Verified Answer :
- Identify the conjugate of the denominator: \(\sqrt{5} + \sqrt{3}\).
- Multiply both the numerator and denominator by the conjugate to rationalize: \(\frac{1}{\sqrt{5}-\sqrt{3}} \cdot \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\).
- Simplify the resulting denominator using the difference of squares identity: \((\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2\).
- State the final rationalized expression: \(\boxed{\frac{\sqrt{5}+\sqrt{3}}{2}}\).
Explanation
- Analyzing the Problem and Determining the Strategy
The expression we want to simplify is \(\frac{1}{\sqrt{5}-\sqrt{3}}\). Our objective is to rationalize the denominator, which means we want to rewrite the fraction so that there are no square roots in the denominator. A common and effective method to achieve this is to multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of a expression in the form \(a - b\) is \(a + b\). In our case, the denominator is \(\sqrt{5} - \sqrt{3}\), so its conjugate is \(\sqrt{5} + \sqrt{3}\).
- Applying the Conjugate Multiplication
To rationalize the denominator, we multiply the original fraction by \(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}\):
\[\frac{1}{\sqrt{5}-\sqrt{3}} \cdot \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\]
Multiplying these gives us:
\[\frac{1 \cdot (\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3}) \cdot (\sqrt{5}+\sqrt{3})}\]
This simplifies to:
\[\frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}\]
- Simplifying the Denominator and Final Result
We use the algebraic identity for the difference of squares: \((a-b)(a+b) = a^2 - b^2\). Applying this to the denominator, where \(a = \sqrt{5}\) and \(b = \sqrt{3}\):
\[(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2\]
Calculating the squares:
\[(\sqrt{5})^2 = 5 \quad \text{and} \quad (\sqrt{3})^2 = 3\]
Thus, the denominator becomes:
\[5 - 3 = 2\]
Substituting this back into our fraction, we get:
\[\frac{\sqrt{5}+\sqrt{3}}{2}\]
Examples
Rationalizing denominators is a fundamental skill in algebra and physics. For example, when calculating the magnitude of vectors, impedance in electrical circuits, or when working with geometric properties like the side lengths of triangles involving square roots, expressing the answer with a rational denominator makes it significantly easier to compare values or perform further calculations, such as summing or subtracting terms. Having a common denominator, in this case, a rational number, simplifies complex expressions.