Rationalizing the Denominator for Square Root Expressions

The problem asks us to simplify the expression \( \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \). To simplify this fraction involving square roots in the denominator, we use a technique called rationalizing the denominator. This involves multiplying both the numerator and the denominator by the conjugate of the denominator.

Understanding the Conjugate

The denominator is \( \sqrt{5}+\sqrt{3} \). Its conjugate is \( \sqrt{5}-\sqrt{3} \). Multiplying an expression by its conjugate helps eliminate the square roots from the denominator because it follows the difference of squares pattern: \( (a+b)(a-b) = a^2 - b^2 \).

Step-by-Step Simplification

We multiply the given expression by \( \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \) (which is equal to 1, so it doesn't change the value of the expression):

$$ \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} $$

Calculating the Numerator

The numerator is \( (\sqrt{5}-\sqrt{3}) \times (\sqrt{5}-\sqrt{3}) \). This is a perfect square: \( (a-b)^2 = a^2 - 2ab + b^2 \). Here, \( a = \sqrt{5} \) and \( b = \sqrt{3} \).

• \( a^2 = (\sqrt{5})^2 = 5 \)
• \( 2ab = 2(\sqrt{5})(\sqrt{3}) = 2\sqrt{5 \times 3} = 2\sqrt{15} \)
• \( b^2 = (\sqrt{3})^2 = 3 \)

So, the numerator becomes \( 5 - 2\sqrt{15} + 3 = 8 - 2\sqrt{15} \).

Calculating the Denominator

The denominator is \( (\sqrt{5}+\sqrt{3}) \times (\sqrt{5}-\sqrt{3}) \). Using the difference of squares formula \( (a+b)(a-b) = a^2 - b^2 \):

• \( a^2 = (\sqrt{5})^2 = 5 \)
• \( b^2 = (\sqrt{3})^2 = 3 \)

So, the denominator becomes \( 5 - 3 = 2 \).

Final Result

Now, we combine the simplified numerator and denominator:

$$ \frac{8 - 2\sqrt{15}}{2} $$

We can simplify this further by factoring out a 2 from the numerator:

$$ \frac{2(4 - \sqrt{15})}{2} $$

Canceling the common factor of 2, we get:

$$ 4 - \sqrt{15} $$

Matching the Option

The simplified expression is \( 4 - \sqrt{15} \), which corresponds to the second option provided.