$${{\sqrt{{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}}}}^{\,{\mathtt{2}}} = {\left({\mathtt{4}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}}}\right)}^{{\mathtt{2}}} \Rightarrow {\mathtt{x}} = {\mathtt{8}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{13}}$$
Is it possible to simplfy more the result?
+130511 So, squaring both sides, we have
x + 1 = 16 - 8√(2x -2) + 2x - 2 simplify
x+ 1 = 14 + 2x - 8√(2x -2)
- x - 13 = -8√(2x -2) multiply through by -1
x + 13 = 8√(2x -2) square both sides
(x +13)^2 = 64(2x -2) expand
x^2 + 26x + 169 = 128x - 128 rearrange
x^2 - 102x - 297 = 0 factor
(x -3) (x - 99) = 0
So x = 3 or x = 99
Both answers will solve your given situation......however..if the original problem was:
√(x + 1) = 4 - √(2x - 2) (as I suspect it was)
Then only x = 3 actually "works"
+130511 So, squaring both sides, we have
x + 1 = 16 - 8√(2x -2) + 2x - 2 simplify
x+ 1 = 14 + 2x - 8√(2x -2)
- x - 13 = -8√(2x -2) multiply through by -1
x + 13 = 8√(2x -2) square both sides
(x +13)^2 = 64(2x -2) expand
x^2 + 26x + 169 = 128x - 128 rearrange
x^2 - 102x - 297 = 0 factor
(x -3) (x - 99) = 0
So x = 3 or x = 99
Both answers will solve your given situation......however..if the original problem was:
√(x + 1) = 4 - √(2x - 2) (as I suspect it was)
Then only x = 3 actually "works"
