What is the least real value of the expression $\sqrt{x^2-6x+13} + \sqrt{x^2-14x+58}$ where $x$ is a real number?

$ 
\textbf{(A)}\ \sqrt {39}
\qquad\textbf{(B)}\ 6
\qquad\textbf{(C)}\ \frac {43}6
\qquad\textbf{(D)}\ 2\sqrt 2 + \sqrt {13}
\qquad\textbf{(E)}\ \text{None of the above}
$
1. Rewrite the given expression in a more convenient form:
   \[
   \sqrt{x^2 - 6x + 13} + \sqrt{x^2 - 14x + 58}
   \]
   We can complete the square for each quadratic expression inside the square roots:
   \[
   x^2 - 6x + 13 = (x-3)^2 + 4
   \]
   \[
   x^2 - 14x + 58 = (x-7)^2 + 9
   \]
   Thus, the expression becomes:
   \[
   \sqrt{(x-3)^2 + 4} + \sqrt{(x-7)^2 + 9}
   \]

2. Interpret this geometrically. The expression represents the sum of the distances from the point \((x, 0)\) on the x-axis to the points \((3, 2)\) and \((7, -3)\).

3. To minimize the sum of these distances, we can use the fact that the shortest path between two points is a straight line. Therefore, the minimum value of the sum of the distances from \((x, 0)\) to \((3, 2)\) and \((7, -3)\) is the direct distance between \((3, 2)\) and \((7, -3)\).

4. Calculate the distance between the points \((3, 2)\) and \((7, -3)\):
   \[
   \sqrt{(7 - 3)^2 + (-3 - 2)^2} = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}
   \]

5. Therefore, the least real value of the given expression is:
   \[
   \boxed{\sqrt{41}}
   \]