Originally Posted by
OReilly $\displaystyle \begin{array}{l}
\sqrt {4 - 2x} = 1 + \sqrt {x - 3} \\
\left( {\sqrt {4 - 2x} } \right)^2 = \left( {1 + \sqrt {x - 3} } \right)^2 \\
4 - 2x = 1 + 2\sqrt {x - 3} + x - 3 \\
6 - 3x = 2\sqrt {x - 3} \\
\left( {6 - 3x} \right)^2 = \left( {2\sqrt {x - 3} } \right)^2 \\
9x^2 - 36x + 36 = 4x - 12 \\
9x^2 - 40x + 48 = 0 \\
$
$\displaystyle x_{1/2} = \frac{{40 \pm \sqrt {1600 - 1728} }}{{18}} = \frac{{40 \pm \sqrt { - 128} }}{{18}}$
$\displaystyle x$ can't be real number since under square root is negative number.