just to clarify...

[deederz]

im not sure how to type the actual problem using symbols but i need clarification on a problem.
index16 sqrt(-2)^16. There is no solution because the root is not a real number. Is that correct?

 

[pka]

im not sure how to type the actual problem using symbols but i need clarification on a problem.
index16 sqrt(-2)^16. There is no solution because the root is not a real number. Is that correct?

Is it \(\displaystyle \sqrt[{16}]{{\left( { - 2} \right)^{16} }}\text{ or }\left( {\sqrt[{16}]{{ - 2}}} \right)^{16}~? \)

 

[mmm4444bot]

There is no solution because the root is not a real number. Is that correct?

There is no way for us to answer your question because you did not include the instructions that come with this exercise.

Do the instructions restrict you to the set of Real numbers? What exactly does "solution" indicate, in this exercise?

(Solutions generally involve equations.)

Please be complete. :cool:

 

[deederz]

To PKA...it is the first one...to MMM, it only says find the following then gives the problem.

 

[mmm4444bot]

When the index of a radical is an even number (greater than zero, of course), we have this property of radicals:

\(\displaystyle \sqrt[{n}]{{\left( {x} \right)^{n} }} = |x|, \text{for all Real numbers x}\)

This is true because, when n is even, x^n is never negative.

 

[deederz]

Since the index if even, then the answer is -2, correct?

 

[pka]

When the index of a radical is an even number, we have this property of radicals:

\(\displaystyle \sqrt[{n}]{{\left( {x} \right)^{n} }} = x, \text{for all Real numbers x}\)

This is true because, when n is even, x^n is never negative.

Actually this is true: If n is an even positive integer
\(\displaystyle \sqrt[{n}]{{\left( {x} \right)^{n} }} = |x|\)

Just as this question is designed to show.
\(\displaystyle \sqrt[{16}]{{\left( { - 2} \right)^{16} }}=|-2|=2.\)

 

[mmm4444bot]

Since the index if even, then the answer is -2, correct?

I made a sloppy mistake (fixed) in my first response, by forgetting to type the absolute-value symbols.

The radical equals |-2|.

I apologize for any confusion or frustration.