In the realm of 6 Dimensions-2

Algebra Level 5

The polynomial P ( x ) = x 6 + ( x r ) 6 + ( x 2 r ) 6 3 r \large P(x)=x^6+(x-r)^6+(x-2r)^6-3r has 6 complex non real zeros for real r r . Find the value of r r if it is given that the real part of one of it's imaginary zero is 4 4 .

This problem is inspired by Joel Tan's comment on its sister problem . For more of my problems check the set Questions I've Made ©


The answer is 4.0.

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1 solution

Sanjeet Raria
Mar 11, 2015

We observe that P ( r + x ) = P ( r x ) \large P(r+x)=P(r-x) for all x x .That means if one zero is ( r + S ) (r+S) the corresponding other would be ( r S ) (r-S) . We also know that complex zeros always occurs in pair. Now

Let α + ι β \alpha+\iota \beta be a root of the equation P ( x ) = 0 P(x)=0 .

α ι β \Rightarrow \alpha-\iota \beta is the other root.

Now let α + ι β = r + x r = α s , x = s + ι β \large \alpha+\iota \beta=r+x \Rightarrow r=\alpha-s, x=s+\iota \beta for a real s s .

r x = α 2 s ι β \large \Rightarrow r-x=\alpha-2s-\iota \beta is also a root. Now since there are only two complex roots we can say that α 2 s ι β = α ι β s = 0 \large \alpha-2s-\iota \beta=\alpha-\iota \beta \Rightarrow s=0 α = r \Large \Rightarrow \alpha=r

And hence r = 4 \large r=4

After being told I was correct for my answer 4 \boxed {4} I put 4 into P ( x ) P(x) and solved for the roots. There were SIX complex roots NOT "only two complex roots". Something is wrong. The question has to be changed and this solution is invalid. All SIX complex roots have the Real Component of 4. They are:

4 + 11.9215 i , 4 11.9215 i , 4 + 4.0903 i , 4 4.0903 i , 4 + 1.0708 i , 4 1.0708 i 4+11.9215i , 4-11.9215i, 4+4.0903i, 4-4.0903i, 4+1.0708i, 4-1.0708i

If my analysis is wrong, please tell me the value of β \beta in the solution above where α + ι β \alpha +\iota\beta is one of the conjugates in the 'only two complex solutions'

Bob Kadylo - 4 years, 3 months ago

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