Chloe W.

asked • 12/30/15

Explain how a polynomial can have roots that are not complex conjugates

Question: The roots of (x-1)(x+i) = 0 are obviously 1, -i. These are not complex conjugates. How do you explain this? Note that 'i' represents the imaginary number i.

2 Answers By Expert Tutors

By:

Chloe W.

This answer clearly answers the question. Thanks!
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12/30/15

Saleel M. answered • 12/31/15

Tutor
5 (1)

Fun and Creative Math / English Tutor with Management Degree

Mark M.

tutor
-(x2-x)2 = -(x4-2x3+x2) = -x4+2x3-x2
 
So, -(x2-x)2 = x2-2x+1 simplifies to x4-2x3+2x2-2x+1=0.
 
This equation does in fact have 1 and -i as two of its roots.  
 
 
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12/31/15

Mark M.

tutor
x4-2x3+2x2-2x+1 factors as (x-1)2(x2+1), which has roots 1 (with multiplicity 2), i, and -i.
 
1 and -i are the roots of the original equation (x-1)(x+i)=0 and i is an extraneous root of (x-1)(x+i)=0 that arose from squaring both sides of the given equation.  
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12/31/15

Hilton T.

tutor
Saleel,
There is an error in your analysis.
I extract below from your answer.
[-(x2 - x)2 = x2 - 2x + 1

Finally we reduce both sides of the equation in order to get a polynomial on one side and zero on the other:

-(x4 - 2x3 - x2) = x2 - 2x + 1]
 
Your error is that you substitute x4 - 2x3 - x2 for (x2 - x)2, when it should have been x4 - 2x3 + x2
As such, you end up with erroneous conclusions.
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12/31/15

Mark M.

By the Product Rule of Zero:
If (x - 1)(x + i) = 0
The x - 1 = 0
Or x + i = 0
Two roots: {-i, 1} 
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04/20/16

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