\[= 18 + 8\sqrt{5} - 9\sqrt{5} - 20\]After simplification: \[= -2 - \sqrt{5}\]

Step 4: Finalize the expression

Complete the simplification by combining like terms. The expression reduces to:\(-2 - \sqrt{5}\)

Key Concepts

Distributive Property

Understanding the Distributive Property is essential when dealing with expressions like \((2 - \sqrt{5})(2 + \sqrt{5})^2\). This property helps to spread out or distribute a multiplying number or variable across terms inside a parentheses. It's a cornerstone concept in algebra and simplifies the process of multiplying expressions.

When we apply the Distributive Property to an expression like \((a + b)(c + d)\), what we are essentially doing is multiplying each term in the first parentheses by each term in the second. This results in four products:

• \(a \times c\)
• \(a \times d\)
• \(b \times c\)
• \(b \times d\)

You then add these products together to get the expanded form of the expression.

In the original solution, the Distributive Property was first applied to expand \((2 + \sqrt{5})^2\) and then again in the next steps of calculations. This ensures that every component is accounted for, creating a clear path to simplifying complex expressions.

Simplifying Expressions

Once you've expanded an expression using the Distributive Property, the next step is simplifying, which means to make the expression as straightforward as possible. In the exercise provided, simplifying took place after the expansion of \((2 + \sqrt{5})(2 + \sqrt{5})\), where our aim was to combine like terms.

For example, when dealing with the term \(2\sqrt{5} + 2\sqrt{5}\), we add the coefficients of the \(\sqrt{5}\) terms:

• \(2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5}\)

The whole expression \(2 \times 2 + 2 \times \sqrt{5} + \sqrt{5} \times 2 + \sqrt{5} \times \sqrt{5}\) simplifies to \(9 + 4\sqrt{5}\). This involves adding integer terms together and radical terms separately.

Next, combining this with \((2 - \sqrt{5})\), further distribution is applied, followed by combining like radical and integer terms. Both processes depend heavily on organizing similar terms, leading to the expression: \(-2 - \sqrt{5}\). Ensuring that each term is correctly combined results in the simplest form of any algebraic expression.

Radical Equations

Radical equations involve terms with roots, like the square root \(\sqrt{5}\) in our expression, \((2 - \sqrt{5})(2 + \sqrt{5})^2\). Learning to work with radicals is vital because they often appear in various algebraic problems.

A radical term, such as \(\sqrt{5}\), can make an equation seem complicated, but it's more manageable once you understand that similar radical terms can be treated like regular variables.
For instance, combining \(2\sqrt{5}+4\sqrt{5}\) results in \(6\sqrt{5}\), just like combining \(2x + 4x = 6x\).

One important aspect of working with radicals, seen in our exercise, is recognizing how to simplify or eliminate them through multiplication. Multiplying a radical by itself, such as \(\sqrt{5} \times \sqrt{5}\), results in \(5\), turning a radical product into an integer which simplifies the expression significantly. By mastering these strategies, you can tackle complex radical equations with more confidence and ease.