($\Leftarrow$) Suppose every root of $f(x)$ is in $K$. Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$. Then $f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$, showing that $f(x)$ splits in $K$.
Solution: ($\Rightarrow$) Suppose $f(x)$ splits in $K$. Then $f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$ for some $\alpha_1, \ldots, \alpha_n \in K$. Hence, every root of $f(x)$ is in $K$. abstract algebra dummit and foote solutions chapter 4
You're looking for solutions to Chapter 4 of "Abstract Algebra" by David S. Dummit and Richard M. Foote! ($\Leftarrow$) Suppose every root of $f(x)$ is in $K$
Solution: Let $\alpha$ and $\beta$ be roots of $f(x)$. Since $f(x)$ is separable, there exists $\sigma \in \operatorname{Aut}(K(\alpha, \beta)/K)$ such that $\sigma(\alpha) = \beta$. By the Fundamental Theorem of Galois Theory, $\sigma$ corresponds to an element of the Galois group of $f(x)$, which therefore acts transitively on the roots of $f(x)$. Solution: ($\Rightarrow$) Suppose $f(x)$ splits in $K$
Exercise 4.1.1: Let $K$ be a field and $\sigma$ an automorphism of $K$. Show that $\sigma$ is determined by its values on $K^{\times}$.
Chapter 4 of Dummit and Foote covers "Galois Theory". Here are some solutions to the exercises:
Exercise 4.2.2: Let $K$ be a field, $f(x) \in K[x]$, and $L/K$ a splitting field of $f(x)$. Show that $L/K$ is a finite extension.