Răspuns:
Explicație pas cu pas:
a) x²-x-90 > 0
Calculam mai intai x²-x-90 = 0
a = 1 ; b = -1 ; c = -90 ; Δ=b²-4ac = 1-4·(-90) = 361 ; √Δ = √361 = 19
x₁,₂ = (-b±√Δ)/2a = (1±19)/2 => x₁=-9 ; x₂ = 10 =>
x ∈ (-∞ ; -9) ∪ (10 , +∞)
b) 4x²-12x+9 ≤ 0 <=> (2x-3)² ≤ 0 ..... orice patrat perfect este ≥ 0 =>
ramane doar varianta 2x-3 = 0 => x = 3/2
c) x²-6x < 0 <=> x(x-6) < 0 x(x-6) = 0 => x₁ = 0 ; x₂ = 6
x I -∞ 0 6 +∞
x I --------------0+++++++++++++++++
x-6 I---------------------------0++++++++++
x(x-6) I +++++++++0----------0++++++++++
x ∈ (0 ; 6)
d) x²-x+2 > 0 ; x²-x+2 = 0 => Δ = 1-8 < 0 =>
x²-x+2 > 0 pentru (∀) x ∈ R
e) -6x²+5x-1 ≤ 0 <=> 6x²-5x+1 ≥ 0 ; a = 6 ; b = -5 ; c = 1
Δ = 25-24 = 1 => √Δ=1 ; x₁,₂ = (5±1)/12 => x₁ = 1/3 ; x₂ = 1/2
x I -∞ 1/3 1/2 +∞
6x²-5x+1 I ++++++++++++0------0++++++++++ =>
x ∈ (-∞ ; 1/3) ∪ (1/2 ; +∞)
f) 4x²-x+5 ≤ 0 ; 4x²-x+5 = 0 ; a = 4 ; b = -1 ; c = 5 ; Δ = 1-20 < 0 =>
4x²-x+5 > 0 pentru (∀) x ∈ R => solutia x = {Ф}
