Will making even both sides present a $x ^ 4$ term that may make complex the formula?
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 $v > 0$
. Settling both sides we have $.