Question

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

Am inceput asa: $<var>x_{1}>0, x_{2}<0 => P<0, D\geq0</var>$

Answer 1

Ai inceput aproape bine. Trebuie ca $\Delta>0$ (fara =)

Answer 2

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