Show that $Gal(X^4+aX^2+b)\cong V_4$ if and only if $b$ is a square in $\mathbb{Q}$

Question

From right to left I showed that the discriminant is a square \begin{align} D&=16a^4b-128a^2b^2+256b^3=16b(a^4-8a^2b+16b^2)\\ &=16b(4b-a^2)^2 \end{align} So now I have to show that the cubic resolvent is reducible, but I tried a lot of writing stuff trying to show that $g=(X-\lambda_1)(X-\lambda_2)(X-\lambda_3)$ has a root in $\mathbb{Q}$ where $\lambda_1=\alpha_1\alpha_3+\alpha_2\alpha_4,\lambda_2=\alpha_1\alpha_2+\alpha_3\alpha_4$ and $\lambda_3=\alpha_1\alpha_4+\alpha_2\alpha_3$ where $\alpha_i$ are the four roots of $X^4+aX^2+b\in\mathbb{Q}[X]$. I need help with this step. Also I don't know how to start the implication from left to right. Thanks in advance.