What is the following quotient?

[tex]\[ \frac{1}{1+\sqrt{3}} \][/tex]

A. [tex]\(\frac{\sqrt{3}}{4}\)[/tex]

B. [tex]\(\frac{1+\sqrt{3}}{4}\)[/tex]

C. [tex]\(\frac{1-\sqrt{3}}{4}\)[/tex]

D. [tex]\(\frac{-1+\sqrt{3}}{2}\)[/tex]



Answer :

To solve the quotient [tex]\(\frac{1}{1+\sqrt{3}}\)[/tex], we need to rationalize the denominator. Rationalizing means getting rid of the square root in the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.

The original expression is:
[tex]\[ \frac{1}{1 + \sqrt{3}} \][/tex]

The conjugate of [tex]\(1 + \sqrt{3}\)[/tex] is [tex]\(1 - \sqrt{3}\)[/tex]. We multiply the numerator and the denominator by this conjugate:

[tex]\[ \frac{1 \cdot (1 - \sqrt{3})}{(1 + \sqrt{3}) \cdot (1 - \sqrt{3})} \][/tex]

Let's calculate the numerator and denominator separately.

1. Numerator:
[tex]\[ 1 \cdot (1 - \sqrt{3}) = 1 - \sqrt{3} \][/tex]

2. Denominator:
[tex]\[ (1 + \sqrt{3}) \cdot (1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2 \][/tex]

So the expression simplifies to:

[tex]\[ \frac{1 - \sqrt{3}}{-2} \][/tex]

By simplifying this fraction, we get:

[tex]\[ \frac{1 - \sqrt{3}}{-2} = \frac{-1 + \sqrt{3}}{2} \][/tex]

Therefore, the equivalent fractional form of [tex]\(\frac{1}{1 + \sqrt{3}}\)[/tex] is:

[tex]\[ \boxed{\frac{-1+\sqrt{3}}{2}} \][/tex]