2023/020) Find the minumum of $\sqrt{(x^2 + 4x + 13)} + \sqrt{(x^2 -8x + 41)}$

 We have 

 $\sqrt{(x^2 + 4x + 13)}$ Completing the square we get

$= \sqrt{x^2+ 4x+ 4 + 9} = \sqrt{(x+2)^2 + 3^2}$

The above in a plane is the distance from $(-2,-3)$ to $(x,0)$

 $\sqrt{(x^2 - 8x + 41)}$ Completing the square we get

$= \sqrt{x^2 - 8x+ 16 + 25} = \sqrt{(x-4)^2 + 5^2}$

The above in a plane is the distance from $(4,-5)$ to $(x,0)$ also from (4,5) to $(x,0)$ and because of symmetry as the line has to be to $(x,0)$ we can choose $(4,-5)$

The sum of the 2 expression is the distance from $(-2,3)$ to $(x,0)$ and from $(x,0)$ to $(4,5)$ or $(4,-5)$ as the 2 distances are same. it is better to chose a point on the other side of x axis so (4,5).

The distance ins minimum if  $(-2,3),(x,0)$ and $(4,5)$ are in same line and so minimum distance is distance from $(-2,3)$ to $(4,5)$ $= \sqrt{(4+2)^2 + (5+3)^2} = 10$