Problem 50 - Intermediate Algebra

Question

Rationalize each denominator. See Example 4. $$ \frac{-3}{\sqrt{6}-2} $$

Step-by-Step Solution

Answer

The rationalized expression is \( \frac{-3\sqrt{6}}{2} - 3 \).

Step 1: Identify the Conjugate

The given expression is \( \frac{-3}{\sqrt{6} - 2} \). The denominator is \( \sqrt{6} - 2 \). To rationalize the denominator, we need to multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \( \sqrt{6} - 2 \) is \( \sqrt{6} + 2 \).

Step 2: Multiply by the Conjugate

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

Step 3: Simplify the Denominator

The denominator \((\sqrt{6} - 2)(\sqrt{6} + 2)\) is a difference of squares, which simplifies to:\[(\sqrt{6})^2 - (2)^2 = 6 - 4 = 2\]

Step 4: Simplify the Numerator

Distribute \(-3\) in the numerator:\[-3(\sqrt{6} + 2) = -3\sqrt{6} - 6\]

Step 5: Write the Simplified Expression

The expression is now:\[\frac{-3\sqrt{6} - 6}{2}\]Divide both terms in the numerator by 2:\[= \frac{-3\sqrt{6}}{2} - 3\]

Step 6: Conclusion

The rationalized expression for \( \frac{-3}{\sqrt{6}-2} \) is \( \frac{-3\sqrt{6}}{2} - 3 \).

Key Concepts

Conjugate of a Binomial

When we talk about the conjugate of a binomial, we're referring to a useful mathematical tool that helps rationalize denominators. Typically, a binomial is an expression that consists of two terms. In rationalizing the denominator, we make use of the conjugate because it allows us to remove square roots or other irrational numbers from the denominator.

• The conjugate of a binomial of the form \(a - b\) is \(a + b\).
• The conjugate of \(a + b\) is \(a - b\).

For the exercise \(\frac{-3}{\sqrt{6} - 2}\), the denominator is a binomial \(\sqrt{6} - 2\). We use its conjugate, \(\sqrt{6} + 2\), and multiply both the numerator and the denominator by this conjugate. This approach leverages the concept called the 'difference of squares,' which simplifies expressions by eliminating square roots.

Difference of Squares

The 'difference of squares' is a special algebraic formula that simplifies expressions by recognizing a pattern. This formula helps simplify expressions like \((a - b)(a + b)\), which equal \(a^2 - b^2\). It plays a big role in rationalizing denominators.
For our problem, when we multiply the denominator \(\sqrt{6} - 2\) by its conjugate \(\sqrt{6} + 2\), it creates a difference of squares:

• The expression becomes \((\sqrt{6})^2 - (2)^2\).
• This simplifies to \(6 - 4\) or just \(2\).

This simplification helps us clear radicals from the denominator, allowing us to express the fraction in a more straightforward form. By simplifying to 2, we've effectively rationalized the denominator.

Simplifying Radicals

Simplifying radicals is another essential step in rationalizing denominators. It involves expressing square roots in their simplest form, sometimes requiring the use of the conjugate or the difference of squares.
In our exercise, simplifying is more about using these previous concepts to rewrite the original expression in an easier-to-manage form:

• Multiplying the denominator by its conjugate changed the denominator to a simpler integer \(2\).
• The numerator, after simplifying \(-3(\sqrt{6} + 2)\), became \(-3\sqrt{6} - 6\).
• The entire rationalized expression is \(\frac{-3\sqrt{6} - 6}{2}\).

Further simplification involves dividing each term in the numerator by 2, resulting in the rational expression \(\frac{-3\sqrt{6}}{2} - 3\). Through these steps, the expression is in a form where radicals are removed from the denominator, enhancing clarity.