Question

Como resolvo ..
y²-4y+3=0 ?

Answer 1

Olá, Leticiellenbitten!

y² - 4y + 3 = 0
a = 1
b = -4
c = 3

Δ = (-4)² - 4.1.3
Δ = 16 - 12
Δ = 4

-b +- √Δ / 2.a
-(-4) +- √4 / 2.1
4 +- 2 / 2

x' = 4-2/2=2/2=1
x'' = 4+2/2=6/2=3

Answer 2

Identificando os termos a = 1 ; b = - 4 e c= 3

Δ = b² - 4.a.c
Δ = (- 4)² - 4 . 1 . 3
Δ = 16 - 12
Δ = 4

daí
 [tex] x_{1}  =( - b ÷ √Δ) / 2a  =( -(-4) + √4)/2 .1 = (4 + 2) / 2 = 6/ 2 = 3
 [tex] x_{2}  =( - b - √Δ) / 2a  =( -(-4) - √4)/2 .1 = (4 - 2) / 2 = 2/ 2 = 1

S = { 1 ; 3}