Problem 47 - Contemporary Precalculus

Question

Rationalize the denominator and simplify your answer. $$\frac{3}{2+\sqrt{12}}$$

Step-by-Step Solution

Answer

Answer: \(\frac{3 - \sqrt{12}}{-4}\)

Step 1: Identify the conjugate of the denominator

The conjugate of the given denominator (2 + \(\sqrt{12}\)) is (2 - \(\sqrt{12}\)).

Step 2: Multiply both numerator and denominator with the conjugate

Multiply the given expression by \(\frac{2 - \sqrt{12}}{2 - \sqrt{12}}\): $$\frac{3}{2+\sqrt{12}} \times \frac{2 - \sqrt{12}}{2 - \sqrt{12}}$$

Step 3: Simplify the numerator

To simplify the numerator, distribute the number 3 to both terms in the conjugate: $$3\times (2 - \sqrt{12}) = 6 - 3\sqrt{12}$$

Step 4: Simplify the denominator

To simplify the denominator, calculate the product of (2 + \(\sqrt{12}\)) and (2 - \(\sqrt{12}\)) using the difference of squares formula (\(a^2 - b^2 = (a+b)(a-b)\)): $$(2+\sqrt{12})(2-\sqrt{12}) = 2^2 - (\sqrt{12})^2 = 4 - 12 = -8$$

Step 5: Rewrite the expression with simplified numerator and denominator

Replace the simplified numerator and denominator to get the new expression: $$\frac{6 - 3\sqrt{12}}{-8}$$

Step 6: Simplify the expression further if possible

In this case, both the numerator and denominator have a common factor of 3, so divide both by 3: $$\frac{6 - 3\sqrt{12}}{-8} = \frac{2(3 - \sqrt{12})}{-2(4)} = \frac{3 - \sqrt{12}}{-4}$$ So, after rationalizing the denominator and simplifying, the given expression becomes: $$\frac{3 - \sqrt{12}}{-4}$$

Key Concepts

Simplify Radical Expressions

Simplifying radical expressions involves the process of making the radicals as straightforward as possible. In the context of our exercise, the term \(\sqrt{12}\) can be broken down. Since 12 is equal to 4 times 3, and 4 is a perfect square, this radical can be simplified to \(2\sqrt{3}\). Thus, you often begin by identifying perfect squares within the radical to simplify the expression further.

Step 6 of the given solution demonstrates simplification through common factors. By factoring out a 3 from the numerator and a 2 from the denominator, we arrive at a more simplified version of the expression. Always look for such factors to reduce expressions to their most elemental form.

Conjugate Pairs

Conjugate pairs in mathematics are a duo where two binomials differ only by the sign between their terms. In relation to our exercise, the conjugate of the denominator \(2 + \sqrt{12}\) is \(2 - \sqrt{12}\). When you multiply a binomial by its conjugate, the product is always a difference of squares, which significantly simplifies the expression. This technique is essential when rationalizing the denominator because it removes the radical from the denominator, ensuring the rationality of the expression.

Using conjugate pairs is a cornerstone strategy in precalculus when dealing with rational expressions that contain square roots. Conjugates also have wide applications in complex number arithmetic where they help in simplification and division.

Difference of Squares Formula

The difference of squares formula is a key algebraic identity which states that for any two terms \(a\) and \(b\), the product of their sum and difference yields a difference of their squares: \(a^2 - b^2 = (a+b)(a-b)\). This property can incredibly simplify the multiplication of binomials, particularly when involving radicals.

In our exercise, we apply this formula during the rationalization process (Step 4). Multiplying \(2 + \sqrt{12}\) by \(2 - \sqrt{12}\) gives us \(4 - 12\), eliminating the radical in the denominator, which is a desired outcome when manipulating fractions. Understanding this formula is vital for simplification tasks and is found often in the realm of precalculus problems.

Precalculus

Precalculus is a mathematical course that prepares students for the study of calculus. It encompasses a variety of topics such as functions, algebraic structures, and trigonometry, and it builds a foundation for understanding the complex concepts encountered in calculus.

The problem we're exploring with rationalizing the denominator is a classic example of a precalculus exercise. It draws upon a student's knowledge of algebraic identities, radical simplification, and conjugate pairs – all fundamental concepts that are deeply explored precalculus. Mastering these concepts in precalculus sets the stage for future success in higher-level mathematics courses.