1. Rationalization

L3CALC

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Rationalization is the process of making the denominator into a rational whole number. For example, when we rationalize:

$\frac{2}{\sqrt{5}}$

…we try to remove the ​ in the denominator.

To do this, we can multiply the fraction by 1. However, we don’t multiply it by $\frac{1}{1}$, we do $\frac{\sqrt{5}}{\sqrt{5}}$, since $\sqrt{5}\times \sqrt{5}=5$. Therefore:

$\frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}$

$\therefore \frac{2\sqrt{5}}{5}$

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Using Conjugates

(out of context) When we solve a quadratic equation, and the answer is irrational, it will ALWAYS be in the form:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

… even if it returns the same number, the sign will be reversed, because of the $\pm$. We call this the ‘pair’.

For instance, if $a+\sqrt{b}$ is one solution, the other solution is $a-\sqrt{b}$. We call this the conjugate — the opposite sign pairs.

Similarly, if $a-\sqrt{b}$ is a surd, then the conjugate is $a+\sqrt{b}$.

(back to context) When we rationalize, we might have to rationalize the fraction when the denominator is in the form of $a+\sqrt{b}$. We multiply the denominator by the conjugate. This will result in the form of:

$(a+\sqrt{b})(a-\sqrt{b})$

This is the ‘difference of two squares’ pattern, which means that the expanded form of the fraction above will be:

$a^2-b^2$

This removes the ​ in the denominator, rationalizing the fraction.

For example, when we rationalize:

$\frac{2-\sqrt{3}}{1-\sqrt{2}}$

We multiply by the conjugate:

$\frac{2-\sqrt{3}}{1-\sqrt{2}}\times \frac{1+\sqrt{2}}{1+\sqrt{2}}$

$\frac{(2-\sqrt{3})(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}$

$\frac{2-\sqrt{3}+2\sqrt{2}-\sqrt{6}}{-1}$

$\therefore -2+\sqrt{3}-2\sqrt{2}+6$