Question

3r ²+1=4r
4a2=2a+3

método general porfavor

Answer 1

Para estos problemas utilizaremos la formula general para ecuaciones de 2° grado

[-b±√(b²-4ac)]/2a

1)
3r²+1 = 4r
3r²-4r+1 = 0

a = 3
b = -4
c = 1

Reemplazando en la formula

[4±√(16-12)]/6
[4±√4]/6
[4±2]/6

r1 = (4+2)/6 = 6/6 => 1
r2 = (4-2)/6 = 2/6 => 1/3

2)
4a² = 2a+3
4a²-2a-3 = 0

a = 4
b = -2
c = -3

Reemplazamos en la formula

[2±√(4+48)]/8
[2±√52]/8
[2±2√13]/8
[1+√13]/4

a1 => (1+√13)/4
a2 => (1-√13]/4

Saludos Ariel