How do you find the number of complex, real and rational roots of #7x^6+3x^4-9x^2+18=0#?

1 Answer
Sep 13, 2016

Answer:

This sextic equation has #6# non-Real Complex roots, two of which are pure imaginary.

Explanation:

Consider the cubic equation:

#7y^3+3y^2-9y+18 = 0#

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Descartes' rule of signs

The signs of the coefficients have the pattern: #+ + - +#

Since there are two changes of sign, this cubic has #2# or #0# positive Real solutions.

It has exactly one negative Real solution.

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Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ay^3+by^2+cy+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=7#, #b=3#, #c=-9# and #d=18#, so we find:

#Delta = 729+20412-1944-428652-61236 = -470691#

Since #Delta < 0# this cubic has #1# Real zero (which we have already ascertained is negative) and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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Conclusion

Putting #y = x^2#, we arrive at our original sextic equation:

#7x^6+3x^4-9x^2+18 = 0#

So either #x^2# is negative or non-Real Complex. In the first case, #x# is pure imaginary. In the latter it is not pure imaginary but it is non-Real Complex.

So our sextic equation has #6# non-Real Complex roots, two of which are pure imaginary.