Finding The Minimum Value

$\begin{aligned} L & = \sqrt{x^4-x^2-24x+145} + \sqrt{x^4-23x^2-2x+145} \\ & = \sqrt{x^4-2x^2+1 + x^2 -24x+144} + \sqrt{x^4-24x^2 + 144 + x^2 -2x+1} \\ & = \sqrt{({\color{#3D99F6}x^2}-1)^2 + (x -12)^2} + \sqrt{({\color{#3D99F6}x^2}-12)^2 + (x-1)^2} & \small \color{#3D99F6} \text{Let }y = x^2 \\ & = \sqrt{({\color{#3D99F6}y}-1)^2 + (x -12)^2} + \sqrt{({\color{#3D99F6}y}-12)^2 + (x-1)^2} \end{aligned}$

We note that $L$ is the sum of the distances from a point $P(x,y)$ on the curve $y=x^2$ to $Q(1,12)$ and $R(12,1)$ . Then, the smallest $L$ is when $Q$ , $P$ and $R$ are on a collinear. Therefore,

$L_{\text{min}} = \sqrt{(12-1)^2+(1-12)^2} = 11\sqrt 2$ , $\implies a+b = 11+2 = \boxed{13}$