Question

Aflati valorile parametrului real "a" pentru care ecuatia: ax^2-(a+1)x+2a-1=0 are o solutie unica

Answer 1

Δ=0
(a+1)²-4a(2a-1)=0
a²+2a+a-8a²+4a=-7a²+7a=0, a(a-1)=0, a=0, a=1
dar a≠0 pt ca daca e 0 ecuatia nu mai este de gradul 2
a=1

Answer 2

ax²-(a+1)x-2a-1=0
a=a
b=(a+1)
c=2a+1
Δ=b²-4ac⇒(a+1)²-4×a×(2a+1)⇒
Δ=(a²+2a+1)-4a×(2a+1)⇒
Δ=a²+2a+1-8a²-4a⇒
Δ=-7a²-2a+1=0⇒
a=-7
b=-2
c=1
Δ=b²-4ac⇒Δ=(-2)²-4×(-7)×1⇒4+28⇒Δ=32⇒√Δ=√32⇒Δ=2√6
x1x2=-b+-√Δ/2a
x1=2+2√6/2⇒x1=2√6
x2=2-2√6/2=x2=-2√6
suces