Question about existence of root of a polynomial

Question

Is the below true or false?

If $f(x)$ is a non-constant real polynomial for which $g(x)=f(x^2)+1$ has a real root then $f(x)$ also has a real root?

If $f(x)=\sum_{n=0}^m a_nx^n$ with $\deg(f)=m$, then $f(x^2)=\sum_{n=0}^m a_nx^{2n}$ , $\deg(g)=\deg(f(x^2))=2m$ ($g(x)$ has only even powers) and if $x_0$ the real root of $g(x)$ then: $g(x_0)=0 \Rightarrow f(x_0^2)+1=0$, but I don't have anything else. What do I miss?

Answer 1

FALSE

Counterexample To The Claim: $$f(x)=-x^2-\frac{1}{2} \implies g(x)=f(x^2)+1=-x^4+\frac{1}{2}$$

$g(x)$ has a real root, but $f(x)$ does not.