Question 1196478
<font color=black size=3>
The smallest {{{x^2}}} can get is 0, assuming x is a real number. Squaring a negative gets us a positive
Eg: {{{x^2 = (-7)^2 = (-7)*(-7) = 49}}}


The same goes for {{{x^4}}} and {{{x^6}}}
because {{{x^4 = (x^2)^2}}} and {{{x^6 = (x^3)^2}}}
We can rewrite those terms to be in the form of {{{(something)^2}}}


This means the smallest {{{x^6+3x^4+30x^2}}} can get is {{{0+0+0=0}}}


By extension, the smallest {{{x^6+3x^4+30x^2+6}}} can get is {{{0+6 = 6}}}. This is the lower bound of the range. There is no upper bound.


The range of {{{y = x^6+3x^4+30x^2+6}}} is the the inequality *[tex \Large 6 \le y < \infty] which gives the interval notation [6, ∞)
Or put simply: the range is {{{y >= 6}}}


As you can see, the output {{{y = 0}}} is not possible. 
There are no real number solutions, or roots, for {{{x^6+3x^4+30x^2+6 = 0}}}
A root is an x input that produces the output {{{y = 0}}}
Real numbered roots correspond to the x intercepts.


Visual confirmation with a graph
<a href = "https://www.desmos.com/calculator/tcrnn8advc">https://www.desmos.com/calculator/tcrnn8advc</a>
The curve does not cross the x axis, so it doesn't have any x intercepts.
The curve may look like a parabola, but it's not.
</font>