Question

Rezolvați ecuația: (x^2-2x)(x^2-2x+5)+4=0

Answer 1

$x^{2} -2x=y$
$y(y+5)+4=0$
$y^{2} +5y+4=0 \\ y^{2}+4y+y+4=0 \\ (y+1)(y+4)=0$
y=-1,y=-4
$x^{2} -2x=-1 \\ (x-1)^{2} =0 \\ x=1$
$x^{2} -2x=-4 \\ x^{2} -2x+1+3=0 \\ (x-1)^{2} +3=0$
 imposibil

Answer 2

Notez $a= x^{2} -2x.$

$( x^{2} -2x)( x^{2} -2x+5)+4=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ a(a+5)+4=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ a^{2} +5a+4=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ a^{2} +a+4a+4=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ a(a+1)+4(a+1)=0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (a+1)(a+4)=0=\ \textgreater \ a=-1~sau~a=-4. \\ \\ a=-1=\ \textgreater \ x^{2} -2x=-1\ \textless \ =\ \textgreater \ x^{2} -2x+1=0\ \textless \ =\ \textgreater \ (x-1) ^{2} =0=\ \textgreater \ \\ =\ \textgreater \ x-1=0=\ \textgreater \ \boxed{x=1}. \\ \\ a=-4=\ \textgreater \ x^{2} -2x=-4\ \textless \ =\ \textgreater \ x^{2} -2x+1=-3\ \textless \ =\ \textgreater \ (x-1) ^{2} =-3,~imposibil }.$

Solutia ecuatie este x=1.