#### Ways to Rationalizing Binomial Denominators, explanation and examples about Rationalizing Binomial Denominators.

Rationalizing Binomial Denominators;

lf we expand $\displaystyle \left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)$ we get:

$\displaystyle \left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)=3-\sqrt{6}+\sqrt{6}-2=1$

and the expansion of $\displaystyle \left( 3\sqrt{6}-4\sqrt{3} \right)\left( 3\sqrt{6}+4\sqrt{3} \right)$ gives us:

$\displaystyle \left( 3\sqrt{6}-4\sqrt{3} \right)\left( 3\sqrt{6}+4\sqrt{3} \right)$

$\displaystyle ={{\left( 3\sqrt{6} \right)}^{2}}+\left( 3\sqrt{6} \right)\left( 4\sqrt{3} \right)-\left( 4\sqrt{3} \right)\left( 3\sqrt{6} \right)-{{\left( 4\sqrt{3} \right)}^{2}}$

$\displaystyle =54-48=6$

Notice that in each of these examples the results are rational numbers, 1 and 6. In general, if we expand a pair of binomials, or expressions with two terms, of the form $\displaystyle \left( \sqrt{a}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)$ where a and b are rational numbers we obtain :

$\displaystyle \left( \sqrt{a}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)=a-\sqrt{ab}+\sqrt{ab}-b=a-b$

which is a rational number since the two irrational parts of the expansion, $\displaystyle -\sqrt{ab}$ and $\displaystyle \sqrt{ab}$ , sum to zero.

We call a pair of binomials of the form $\displaystyle \left( \sqrt{a}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)$ conjugates.

So $\displaystyle \left( \sqrt{3}+\sqrt{2} \right)$ and $\displaystyle \left( \sqrt{3}-\sqrt{2} \right)$ are a pair of conjugates.

Also $\displaystyle \left( 3\sqrt{6}-4\sqrt{3} \right)$ and $\displaystyle \left( 3\sqrt{6}+4\sqrt{3} \right)$ are a pair of conjugates.

We say that $\displaystyle \left( \sqrt{3}+\sqrt{2} \right)$ is the conjugate of $\displaystyle \left( \sqrt{3}-\sqrt{2} \right)$ and that $\displaystyle \left( \sqrt{3}-\sqrt{2} \right)$ is the conjugate of $\displaystyle \left( \sqrt{3}+\sqrt{2} \right)$.

Also $\displaystyle \left( 3\sqrt{6}+4\sqrt{3} \right)$ and $\displaystyle \left( 3\sqrt{6}-4\sqrt{3} \right)$ are conjugates of each other.

We can use the fact that, when we expand a pair of conjugates of the form $\displaystyle \left( \sqrt{a}+\sqrt{b} \right)\left( \sqrt{a}-\sqrt{b} \right)$ we always obtain a rational number, to rationalize the denominators of fractions having irrational binomial denominators. By multiplying both the numerator and denominator of the fraction by the conjugate of the denominator we obtain a rational denominator.

Example 1 : Rationalize the denominator and simplify :

$\displaystyle \frac{\sqrt{6}+3}{\sqrt{3}+\sqrt{2}}$

Multiplying the numerator and denominator by the conjugate of the denominator:

$\displaystyle \frac{\left( \sqrt{6}+3 \right)\left( \sqrt{3}-\sqrt{2} \right)}{\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)}$

$\displaystyle \frac{3\sqrt{2}-2\sqrt{3}+3\sqrt{3}-3\sqrt{2}}{3-2}=\frac{\sqrt{3}}{1}=\sqrt{3}$

Example 2 : Rationalize the denominator and simplify :

$\displaystyle \frac{\sqrt{96}-5}{\sqrt{18}-\sqrt{12}}$

Simplifying each irrational number first:

$\displaystyle \frac{4\sqrt{6}-5}{3\sqrt{2}-2\sqrt{3}}$

Multiplying the numerator and denominator by the conjugate of the denominator:,

$\displaystyle \frac{\left( 4\sqrt{6}-5 \right)\left( 3\sqrt{2}+2\sqrt{3} \right)}{\left( 3\sqrt{2}-2\sqrt{3} \right)\left( 3\sqrt{2}+2\sqrt{3} \right)}$

$\displaystyle \frac{24\sqrt{3}+24\sqrt{2}-15\sqrt{2}-10\sqrt{3}}{18-12}=\frac{14\sqrt{3}+9\sqrt{2}}{6}$