Explicație pas cu pas:
Avem ecuatia [tex] f(x)=0 [/tex].
O rescriem:
[tex] (a+1)x^2+(2a+3)x+(a+1)=0 [/tex]
Scriem relatiile lui Viete:
[tex] S=x_1+x_2=\frac{-b}{a}=\frac{-(2a+3)}{a+1}=\frac{-2a-3)}{a+1}=\frac{-2a-2-1}{a+1}=\frac{-2a-2}{a+1}+\frac{-1}{a+1}=\frac{-2(a+1)}{a+1}+\frac{-1}{a+1}=-2+\frac{-1}{a+1} [/tex]
[tex] P=x_1x_2=\frac{c}{a}=\frac{a+1}{a+1}=1 [/tex]
Rezolvam ecuatia [tex] x_1+x_2=-3x_1x_2 [/tex] si aflam a:
[tex] -2+\frac{-1}{a+1}=-3\\ \frac{-1}{a+1}=-1\\ \frac{1}{a+1}=1 \\a+1=1\\a=0 [/tex]
