Question #586a5

2 Answers
Aug 30, 2017

Your question doesn’t seem logical.

Explanation:

Neither #3sqrt(-64)# or #sqrt(-64)# can be simplified to real numbers.
The square root of a negative number is always an imaginary number. The examples can be simplified to have the lowest value radicand, but this will still be an imaginary number.

From examples:

#3sqrt(-64) => # #3sqrt(64 *( -1) )## =># #24sqrt(-1)#

#sqrt(-64) =>## sqrt(64*(-1)# #=> 8sqrt(-1)#

These can also be expressed as #24i# and #8i# or #(0 + 24i)# and #(0 + 8i)#, where #i# is the imaginary unit #i =# #sqrt(-1)#.
These are known as pure imaginary numbers.

Aug 30, 2017

Because a negative number raised to an odd power gives a negative result.

Explanation:

#(-3)^3 = -27#, so #root(3)(-27) = -3#

#(-2)^5 = -32#, so #root(5)(-32) = -2#

But for even powers, both positives and negatives raised to an even power give a positive result.

#(5)^2 = 25# and also #(-5)^2 = 25#

And any other number raised to the power #2# will give a positive answer (except #0^2 = 0#).

So no number can be raised to the power #2# and give a negative result.

(Until we introduce a new kind of numbers that lets us do that.)