What is #sqrt(6x) - 6sqrt(96x) + 2sqrt(216x)#?

1 Answer
A. S. Adikesavan
Apr 3, 2016

#-23sqrt(6x) and 25sqrt(6x)#. In effect, there are four values# +-23sqrt6sqrtx and +-25sqrt6sqrtx##

Explanation:

#sqrt (a^2)= +-a#. This applies to any #sqrt#(positive number, including #sqrt6#
#sqrt96 = sqrt(6X16)=+-4sqrt6#
#sqrt216=sqrt(6X36)=+-6sqrt6#.
So, the given expression is #sqrt(6x)(1+-12+-12)#.
Here again, #sqrt(6x) = sqrt6sqrtx# has two values #+-sqrt6 sqrtx#.

#a^(1/N)# has N values, including complex conjugate pairs, if any. In the case N=2 and a>0, the two values are #+-sqrta#.

I regret for making the answer heavy. My intention is to read 'Mathematics' as a synonym for 'Exactitude'.. In applying Mathematics to other subjects, including science, this aspect is important for analysis.

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