Problem 43 - Algebra A Combined Function

Question

Rationalize each denominator. Assume that all variables represent positive real numbers. \(\frac{2 \sqrt{3}+\sqrt{6}}{4 \sqrt{3}-\sqrt{6}}\)

Step-by-Step Solution

Answer

\(\frac{11}{7}\).

Step 1: Identify the Conjugate

The expression given is \(\frac{2 \sqrt{3} + \sqrt{6}}{4 \sqrt{3} - \sqrt{6}}\). To rationalize the denominator, we will use its conjugate. The conjugate of the denominator \(4 \sqrt{3} - \sqrt{6}\) is \(4 \sqrt{3} + \sqrt{6}\).

Step 2: Multiply Numerator and Denominator by the Conjugate

Multiply both the numerator and the denominator by the conjugate of the denominator: \[\frac{(2 \sqrt{3} + \sqrt{6})(4 \sqrt{3} + \sqrt{6})}{(4 \sqrt{3} - \sqrt{6})(4 \sqrt{3} + \sqrt{6})}.\]

Step 3: Simplify the Denominator

Apply the difference of squares formula to the denominator: \((a - b)(a + b) = a^2 - b^2\). Here, \(a = 4 \sqrt{3}\) and \(b = \sqrt{6}\). This gives: \[(4 \sqrt{3})^2 - (\sqrt{6})^2 = 48 - 6 = 42.\]

Step 4: Expand and Simplify the Numerator

Expand the numerator using distribution:\[(2 \sqrt{3} + \sqrt{6})(4 \sqrt{3} + \sqrt{6}) = 2 \sqrt{3} \times 4 \sqrt{3} + 2 \sqrt{3} \times \sqrt{6} + \sqrt{6} \times 4 \sqrt{3} + \sqrt{6} \times \sqrt{6}.\]This simplifies to:\[24 + 2 \times 6 + 4 \times 6 + 6 = 24 + 12 + 24 + 6 = 66.\]

Step 5: Form the Simplified Expression

We now form the new expression by combining simplified numerator and denominator: \[\frac{66}{42} .\]

Step 6: Reduce the Expression to Simplest Form

Simplify the fraction \(\frac{66}{42}\) by dividing both the numerator and the denominator by their greatest common divisor, which is 6:\[\frac{66}{42} = \frac{11}{7}.\]

Key Concepts

Conjugate of Radicals

When working with expressions involving radicals, the term "conjugate" refers to a paired expression where the sign between two terms is flipped. For example, the conjugate of \(4 \sqrt{3} - \sqrt{6}\) is \(4 \sqrt{3} + \sqrt{6}\). The primary reason for using conjugates when rationalizing denominators is that it allows us to eliminate radicals from the denominator by multiplying the expression by a cleverly chosen form of 1.

• The conjugate of \(a + b\) is \(a - b\), and vice versa.
• Multiplying by the conjugate helps in utilizing the difference of squares formula effectively.
• This technique is crucial when dealing with irrational denominators.

In our example, to rationalize \(\frac{2 \sqrt{3} + \sqrt{6}}{4 \sqrt{3} - \sqrt{6}}\), we multiplyboth the numerator and the denominator by \(4 \sqrt{3} + \sqrt{6}\). This process results in a simplified form where the denominator no longer contains radical terms.

Simplifying Radical Expressions

Simplifying radical expressions involves combining like terms or performing arithmetic operations to express the radicals in a more manageable form. In the context of rationalizing the denominator, simplifying means applying properties of square roots and basic arithmetic to transform a complex radical expression into something simpler. In the example \((2 \sqrt{3} + \sqrt{6})(4 \sqrt{3} + \sqrt{6})\), each term is multiplied individually, often using the distributive property. Here’s how it looks:

• \(2 \sqrt{3} \times 4 \sqrt{3} = 8 \times 3 = 24\)
• \(2 \sqrt{3} \times \sqrt{6} = 2 \times \sqrt{18} = 2 \times 3 \sqrt{2} = 6 \sqrt{2}\)
• \(\sqrt{6} \times 4 \sqrt{3} = 4 \times \sqrt{18} = 12 \sqrt{2}\)
• \(\sqrt{6} \times \sqrt{6} = 6\)

By adding all these, we simplify and combine the terms to get a neat expression that no longer has unnecessary complexities. It is through this process that the original complicated expressions become easier to handle and to further simplify.

Difference of Squares Formula

The difference of squares formula is a mathematical identity that simplifies expressions involving squares. It's represented as \((a - b)(a + b) = a^2 - b^2\). This formula helps in eliminating terms when multiplying conjugates, which is pivotal when rationalizing denominators.Applying this formula to the denominator \((4 \sqrt{3} - \sqrt{6})(4 \sqrt{3} + \sqrt{6})\), we identify \(a = 4 \sqrt{3}\) and \(b = \sqrt{6}\). Here's how it simplifies:

• \((4 \sqrt{3})^2 = 48\)
• \((\sqrt{6})^2 = 6\)
• Resulting expression: \(48 - 6 = 42\)

By using this formula, the product simplifies neatly, removing the radicals. This is the power of the difference of squares: it turns a seemingly complex multiplication problem involving radicals into a straightforward subtraction. In rationalizing denominators, employing the difference of squares formula is integral for achieving expressions without radicals.