Question

$x^{2} -x-1=0 are    radacinile       x_{1} si x x_{2}     calculati ( x_{1} -1)(x {2} -1)$

Answer 1

[tex] (x_{1}-1)( x_{2}-1) = x_{1} x_{2} - (x_{1} + x_{2}) +1=P-S+1,unde,S=- \frac{b}{a} [/tex]=$- \frac{-1}{1}=1,si,P= \frac{c}{a}= \frac{-1}{1}=-1.$
$Deci( x_{1}-1 )( x_{2}-1)=-1-1+1=-1$