 Solution

The genus-degree formula gives arithmetic genus three for a plane quartic. Each of its three double points has multiplicity of a plane curve at a point two, so the arithmetic genus drop under a point blowup is one at each point. The resulting integral projective curve $C^{\prime }$ hasFor its finite normalization of an algebraic curve $\nu :\tilde{C}\to C^{\prime }$, the arithmetic genus identity isBoth terms are nonnegative, so $g(\tilde{C})=0$. Over the algebraically closed field, the rationality of a smooth projective genus-zero curve identifies $\tilde{C}$ with the projective line: a point $P$ exists, and Riemann-Roch theorem gives a degree-one map from the two-dimensional space $L(P)$. Consequently $C$ is birational to ${\mathbb{P}}^1$. This proves the rationality of a plane quartic with three double points even when a double point is not an ordinary node.