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

##### 1 Answer

#### Answer:

This sextic equation has

#### Explanation:

Consider the cubic equation:

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

**Descartes' rule of signs**

The signs of the coefficients have the pattern:

Since there are two changes of sign, this cubic has

It has exactly one negative Real solution.

**Discriminant**

The discriminant

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

In our example,

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

Since

**Conclusion**

Putting

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

So either

So our sextic equation has