How do you simplify radical expressions?

1 Answer
Feb 1, 2015

There are two common ways to simplify radical expressions, depending on the denominator.

Using the identities #\sqrt{a}^2=a# and #(a-b)(a+b)=a^2-b^2#, in fact, you can get rid of the roots at the denominator.

Case 1: the denominator consists of a single root. For example, let's say that our fraction is #{3x}/{\sqrt{x+3}}#. If we multply this fraction by #{\sqrt{x+3}}/{\sqrt{x+3}}#, we won't change its value (since of course #{\sqrt{x+3}}/{\sqrt{x+3}}=1#, but we can rewrite it as follows:
#{3x}/{\sqrt{x+3}} \cdot {\sqrt{x+3}}/{\sqrt{x+3}}= \frac{3x\sqrt{x+3}}{\sqrt{x+3}^2}#, and finally obtain
#\frac{3x\sqrt{x+3}}{x+3}#

Case 2: the denominator consists of a sum/difference of roots. If we multiply by the difference/sum of the roots, we'll have the same result as above. For example, if you have

#\frac{\cos(x)}{\sqrt{x}+\sqrt{\sin(x)}#

You'll multiply numerator and denominator by the difference #\sqrt{x}-\sqrt{\sin(x)#, and obtain

#\frac{\cos(x)}{\sqrt{x}+\sqrt{\sin(x))} \cdot \frac{\sqrt{x}-\sqrt{\sin(x)}}{\sqrt{x}-\sqrt{\sin(x)}}# which is

#\frac{\cos(x)(\sqrt{x}-\sqrt{\sin(x)})}{\sqrt{x}^2-\sqrt{\sin(x)}^2}#

which finally equals

#\frac{\cos(x) (\sqrt{x}-\sqrt{\sin(x)})}{x-\sin(x)}#

Of course, when working with radicals, you always need to pay attention and make sure that the argument of the root is positive, otherwise you will write things that have no meaning!