Question

cine mă ajuta la exercițiul asta?​

Answer 1

a) x²-6x+5= 0

∆= b²-4ac => ∆= (-6)²-4×1×5 => ∆= 36-20 => ∆= 16 => √∆= 4

x1= (-b+√∆)/2a => x1= (6+4)/2 => x1= 10/2 => x1= 5

x2= (-b-√∆)/2a => x2= (6-4)/2 => x2= 2/2 => x2= 1

b) x²+4x+4= 0

∆= b²-4ac => ∆= 4²-4×1×4 => ∆= 16-16 => ∆= 0

x1= x2= -b/2a => x= -4/2 => x= -2

c) x²-12x+27= 0

∆= b²-4ac => ∆= (-12)²-4×1×27 => ∆= 144-108 => ∆= 36 => √∆= 6

x1= (-b+√∆)/2a => x1= (12+6)/2 => x1= 18/2 => x1= 9

x2= (-b-√∆)/2a => x2= (12-6)/2 => x2= 6/2 => x2= 3

------------------ Formule ------------------

→ ∆= b²-4ac

→ x1, x2= (-b±√∆)/2a