Răspuns:
pentru a rezolva ecuatiile de gradul 2 aplicam formula
x₁ ₂ = [- b ± √(b² - 4ac)]/2a
3x² - 2x + 1 = 0
a = 3 ; b = -2 ; c = 1
x₁ ₂ = (2 ± √(4 - 4·3·1)/2·3
= (2 ± √(4 - 12)/6
= (2 ± √-8)/5
x ∉ R deoarece avem √-8
-4x² + 9x - 2 = 0
4x² - 9x + 2 = 0
a = 4 ; b = - 9 ; c = 2
x₁ ₂ = (9 ± √(-9)² - 4·4·2)/2·4
= (9 ± √49)/8
= (9 ± 7)/8
x₁ = (9+7)/8 = 16/8 = 2
x₂ = (9-7)/8 = 2/8 = 1/4
4x² - 2x - 3 = 0
a = 4 ; b = -2 ; c = -3
x₁ ₂ = (2 ± √4+48)/8
= (2 ± √52)/8
= (2 ± 2√13)/8
= 2(1 ± √13)/8
= (1 ± √13)/4
x₁ = (1 + √13)/4
x₂ = (1 - √13)/4
ti-am facut 3 exemple, cu celelalte 2 exersezi singur