Problem 36 - 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. $$ (\sqrt{6}+6 c)(\sqrt{6}-6 c) $$

Step-by-Step Solution

Answer

The simplest form is \(6 - 36c^2\).

Step 1: Recognize the Structure

The given expression \((\sqrt{6}+6c)(\sqrt{6}-6c)\) resembles the formula for the difference of squares: \((a+b)(a-b) = a^2 - b^2\). Identify \(a = \sqrt{6}\) and \(b = 6c\).

Step 2: Apply the Difference of Squares Formula

Using the difference of squares, express the product:\[(\sqrt{6})^2 - (6c)^2\].

Step 3: Simplify Each Term

Calculate \((\sqrt{6})^2\) which equals 6, and \((6c)^2\) which equals \(36c^2\).

Step 4: Substitute Back into the Expression

Substitute the simplified terms back into the expression from Step 2: \[6 - 36c^2\].

Step 5: Write the Final Expression

The simplest form of the given product is \[6 - 36c^2\].

Key Concepts

Simplifying Expressions

Simplifying expressions in algebra involves reducing them to their simplest form, making calculations more manageable. This practice helps in recognizing patterns and transforming complex algebraic combinations into easier components. Let's take the expression \((\sqrt{6}+6c)(\sqrt{6}-6c)\) as an example. At first glance, it might seem complicated, but by applying the difference of squares formula, we can simplify it effectively.

• Recognize familiar patterns, like \((a+b)(a-b)\) which simplifies to \(a^2 - b^2\).
• Calculate each component; the square of a square root or a variable to untangle complexity.

By converting the initial product to \(6 - 36c^2\), we not only simplify it, but make it much easier to work with in future calculations.

Radicand

The term 'radicand' refers to the number or expression inside a radical symbol. In the equation \((\sqrt{6} + 6c)(\sqrt{6} - 6c)\), the radicand is 6, found inside the square root symbol. When simplifying an expression involving radicals, understanding the radicand is critical.

• A radicand under a square root symbol implies it is being raised to the power of 1/2.
• The act of squaring the square root effectively 'cancels out' the radical, simplifying calculations directly.

By identifying and working through the radicand, \((\sqrt{6})^2\) becomes 6, demonstrating how targeting the radicand facilitates the simplification process.

Algebraic Formulas

Algebraic formulas provide the foundation for a wide array of operations in algebra and are particularly critical for simplifying expressions. The difference of squares formula, \((a+b)(a-b) = a^2 - b^2\), is a classic example.
Using this formula in the expression \((\sqrt{6}+6c)(\sqrt{6}-6c)\) allows us to skip multiple steps of distribution straight to the simplified result, \(6 - 36c^2\).

• These formulas help to identify shortcuts in computation.
• Remember formulas for various scenarios to tackle diverse algebraic challenges effectively.

Knowing and applying such formulas streamlines complex problems into manageable solutions, saving time and reducing errors.

Algebraic Expressions

Algebraic expressions form the basis of algebraic problem solving, consisting of terms formed by variables and constants. In our example, the expression \((\sqrt{6} + 6c)(\sqrt{6} - 6c)\) is an algebraic expression that challenges the solver to simplify it using standard procedures.
Understanding the nature and properties of variables (like \(c\)) and constants (such as 6) is essential:

• Variables represent unknown values and are manipulated algebraically to reach solutions.
• Constants have fixed values and serve as reference points in computations.

By mastering algebraic expressions and their components, we gain the competence to transform complex mathematical ideas into simplified expressions like \(6 - 36c^2\), clearly expressing the product of the given original form.