The equation $\sqrt {x + 4} - \sqrt {x - 3} + 1 = 0$ has: 
$\textbf{(A)}\ \text{no root} \qquad  \textbf{(B)}\ \text{one real root} \\ \textbf{(C)}\ \text{one real root and one imaginary root} \\ \textbf{(D)}\ \text{two imaginary roots} \qquad \qquad\textbf{(E)}\ \text{two real roots}$
Simplify.
\[\sqrt {x + 4} = \sqrt{x-3} -1\]
\[x + 4 = x - 3 -2\sqrt{x-3} + 1\]
\[6 = -2\sqrt{x-3}\]
We don't need to solve any further. The square root term is negative, therefore there will be no real roots. The answer is $\boxed{A}.$
-coolmath34