pentru rezolvarea acesteia se calculeaza Δ=b²-4ac
daca Δ>0, ecuatia are doua solutii reale distincte x1=-b-√Δ/2a si x2=-b+√Δ/2a
dacaΔ=0 , ecuatia are o solutie reala x=-b/2a
dacaΔ<0, ecuatia nu are solutii reale.
exemple
1) rezolvati ecuatia: 2x²-x-6=0
a=2 b=-1 c=-6 Δ-inseamna delta
si Δ=(-1)²-4*2*(-6)=49
x1=-b+√Δ/2a=1+7/4=2
x2=-b-√Δ/2a=1-7/4=-3/2
asadar x∈{2, -3/2}
2) determinati m∈R astfel incat ecuatia x²-mx+1=0 sa aiba solutii egale.
rezolvare. trebuie caΔ=0=>m²-4=0⇔m²=4 deci m=2⇔m∈{-2, 2}
3) aratati ca ecuatia nu are solutii reale, oricare ar fi m∈R.
x²+(m+2)x+m²+m+2=0
rezolvare.
a=1 b=m+2 c=m²+m+2
Δ=(m+2)²-4(m²+m+2)=m²+4m+4-4m²-8=-4-3m²<0 pentru orice m∈R, asadar ecuatia nu are solutii reale.
calculati
x²-8x+7=0
a=1 b=-8 ²c=7 afli delta care este Δ=b²-4ac
Δ=(-8)²-4*1*7
Δ=64-28
Δ=36
x1=-b-√Δ/2a=-(-8)-√36/2*1=8-6/2=2/2=1
x2=-b+√Δ/2a=-(-8)+√36/2*1=8+6/2=14/2=7
