\(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\)

Squaring both sides, we have

\((\sqrt{x} + \sqrt{x + 1})^{2} = (\sqrt{2x + 1})^{2}\)

\(x + 2\sqrt{x(x + 1)} + x + 1 = 2x + 1\)

\(2x + 1 + 2\sqrt{x(x+1)} - (2x + 1) = 0\)

\((2\sqrt{x(x + 1)})^{2}= 0^{2}  \implies 4(x(x + 1)) = 0\)

\(\therefore x(x + 1) = 0\)

\(x = \text{0 or -1}\)