I'm looking at the following example on page 8 of Algebraic Geometry V by Iskovskikh.
I've been trying to show that $-K_X = 3E$. I'll outline my work so far. Let the vertex of $X$ be $p$. Then $X \backslash p$ is equipped with a projection map $\pi$ to $F_4$, which gives $X \backslash p$ the structure of an $\mathbb A^1$ bundle. Thus $Cl X \cong Cl (X \backslash p) \cong Cl (F_4) \cong \mathbb Z$. Examination of the isomorphism tells us that $Cl(X)$ is generated by a hyperplane section. We also have a map $\iota: Cl(X) \to Cl(F_4)$ given by restriction to the copy of $F_4$ at infinity (that is, at $Z(x_6)$). This map need only be nonzero to be injective. In fact it takes a general hyperplane section to twice a generator of $CL(F_4)$, which we can see by first intersection with $Z(x_6) \cong \mathbb P^5$ and observing that $F_4 \hookrightarrow \mathbb P^5$ is induced by $\mathcal O(2)$.
My idea now is to calculate $K_X$ and show that $\iota(K_X) = -3\iota(E)$, which by injectivity will yield the result. By a slight abuse of notation and the adjunction formula $\mathcal O_{\mathbb P^2}(-3) \cong K_{F_4} \cong K_X|_{F_4} \otimes \mathcal O(X \cap Z(x_6))|_{F_4}^\vee$. Replacing $X \cap Z(x_6)$ by a general hyperplane section, we find that $\mathcal O(-5) \cong K_X|_{F_4}$. I now seem to have contradicted myself: if $\iota(K_x) = -3\iota(E)$ we would have $-3 | 5$. What is my mistake? How does one demonstrate that $-K_X = 3E$?
There is a typo in the book. In fact, $K_X = -5E$.
One way to see it is with your argument: $F_4$ is a divisor of class $H = 2E$, so if $K_X = -rE$ then by adjunction $$ K_{F_4} = (K_X + F_4)\vert_{F_4} = (-rE + 2E)\vert_{F_4} = -(r-2)E\vert_{F_4}, $$ which gives $r - 2 = 3$, so $r = 5$.
Another way to see this is by identifying $X$ with the weighted projective space $\mathbb{P}(1,1,1,2)$, which gives $r = 1 + 1 + 1 + 2 = 5$.