- Simplify the radicals: \(\sqrt{8} = 2\sqrt{2}\) and \(\sqrt{12} = 2\sqrt{3}\).
- Simplify the fraction by canceling a factor of 2: \(\frac{2 - 2\sqrt{2}}{4 + 2\sqrt{3}} = \frac{1 - \sqrt{2}}{2 + \sqrt{3}}\).
- Rationalize the denominator by multiplying by the conjugate \((2 - \sqrt{3})\).
- Final simplified result: \(\boxed{2-\sqrt{3}-2\sqrt{2}+\sqrt{6}}\).
Explanation
- Problem Analysis
The problem requires us to simplify the numerical expression \(\frac{2-\sqrt{8}}{4+\sqrt{12}}\) and identify which of the provided options it is equivalent to. Our strategy involves simplifying the square roots first, then simplifying the fraction by common factors, and finally rationalizing the denominator.
- Simplifying the Radicals and Fraction
First, simplify the square root terms:
\(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)
Substituting these into the original expression:
\(\frac{2 - \sqrt{8}}{4 + \sqrt{12}} = \frac{2 - 2\sqrt{2}}{4 + 2\sqrt{3}}\)
Now, factor out 2 from both the numerator and the denominator:
\(\frac{2(1 - \sqrt{2})}{2(2 + \sqrt{3})} = \frac{1 - \sqrt{2}}{2 + \sqrt{3}}\)
- Rationalizing the Denominator and Final Result
To rationalize the denominator, multiply both the numerator and denominator by the conjugate of the denominator, which is \(2 - \sqrt{3}\):
\[\frac{1 - \sqrt{2}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{(1 - \sqrt{2})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}\]
Calculate the denominator:
\((2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1\)
Calculate the numerator by expanding the product:
\((1 - \sqrt{2})(2 - \sqrt{3}) = 1 \times 2 - 1 \times \sqrt{3} - \sqrt{2} \times 2 + \sqrt{2} \times \sqrt{3} = 2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}\)
Since the denominator is 1, the entire expression simplifies to \(2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}\). This matches the third option given in the problem.
Examples
In engineering and physics, rationalizing denominators is a fundamental skill for simplifying complex expressions that arise when calculating impedances in electrical circuits or focal lengths in optical systems. For example, when you have a expression like \(\frac{1}{R_1 + iR_2}\) where \(R\) represents resistance, rationalizing the denominator allows you to separate the real and imaginary components, making the formula much easier to use in calculations. Understanding this process helps in simplifying calculations, ensuring precision and readability in final results.