Problem 19 - Algebra 2

Question

Multiply each pair of conjugates. $$ (5-\sqrt{11})(5+\sqrt{11}) $$

Step-by-Step Solution

Answer

The product of the conjugate pairs (5-\(\sqrt{11}\)) and (5+\(\sqrt{11}\)) is 14.

Step 1: Identify the Binomials

Identify the two binomials and recognize them as conjugates. The binomials given are (5-\(\sqrt{11}\)) and (5+\(\sqrt{11}\)), which are conjugates.

Step 2: Apply the Formula

Apply the formula for the product of conjugates, which states that \((a+b)(a-b) = a^2 - b^2\). Here, \(a\) is 5 and \(b\) is \(\sqrt{11}\). So, the formula gives: \((5)^2 - (\sqrt{11})^2\).

Step 3: Simplify the Result

Compute the squares and subtract. \((5)^2 = 25\) and \((\sqrt{11})^2 = 11\). So, 25 - 11 yields 14.

Key Concepts

Understanding Binomials

A binomial is a polynomial with exactly two terms. The word 'binomial' comes from "bi-" meaning two and "-nomial" meaning terms. In this context, each binomial consists of two components that are separated by a '+' or '−' sign. For example, in the expression

• \((5-\sqrt{11})\)
• \(5+\sqrt{11}\)

each part is a binomial. The structure of a binomial makes it versatile and handy in algebra, especially when dealing with operations like multiplication or factoring. Binomials often lead to recognizable patterns, allowing us to simplify expressions or solve problems more easily.

The Magic of Product of Conjugates

When you multiply a pair of conjugates, something interesting happens. Conjugates are formed by changing the sign between two terms in a binomial. For instance,

• \((a + b)\)
• \((a - b)\)

are conjugates. The product of these conjugates follows the formula:\[(a + b)(a - b) = a^2 - b^2\]. This rule is referred to as the "difference of squares." When applied, it effectively eliminates the middle term, making calculations simpler. In our specific problem, where \(a = 5\) and \(b = \sqrt{11}\), multiplying the conjugates gives: \[5^2 - (\sqrt{11})^2\].

Difference of Squares Simplified

The difference of squares is one of the key outcomes when multiplying conjugates. Essentially, it represents the difference between the squares of two numbers. The formula is straightforward:\[a^2 - b^2\] and highlights how multiplying conjugates can lead directly to a simpler expression. This principle proves highly useful in various areas of algebra, helping to simplify expressions, solve equations and factor polynomials. In the exercise, \(5^2\) and \((\sqrt{11})^2\) were calculated as 25 and 11 respectively, leading to the final outcome:

• \[25 - 11 = 14\]

.

Importance of Irrational Numbers

Irrational numbers are numbers that cannot be expressed as a simple fraction or ratio of two integers. Their decimal expansions are infinite and non-repeating. The square root of a non-perfect square is a good example of an irrational number. In our exercise, \(\sqrt{11}\) is irrational, meaning it can't be precisely calculated to a finite decimal. Yet, the multiplication of conjugates effectively simplifies the presence of an irrational number. When \((5 - \sqrt{11})(5 + \sqrt{11})\) is computed using the difference of squares, the irrational components cancel out, resulting in a simple integer, \(14\). This illustrates a valuable algebraic technique—both simplifying complex expressions and removing irrationality without ever needing exact decimal values.