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# Cube root of a negative vs negative to the power of 1/3

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501
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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

#3
+92206
+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
Sort:

#1
+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
+85699
+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

CPhill  Jan 16, 2016
#3
+92206
+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
#4
+85699
0

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

CPhill  Jan 17, 2016
#5
+92206
+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......

Melody  Jan 17, 2016
#6
+85699
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......

CPhill  Jan 17, 2016
#7
+92206
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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