In the following, $X$ is a finite complex. The Adams operator $\psi^{-1}$ (complex conjugation) acts on $K^0(X)$. After inverting $2$, the subgroup $KO^0(X) \otimes \Bbb Z[1/2]$ maps isomorphically to the subset of $K^0(X) \otimes \Bbb Z[1/2]$ fixed by $\psi^{-1}$.

We can then tensor this with $\Bbb Q$. Then $K^0(X) \otimes \Bbb Q \cong \prod_{n \geq 0} H^{2n}(X;\Bbb Q)$. The operator $\psi^{-1}$ acts on this by fixing the factors with $n$ even and negating the factors with $n$ odd. The fixed set is $KO^0(X) \otimes \Bbb Q \cong \prod_{m \geq 0} H^{4m}(X; \Bbb Q)$.

In the following, $X$ is a finite complex. The Adams operator $\psi^{-1}$ (complex conjugation) acts on $K^0(X)$. After inverting $2$, the group $KO^0(X) \otimes \Bbb Z[1/2]$ maps isomorphically to the subset of $K^0(X) \otimes \Bbb Z[1/2]$ fixed by $\psi^{-1}$.

We can then tensor this with $\Bbb Q$. Then $K^0(X) \otimes \Bbb Q \cong \prod_{n \geq 0} H^{2n}(X;\Bbb Q)$. The operator $\psi^{-1}$ acts on this by fixing the factors with $n$ even and negating the factors with $n$ odd. The fixed set is $KO^0(X) \otimes \Bbb Q \cong \prod_{m \geq 0} H^{4m}(X; \Bbb Q)$.