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Why is -5 the answer of the cube root of negative 125 when the answer to the equivalent equation negative 125 to the power of 1/3 is not a real number?

Guest Jan 16, 2016

Best Answer 

 #3
avatar+91024 
+10

Why is -5 the answer of the cube root of negative 125 when the answer to the equivalent equation negative 125 to the power of 1/3 is not a real number?

 

All cube roots have 3 answers.

The calculator only looking for real roots only looks on the positive x axis.

No cubic root of -125 is on the positive x axis so a no real roots answer is returned.  Some things calculators do not do very well.  We have to interpret and understand the answers.

Here are the 3 roots of -125 displayed on a complex number plan.

I thought that you might find it interesting :)

 

 

Melody  Jan 17, 2016
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7+0 Answers

 #1
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+4

It is REAL number just negative. Because -5 X -5=25 X -5=-125

If you state it like (-125i)^1/3=-5 (-1)^(5/6)

Guest Jan 16, 2016
 #2
avatar+78620 
+5

If we assume the real-value root, the answer  is -5  - a real number

 

If we assume the principal root the answer  is    5*cube root (-1)  - not a real number

 

 

 

cool cool cool

CPhill  Jan 16, 2016
 #3
avatar+91024 
+10
Best Answer

Why is -5 the answer of the cube root of negative 125 when the answer to the equivalent equation negative 125 to the power of 1/3 is not a real number?

 

All cube roots have 3 answers.

The calculator only looking for real roots only looks on the positive x axis.

No cubic root of -125 is on the positive x axis so a no real roots answer is returned.  Some things calculators do not do very well.  We have to interpret and understand the answers.

Here are the 3 roots of -125 displayed on a complex number plan.

I thought that you might find it interesting :)

 

 

Melody  Jan 17, 2016
 #4
avatar+78620 
0

Thanks, Melody.......cool graph, too !!!

 

 

cool cool cool

CPhill  Jan 17, 2016
 #5
avatar+91024 
+5

Thanks Chris :)

 

You know, I do not really understand why calculators have such a tough time solving problems like this one.

It is not just one calc that has problems, it is almost universal......      indecision

Melody  Jan 17, 2016
 #6
avatar+78620 
0

I don't know, either......perhaps Alan knows the reason for this.......WolframAlpha will actually give the real solution and the principal root for the non-real solution......

 

 

cool cool cool

CPhill  Jan 17, 2016
 #7
avatar+91024 
0

Is the principal root the one with the smallest argument   (angle) in the complex plane ?

 

I suppose it must be ://

Melody  Jan 17, 2016

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