Question

Sa se rezolve ecuatia :
$2x^{2} +3=5IxI$

Answer 1

|x| -> x daca x>=0
       -x daca x<0

Asadar, presupunem ca x>=0 si obtinem:
2x² + 3 = 5 * x
2x² - 5x + 3 = 0
Rezolvam cu Delta Δ.
Δ=b²-4*a*c
Δ=(-5)²-4*2*3
Δ=25-24 ⇒ Δ=1 si √Δ=√1=1

x1 = (-b+√Δ)/2a = [-(-5)+1]/4 = (5+1)/4 = 6/4
x2 = (-b-√Δ)/2a = (5-1)/4 = 4/4 = 1

Acum rezolvam pentru x<0

2x² + 3 = 5 * (-x)
2x² + 5x + 3 = 0

Δ=b²-4*a*c
Δ=5²-4*2*3
Δ=25-24 ⇒ Δ=1 si √Δ=√1=1

x1 = (-b+√Δ)/2a = [-5+1]/4 = (-4)/4 = -1
x2 = (-b-√Δ)/2a = (-5-1)/4 = -6/4