1. ## Help Math ???'s

solve: square root of 4-2x=1+square root of x-3

2. Please confirm that problem is:
$\displaystyle \sqrt {4 - 2x} = 1 + \sqrt {x - 3}$

3. ## Yes

that is the problem

4. $\displaystyle \begin{array}{l} \sqrt {4 - 2x} = 1 + \sqrt {x - 3} \\ \left( {\sqrt {4 - 2x} } \right)^2 = \left( {1 + \sqrt {x - 3} } \right)^2 \\ 4 - 2x = 1 + 2\sqrt {x - 3} + x - 3 \\ 6 - 3x = 2\sqrt {x - 3} \\ \left( {6 - 3x} \right)^2 = \left( {2\sqrt {x - 3} } \right)^2 \\ 9x^2 - 36x + 36 = 4x - 12 \\ 9x^2 - 40x + 48 = 0 \\$

$\displaystyle x_{1/2} = \frac{{40 \pm \sqrt {1600 - 1728} }}{{18}} = \frac{{40 \pm \sqrt { - 128} }}{{18}}$

$\displaystyle x$ can't be real number since under square root is negative number.

5. 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.
And, just to make things really ugly, you should plug those answers back into the original to see that they actually do solve the problem. I personally wouldn't do this one by hand (uuuuuugly ) but in general, whenever you square both sides of an equation you could be introducing extra solutions, so you need to check those solutions in the original equation.

-Dan