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

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.