👤
a fost răspuns

*[tex](m+1)x^{2}-(2m+1)x+m=0, m apartine R[/tex]m+1)[tex]m=? x_{1}>0, x_{2}<0.[/tex]

Am inceput asa: [tex]x_{1}>0, x_{2}<0 => P<0, D\geq0[/tex]

Răspuns :

Ai inceput aproape bine. Trebuie ca [tex]\Delta>0[/tex] (fara =)
Δ=b²-4ac⇒a=m+1, b= -2m-1,   c=m
Δ=(-2m-1)²-4m(m+1)⇒Δ=4m²+4m+1-4m²-4m⇒Δ=1
x1=(2m+1+1)/2m+2⇒x1=2m+2/2m+2⇒x1=1
x2=2m+1-1/2m+2=2m/2m+2=m/m+1
m/m+1=0 cond.m+1≠0⇒m≠ -1
pt.m=0, x2=0