How do you simplify #sqrt(9e^6)#?

1 Answer
Aug 21, 2016

Answer:

It depends what you mean by #e#...

Explanation:

  • If #e# is the mathematical constant (#~~2.7182818#), then the answer is definitely #3e^3#.

  • If #e# is a Real valued variable, then #3abs(e^3)# covers both positive and negative values of #e#.

  • If #e# is a Complex valued variable then #3sqrt(e^6)# is about the best you can do.

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The square root symbol #sqrt# denotes the principal square root.

We have:

#(3e^3)^2 = 3^2(e^3)^2 = 9e^6#

#(-3e^3)^2 = (-3)^2(e^3)^2 = 9e^6#

So regardless of the value of #e# (Real or Complex), both #3e^3# and #-3e^3# are square roots of #9e^6#, but which is the principal one?

If #3e^3# is a non-negative Real number then it is the principal square root. Hence for #e >= 0# we have #sqrt(9e^6) = 3e^3#

If #3e^3# is a negative Real number then #-3e^3# is positive and #sqrt(9e^6) = -3e^3#.

So if #e# is a Real number then #sqrt(9e^6) = abs(3e^3) = 3abs(e^3)#

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How about the Complex case?

Suppose #e=sqrt(i)=cos(pi/4)+i sin(pi/4)#

Then:

#3e^3 = 3(cos((3pi)/4)+i sin((3pi)/4))#

#sqrt(9e^6) = 3sqrt(cos((6pi)/4)+i sin((6pi)/4))#

#=3sqrt(cos(-pi/2)+i sin(-pi/2))#

#=3(cos(-pi/4)+i sin(-pi/4))#

#=-3(cos((3pi)/4)+sin((3pi)/4))#

#=-3e^3#

So in the general Complex case, about the best we can say is:

#sqrt(9e^6) = +-3e^3#

or better:

#sqrt(9e^6) = 3sqrt(e^6)#

It would be possible to pick out individual cases according to their #Arg# value, but it would hardly be simplifying.