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Section 1.4 : Rationalizing

Rationalize each of the following. Show All Solutions Hide All Solutions

  1. \(\frac{{3xy}}{{\sqrt x + \sqrt y }}\)
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    This is the typical rationalization problem that you will see in an algebra class. In these kinds of problems you want to eliminate the square roots from the denominator. To do this we will use

    \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]

    So, to rationalize the denominator (in this case, as opposed to the next problem) we will multiply the numerator and denominator by \(\sqrt x - \sqrt y \). Remember, that to rationalize we simply multiply numerator and denominator by the term containing the roots with the sign between them changed. So, in this case, we had \(\sqrt x + \sqrt y \) and so we needed to change the “+” to a “-”.

    Now, back to the problem. Here’s the multiplication.

    \[\frac{{3xy}}{{\left( {\sqrt x + \sqrt y } \right)}}\;\;\frac{{\left( {\sqrt x - \sqrt y } \right)}}{{\left( {\sqrt x - \sqrt y } \right)}} = \frac{{3xy\left( {\sqrt x - \sqrt y } \right)}}{{x - y}}\]

    Note that the results will often be “messier” than the original expression. However, as you will see in your calculus class there are certain problems that can only be easily worked if the problem has first been rationalized.

    Unfortunately, sometimes you have to make the problem more complicated in order to work with it.

  2. \(\frac{{\sqrt {t + 2} - 2}}{{{t^2} - 4}}\)
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    In this problem we’re going to rationalize the numerator. Do NOT get too locked into always rationalizing the denominator. You will need to be able to rationalize the numerator occasionally in a calculus class. It works in pretty much the same way however.

    \begin{align*}\frac{{\left( {\sqrt {t + 2} - 2} \right)}}{{\left( {{t^2} - 4} \right)}}\;\;\frac{{\left( {\sqrt {t + 2} + 2} \right)}}{{\left( {\sqrt {t + 2} + 2} \right)}} & = \frac{{t + 2 - 4}}{{\left( {{t^2} - 4} \right)\left( {\sqrt {t + 2} + 2} \right)}}\\ & = \frac{{t - 2}}{{\left( {t - 2} \right)\left( {t + 2} \right)\left( {\sqrt {t + 2} + 2} \right)}}\\ & = \frac{1}{{\left( {t + 2} \right)\left( {\sqrt {t + 2} + 2} \right)}}\end{align*}

    Notice that, in this case there was some simplification we could do after the rationalization. This will happen occasionally.