Question

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

Answer 1

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.