Respostas

2013-12-11T11:33:43-02:00
(x-2)² + a(3a+2)
(x-2) . (x-2) + 3a² + 2a
x² - 2x - 2x + 4 + 3a² + 2a
x² - 4x + 4 + 3a² + 2a


1 2 1
2013-12-11T11:35:42-02:00
(x-2)²+a(3a+2)
x²+2.x.-2+(-2)²+a(3a+2)
x²-4x+4+3a²+2a
x²-4x+4
Δ = b² -4ac
Δ = (-4)² -4 *1 * 4
Δ = 16 -16
Δ = 0
x' = (-b + √∆)/ 2a
(-(-4) + 0)/2 * 1
(4 + 0)/2
4/2= 2
x'' = (-b - √∆)/ 2a
(-(-4) - 0)/2 * 1
(4 - 0)/2
4/2= 2

S={2,2}

Δ = b² -4ac
Δ = 2² -4 *3 * 0
Δ = 4 - 0
Δ = 4
x' = (-b + √∆)/ 2a
(-2 + 2)/2 * 3
(-2 + 2)/6
0/6= 0
x'' = (-b - √∆)/ 2a
(-2 - 2)/2 * 3
(-2 - 2)/6
-4/6

S={0}

Somamos ambas raízes:

2+0 = 2 
Resposta: 2.