Problem 77 - Algebra and Trigonometry

Question

\(77-82\) me Rationalize the denominator. $$ \frac{1}{2-\sqrt{3}} $$

Step-by-Step Solution

Answer

The rationalized fraction is \(2 + \sqrt{3}\).

Step 1: Identify the Conjugate

The first step in rationalizing the denominator is to identify the conjugate of the denominator. The conjugate of a binomial expression like \(2 - \sqrt{3}\) is \(2 + \sqrt{3}\). This is because multiplying by the conjugate will eliminate any irrational numbers in the denominator.

Step 2: Multiply by the Conjugate

Multiply both the numerator and the denominator of the fraction \(\frac{1}{2-\sqrt{3}}\) by the conjugate of the denominator, \(2 + \sqrt{3}\). This gives us: \[\frac{1 \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})}\]

Step 3: Simplify the Denominator

Use the difference of squares formula: \[(a-b)(a+b) = a^2 - b^2\] where \(a = 2\) and \(b = \sqrt{3}\). Thus, \[(2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1\] So the denominator becomes 1.

Step 4: Simplify the Expression

Now, simplify the expression by multiplying out the numerator:\[\frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}\]Since the denominator is 1, the fraction simplifies to just the numerator.

Key Concepts

Conjugate of a Binomial

When you're working with binomials that involve irrational numbers, like the expression \(2 - \sqrt{3}\), the conjugate comes to the rescue for rationalization purposes. The conjugate of a binomial is essentially the same terms, but with the opposite sign between them.
For example, the binomial \(2 - \sqrt{3}\) has a conjugate of \(2 + \sqrt{3}\).

• Bi: Binomial means two terms, like \(2\) and \(-\sqrt{3}\).
• Conjugate: Flip the sign in the middle, changing it to \(2 + \sqrt{3}\).

The reason for using the conjugate is that multiplying a binomial by its conjugate will eliminate any square roots or irrational numbers involved. When these two are multiplied, it leads to expressions that are easily simplified.

Difference of Squares Formula

The difference of squares formula is a powerful algebraic tool, especially when dealing with conjugates and rationalizing denominators. It's given by:

• \((a-b)(a+b) = a^2 - b^2\)

Using this formula, we can handle expressions like \((2 - \sqrt{3})(2 + \sqrt{3})\) efficiently. Here's how it works:

• Identify \(a\) and \(b\). In our case, \(a = 2\) and \(b = \sqrt{3}\).
• Compute \(a^2\) and \(b^2\): \(2^2 = 4\), \((\sqrt{3})^2 = 3\).
• Subtract to find \(a^2 - b^2 = 4 - 3 = 1\).

This results in the time-saving realization that the denominator simplifies to just 1, eliminating its complexity.

Irrational Numbers in Denominator

Having an irrational number in the denominator can complicate numerical expressions. It's common practice, especially in mathematical proofs or when simplifying equations, to 'rationalize' the denominator.
This process involves converting an expression so that the denominator is a rational number. This stems from a desire for precise numerical interpretation and easier understanding:

• A voids infinite, non-repeating decimals that irrational numbers bring.
• Cleans up expressions for clearer communication in mathematics.

Rationalizing a denominator typically uses the conjugate of the denominator binomial, as multiplying by the conjugate results in removing irrational components. This is crucial in blending clarity with complex expressions, making further calculations or interpretations much more straightforward.