Free counters!

Friday, June 21, 2013

Algebraic Operations - Radicals, 6

Category: Algebra

"Published in Newark, California, USA"

Perform the indicated operations


Solution:

Consider the given equation above


The first thing that we have to do is to examine the radicals first if they can simplify or not. As a rule in Mathematics, all radicals must be simplified as much as we can.

At the first term, 162 is not a perfect 4th root. The factors of 162 are 81 and 2. 81 is a perfect 4th root.

At the second term, 32 is a not a perfect 4th root. The factors of 32 are 16 and 2. 16 is a perfect 4th root. Since x6 is not a perfect 4th root, then we can factor x6 into x4 and x2.

At the third term, the denominator contains a radical. We need to eliminate the radical sign at the denominator by rationalization of the denominator. Multiply both the numerator and denominator by x2 so that the denominator becomes a perfect 4th root which is x4

Hence, the given equation above becomes




Take the 4th root of the numbers inside the radicals that are perfect 4th root, we have  



Since all the terms inside the radicals are the same, then we can combine them and therefore, the final answer is