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

Sep 13, 2016

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

#### Explanation:

Consider the cubic equation:

$7 {y}^{3} + 3 {y}^{2} - 9 y + 18 = 0$

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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 $a {y}^{3} + b {y}^{2} + c y + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

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:

$7 {x}^{6} + 3 {x}^{4} - 9 {x}^{2} + 18 = 0$

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.