Question

(x+2).(x-6)=0   obter as raizes reais das seguintes equaçoes

Answer 1

(x + 2)·(x-6) = 0
x² - 6x + 2x - 12 = 0
x² - 4x - 12 = 0
Δ = b² - 4.a.c
Δ = (-4)² -4.1.(-12)
Δ =16 + 48
Δ = 64
Δ = √64 = 8

x, = -b + √Δ / 2a
x, = 4 + 8 / 2.1
x, = 12 / 2
x, = 6

x,, = 4 - 8 / 2
x,, = -4 / 2
x,, = -2

Raízes:
S = {6, -2}
      

Answer 2

(x+2)(x-6)=$x^{2}$-6x+2x-12
$x^{2}$+2x-18
Δ=$(-2)^{2}$±$\frac{ \sqrt{2+4.1.18} }{2}$
X=$\frac{-4+ \sqrt{36} }{2}$
X=$\frac{-4+6}{2} =1$
X"=$\frac{-4-6}{2} =-5 \\ \left \{ {{x=1} \atop {x"=-5}} \right.$