Question

vamos resolver, no conjunto IR, as seguintes equaçoes:
a) x² - 2x= 2x - 4
b) x² - 2x = x + 4
x² + 12X + 36= 0

Answer 1

a) x^2-2x=2x-4
x^2-2x-2x+4
x^2-4x+4=0
delta=b^2-4ac
delta=(-4)^2-4.1.4
delta=16-16
delta=0
x=-b-/+ raiz de delta/2a
x=-(-4)-0/2.1
x=4-0/2
x=4/2
x=2
S={2}
b) x^2-2x=x+4
x^2-2x-x-4
x^2-3x-4=0
delta=b^2-4ac
delta=(-3)^2-4.1.(-4)
delta=9+16
delta=25
x=-b-/+ raiz de delta/2a
x=-(-3)-5/2.1
x=3-5/2
x=-2/2
x=-1
x=-(-3)+5/2.1
x=3+5/2
x=8/2
x=4
S={-1, 4}
c) x^2+12x+36=0
delta=b^2-4ac
delta=12^2-4.1.36
delta=144-144
delta=0
x=-b-/+ raiz de delta/2a
x=-12-0/2.1
x=-12/2
x=-6
S={-6}

Answer 2

a) x² - 2x = 2x - 4
x² - 2x - 2x + 4 = 0
x² - 4x + 4 = 0
Δ= (-4)² - 4 . 1 . 4 
Δ= 16 - 16
Δ= 0
x= $\frac{-(-4)+- \sqrt{0} }{2.1}$
x= $\frac{4+-0}{2}$
x'= $\frac{4+0}{2}$
x'= $\frac{4}{2}$
x'=2
x''= $\frac{4-0}{2}$
x''= $\frac{4}{2}$
x''= 2
S: {2}

b)  x² - 2x = x + 4
   x² - 2x - x - 4 = 0
   x² - 3x - 4 = 0
Δ= (-3)² - 4 . 1 . (-4)
Δ= 9 + 16
Δ= 25
x= $\frac{-(-3)+- \sqrt{25} }{2 .1}$
x= $\frac{3+-5}{2}$
x'= $\frac{3+5}{2}$
x'= $\frac{8}{2}$
x'=4
x''= $\frac{3-5}{2}$
x''= $\frac{-2}{2}$
x''= -1
S: {3,-1)

c) x² + 12x + 36= 0
Δ= 12² - 4 . 1 . 36)
Δ= 144 - 144
Δ= 0 
x= $\frac{-12)+- \sqrt{0} }{2.1}$
x= $\frac{-12+-0}{2}$
x'= $\frac{-12+0}{2}$
x'= $\frac{-12}{2}$
x'= -6
x''= $\frac{-12-0}{2}$
x''= $\frac{-12}{2}$
x''= -6
S: {-6}

Espero que tenha ajudado :)