Question

x2-2mx+m(1+m)=0
m=?
A)sa aiba radacini egale
B)sa aiba radacini reale diferite
C)sa nu aiba radacini reale

x2=x la puterea a doua

Answer 1

a) impunem conditia  Δ=0.

x²-2mx+m(1+m)=0
a=1
b=-2m
c=m(1+m)
Δ= (-2m)²-4m(1+m)
Δ=4m²-4m-4m²
Δ= -4m.
Δ=0.  ⇒ -4m=0⇒m=0

b)impunem conditia Δ>0
Δ=-4m>0 |·(-1)
     4m<0⇒m<0⇒m∈(-inf;  0)

c) impunem conditia ca Δ<0
Δ=-4m<0 ⇒ 4m>0⇒m>0⇒m∈(0; +inf)