Question

Show that the equation
x^9 -5x^5 +4x^4 + 2x^2 + 1=0
has at least 6 imaginary solutions.
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Answer 1

p (x)=x^9 -5x^5 + 4x^4 + 2x^2 + 1=0

no.of sign change=2

max. +ve roots=2

p(-x)=(-x)^9 -5 (-x)^5 +4 (-x)^4 + 2 (-x)^2 +1=0

=) p (-x)= -x^9 +5x^5 + 4x^4 + 2x^2 + 1=0

No.of sign change=1

Max.+ve real roots =1

Max.real roots=2+1=3

Total no.of roots=9

Min.imaginary roots=9-3=6

therefore at least 6 imaginary roots.

Hence, proved.