3. Consideram notatia lg(x) = t, unde t ∈R ,iar prin inlocuire in ecuatia data obtinem
t² +t = 6 <=> t² +t -6 = 0 ⇒ Δ = b² - 4·a·c = 1² -4·1·(-6) = 1-(-24) = 25 > 0, deci avem doua solutii numere reale
t₁ = (-1 -5)/2 = -3 respectiv t₂ = (-1 +5)/2 = 2 => lg(x) ∈{-3;2} <=> x ∈{1/10³;100}
S ={1/10³;100}.
2. x² -m·x +1 -m = 0 ⇒ Δ = b² -4·a·c = (-m)² -4·1·(1-m) = m² -4·(1-m) = m² -4 +4m, iar ecuatia data are doua solutii reale distincte daca si numai daca Δ > 0 <=> m² -4 +4m > 0 => m² +4m +4 -8 > 0 <=> (m+2)² -8 > 0 <=> (m+2)² > 8 <=> m+2 ∈(-∞,-2√2) ∪ (2√2,+∞) <=> m ∈(-∞,-2√2 -2) ∪ (2√2 -2,+∞).
