Minimum value of the given function 
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .

Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$ 
Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.
But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?

If we write this as $$\sqrt{x^2+(x+1)^2}+\sqrt{(x-2)^2+(x-3)^2}$$ then we are searching for a point $T$ on line $y=x$ for which $TA+TB$ takes minumim where $A(0,-1)$ and $B(3,2)$. 
Now this is well know problem froma ancient greek. Reflect $A$ acros this line and get $A'(-1,0)$. By triangle inequality we can see that $TA+TB\geq A'B$ and that minimum is achieved at intersection of lines $y=x$ and line $A'B$ which is $y = {x +1 \over 2}$. 
So $\alpha = A'B^2 = 20$.
