Will making even both sides present a $x ^ 4$ term that may make complex the formula?


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 * * Tip:$$ \ sqrt p>

^ 2 - 4x +3= \ sqrt 2 (x ^ 2 - 2x) + 3 $$ Allow$ v = x ^ 2- 2x $$

$ \ suggests x ^ 2- 2x= \ sqrt 2 (x ^ 2 -2x) +3

\ longrightarrow v = \ sqrt $

$ Square both sides to obtain$$ v ^ 2= 2v + 3 \ \ v ^ 2- 2v- 3= 0$$ Resolve the square formula to discover the origins of $v$ and after that reverse the replacement to locate the worths of $x$ when $v > 0$


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.  * Settling both sides we have $.