If all terms of an equation are positive and there is no term of odd power of $x$... If all terms of an equation are positive and there is no term of odd power of $x$, then all its roots are complex numbers.
As for example, $x^2+2=0$ has roots
$x=\pm\sqrt {-2}$
Similarly, we can have such more examples.
But, what's the general way of proving it?
 A: "Then all it's roots are complex numbers"
That isn't true. Take x^4+x^2. It has an obvious real root at x=0. 
The Fundamental Theorem of Algebra guarantees that you will have roots, but some of them may be real, some of them may be complex. 
A: I think what you're referring to is a polynomial $P$ of degree $n$ 
in which all terms of odd degree have coefficient zero
and all terms of even degree (including the constant term)
have coefficients greater than zero.
The question is whether $P(z) = 0$ has any solutions for real $z,$
or only for complex $z$ that are not real.
What is the minimum value of such a polynomial over the real numbers?
A: If a polynomial can be written as $$p(x)=\sum_{i=0}^nc_{2i}x^{2i}=c_0+\sum_{i=1}^nc_{2i}x^{2i}$$
where $c_{2i} >0$ for all $i \in \{1, \ldots, n\}$,
since $$\forall x \in \mathbb{R}, x^{2i} \geq 0 $$
Hence 
$$\forall x \in \mathbb{R}, \sum_{i=1}^nc_{2i}x^{2i}\geq 0$$
and $$\forall x \in \mathbb{R}, p(x)=c_0+\sum_{i=1}^nc_{2i}x^{2i}\geq c_0 >0.$$
Hence there is no real solution.
