Question

Rationalize the denominator of each expression. Assume that all variables are positive when they appear. $$\frac{\sqrt{3}-1}{2 \sqrt{3}+3}$$

Short answer

The rationalized form is \(3 - \frac{5\root{3}}{3}\).

Step-by-step solution

Identify the Denominator

The denominator of the given expression is \(2\root{3}+3\). To rationalize it, we need to multiply both the numerator and the denominator by a suitable conjugate.

Determine the Conjugate

The conjugate of \(2\root{3}+3\) is \(2\root{3}-3\).

Multiply Numerator and Denominator by the Conjugate

Multiply both the numerator and the denominator by \(2\root{3}-3\): \[\frac{\root{3}-1}{2\root{3}+3} \times \frac{2\root{3}-3}{2\root{3}-3}\]

Apply FOIL Method to the Numerator

Calculate: \[(\root{3}-1)(2\root{3}-3) = \root{3} \times 2\root{3} - \root{3} \times 3 - 1 \times 2\root{3} + 1 \times 3 = 6 - 3\root{3} - 2\root{3} + 3 = 9 - 5\root{3}\]

Apply the Difference of Squares to the Denominator

Calculate: \[(2\root{3}+3)(2\root{3}-3) = (2\root{3})^2 - (3)^2 = 4 \times 3 - 9 = 12 - 9 = 3\]

Simplify the Expression

Divide the numerator by the denominator: \[\frac{9-5\root{3}}{3} = \frac{9}{3} - \frac{5\root{3}}{3} = 3 - \frac{5\root{3}}{3}\]

Key concepts

simplifying radical expressions

Simplifying radical expressions involves making the expression easier to handle while keeping the value intact. Think of it as simplifying fractions but with roots (like square roots or cube roots). For example, instead of using \(\root{3}\times\root{5}\), we simplify it to \(\root{15}\).

Whenever you see a radical, check if you can simplify it. Sometimes, factoring the expression can help. This process often makes it easier to perform more operations later, like adding, subtracting, or rationalizing the denominator.

conjugates

Conjugates are pairs of expressions that you multiply together to eliminate radicals in the denominator. If your original denominator is \(a + b\root{c}\), its conjugate would be \(a - b\root{c}\).

The key reason for using conjugates is to rationalize the denominator, as radicals in the denominator aren't usually considered simplified. By multiplying by the conjugate, you create a situation where the radicals cancel out.

For example, the conjugate of \(2\root{3} + 3\) in our problem is \(2\root{3} - 3\). Multiplying both the numerator and the denominator by this conjugate helps to remove the radical from the denominator.

FOIL method

The FOIL method is a way to multiply two binomials. FOIL stands for First, Outer, Inner, Last, which represents the order in which you multiply the terms.

For example, in \( (\root{3} - 1)(2\root{3} - 3)\), you apply FOIL like this:
1. First: \( \root{3} \times 2\root{3} = 6 \)
2. Outer: \( \root{3} \times -3 = -3\root{3} \)
3. Inner: \( -1 \times 2\root{3} = -2\root{3} \)
4. Last: \( -1 \times -3 = 3 \)
Finally, combine all these results: \(6 - 3\root{3} - 2\root{3} + 3 = 9 - 5\root{3} \). The FOIL method gives a step-by-step way to tackle binomial multiplication efficiently.

difference of squares

The difference of squares is a useful algebraic identity when rationalizing denominators with conjugates. It states that \(a^2 - b^2 = (a + b)(a - b)\). This identity helps eliminate radicals in denominators.

In our problem, the expression \((2\root{3} + 3)(2\root{3} - 3)\) transforms using the difference of squares rule: \(a = 2\root{3}\) and \( b = 3\). So, \((2\root{3})^2 - 3^2 = 4 \times 3 - 9 = 12 - 9 = 3\). This results in a simple, radical-free term in the denominator and completes the rationalization.