Sometimes, a square root is found in the denominator of a fraction. Typically, it is important to remove radicals (e.g., square roots) from the denominator of a fraction. When we do this, we say that we have rationalized the denominator of the fraction.

To rationalize the denominator of a fraction,...

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Sometimes, a square root is found in the denominator of a fraction. Typically, it is important to remove radicals (e.g., square roots) from the denominator of a fraction. When we do this, we say that we have rationalized the denominator of the fraction.

To rationalize the denominator of a fraction, we proceed as follows.

i) If there is only one square root in the denominator of the fraction, multiply the top and bottom of the fraction with the square root. This is similar to multiplying the original fraction with 1, only we also manage to rationalize the denominator.

Example: Rationalize the denominator of `\frac{33}{2\sqrt{3}}` .

We multiply both the numerator and the denominator by `\sqrt{3}` , that is, `\frac{33}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}` = `\frac{33\sqrt{3}}{6}` (note that `\sqrt{3} \times \sqrt{3} = 3^{\frac{1}{2}} \times 3^{\frac{1}{2}} = 3^{\frac{1}{2} + \frac{1}{2}} = 3^{1} = 3` ).

This result can be further simplified by dividing the numerator and denominator by 3 to obtain `\frac{11\sqrt{3}}{2}` .

Thus, the solution of the problem is `\frac{11\sqrt{3}}{2}` .

ii) If the denominator of the fraction is a binomial with a square root, then we proceed as follows: multiply both the numerator and the denominator of the fraction with the conjugate of the binomial. The conjugate here is formed by changing the sign between the two terms of the binomial (e.g., the conjugate of `(x+\sqrt{y})` is `(x-\sqrt{y})` ).

Example: Rationalize the denominator of the fraction `\frac{3}{(2+\sqrt{3})}` .

We proceed by multiplying both the numerator and the denominator by the conjugate of `(2+\sqrt{3})` , which is `(2-\sqrt{3})` .

Thus, we have `\frac{3}{(2+\sqrt{3})} \times \frac{(2-\sqrt{3})}{(2-\sqrt{3})}` = `\frac{3(2-\sqrt{3})}{(2^{2} -(\sqrt{3})^{2})}` (note that `(a+b)(a-b) = a^{2} - b^{2}` ).

Further simplification gives us `\frac{6-3\sqrt{3}}{4 -3}` = `6-3\sqrt{3}` .

Thus, the solution to the problem is `6-3\sqrt{3}` .