- To simplify \(\frac{6}{\sqrt{15}}\), multiply both the numerator and the denominator by \(\sqrt{15}\).
- The expression becomes \(\frac{6\sqrt{15}}{15}\).
- Simplify the coefficient \(\frac{6}{15}\) to \(\frac{2}{5}\) by dividing both parts by their greatest common factor, 3.
- The final simplified result is \(\boxed{\frac{2\sqrt{15}}{5}}\).
Explanation
- Analysis of the Problem
The objective is to simplify the radical expression \(\frac{6}{\sqrt{15}}\) by rationalizing its denominator. Rationalizing the denominator is a process of rewriting an expression so that no square roots remain in the denominator. This makes expressions easier to work with, compare, and manipulate in further calculations. In this case, we have the square root of 15 in the denominator, which we want to eliminate.
- Rationalizing the Denominator
To eliminate the square root from the denominator, we multiply the entire expression by a fraction that is equal to 1, chosen such that the denominator becomes a rational number. Specifically, we multiply both the numerator and the denominator by \(\sqrt{15}\):
\[\frac{6}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = \frac{6 \times \sqrt{15}}{\sqrt{15} \times \sqrt{15}}\]
Multiplying \(\sqrt{15} \times \sqrt{15}\) gives us 15, which is a rational number. This successfully completes the process of rationalizing the denominator.
- Simplifying the Fraction
After multiplying, we have the expression:
\[\frac{6\sqrt{15}}{15}\]
Now, we simplify this fraction by reducing the common factors between the numerator and the denominator. Both 6 and 15 are divisible by 3. We divide both by their greatest common divisor, 3:
\[\frac{6 \div 3}{15 \div 3} \sqrt{15} = \frac{2}{5} \sqrt{15}\]
Thus, the simplified form of the expression is \(\frac{2\sqrt{15}}{5}\).
Examples
Rationalizing denominators is frequently used in engineering and physics when working with complex formulas that involve measurements or geometric proportions. For instance, if you are calculating the side length of a triangle or a specific tension in a structure that results in a value like \(\frac{6}{\sqrt{15}}\) meters, expressing it as \(\frac{2\sqrt{15}}{5}\) meters allows for a more standard way to estimate the value or combine it with other physical constants in a calculation. It streamlines the evaluation of numerical results by standardizing the appearance of such values.