Problem 21 - Algebra 2

Question

Multiply each pair of conjugates. $$ (2 \sqrt{6}+8)(2 \sqrt{6}-8) $$

Step-by-Step Solution

Answer

The multiplication of the pair of conjugates, \((2 \sqrt{6}+8)\) and \((2 \sqrt{6}-8)\), is \(-40\)

Step 1 - Recognize the Conjugates

The given expressions \((2 \sqrt{6}+8)\) and \((2 \sqrt{6}-8)\) are conjugates. When multiplying conjugates, the pattern of a difference of squares occurs. This means \(a^{2} - b^{2} = (a+b)(a-b)\), where a is \((2 \sqrt{6})\) and b is \(8\) in this situation.

Step 2 - Apply the Difference of Squares Formula

We substitute \(a\) and \(b\) into the formula. Thus, we should square \((2 \sqrt{6})\) and \(8\) separately and subtract the squared values. First, \(a^{2} = (2 \sqrt{6})^{2} = 24\). Next, \(b^{2} = 8^{2} = 64\). So, \(a^{2}-b^{2} = 24 - 64 = -40 \)

Step 3 - Final Answer

The result of the multiplication is thus \(-40\).

Key Concepts

Difference of Squares

The difference of squares is a powerful algebraic identity. It describes the multiplication of two conjugates. When you see an expression like \((a+b)(a-b)\), it simplifies to the form \(a^2 - b^2\).
You can visually understand it as subtracting the square of one term from another. This formula is not only quick but helps to avoid lengthy multiplications.

• Identify the numbers or expressions as \(a\) and \(b\).
• Square each term separately.
• Subtract the squares \(a^2 - b^2\).

This operation often results in a simpler expression. For instance, using \((2\sqrt{6}+8)(2\sqrt{6}-8)\) results in \(24 - 64 = -40\).
The pattern shines because it's much quicker and neat!

Multiplication of Radicals

Multiplying radicals might seem tricky, but it's quite systematic.
When you multiply expressions containing square roots, treat the radicals like regular numbers. However, remember some rules:

• Multiply the numbers under the roots directly.
• If possible, simplify the radical after multiplication.

For instance, when multiplying \(2\sqrt{6}\) by itself, you calculate \((2\sqrt{6})^2 = 4 \times 6\). This results in \(24\).
Through following these steps, simplifying radical expressions becomes approachable and clear.

Square Root Simplification

Simplifying square roots is about making them as neat as possible. When you start with a number under a square root,
the goal is to turn it into a product of one or more integers, sometimes making it easier to understand or combine with other terms.

• You look for perfect squares within the number under the root.
• Factorize the number to find squares that you can "take out" from under the radical.

For example, the square root in \(2\sqrt{6}\) consists of 6 which doesn't contain a perfect square,
thus we leave it as it is. If you had \(\sqrt{12}\), you could simplify it by recognizing that \(12 = 4 \times 3\), hence \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\).
Simplifying this way can help solve equations more efficiently.

Algebraic Expressions

Algebraic expressions can contain numbers, variables, and operators such as add, subtract, multiply, and divide.
They often stand for unknown values and require manipulation to solve.
The expression \((2\sqrt{6}+8)(2\sqrt{6}-8)\) is an example where two conjugates are multiplied. It demonstrates how algebra becomes a tool to simplify or solve problems.

• Identify the terms and operations in the expression.
• Look for known patterns like difference of squares.

Using formulas like the difference of squares allows us to approach it systematically. Such tactics simplify complex calculations. So always break down each expression to its core components, use identities and watch your algebraic skills flourish.