f(x) = ax² + bx + c monte um sistema: f(0) = c = 5 (a million) f(a million) = a + b + c = 3 (2) f(-a million) = a - b + c = 9 (3) a + b = 3 - 5 = -2 a - b = 9 - 5 = 4 2a = 2 a = a million b = a - 4 = a million - 4 = -3 f(x) = x² -3x + 5 f(2) = 4 - 6 + 5 = 3 alternativa c)
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Verified answer
{ x - 2y = 0 (-1)
{x + y² = 35
{-x + 2y = 0
{ x + y² = 35
------------------
//////y² + 2y = 35
y² + 2y - 35 = 0
delta--> 4+140=144
Vdelta---> V144 = 12
y' = (-2+12)/2 = 10/2 = 5
y"= (-2-12)/2 = -14/2 = -7
x - 2y = 0
x= 2.5 = 10
x= 2.(-7) = -14
Resposta:
para x = 10 , y = 5
para x = -14 , y = -7
f(x) = ax² + bx + c monte um sistema: f(0) = c = 5 (a million) f(a million) = a + b + c = 3 (2) f(-a million) = a - b + c = 9 (3) a + b = 3 - 5 = -2 a - b = 9 - 5 = 4 2a = 2 a = a million b = a - 4 = a million - 4 = -3 f(x) = x² -3x + 5 f(2) = 4 - 6 + 5 = 3 alternativa c)
x = 2y
2y + y² = 35
y² + 2y - 35 = 0
d² = 4 + 140 = 144
d = 12
y1 = (-2 + 12)/2 = 5 --> x1 = 2y1 = 10
y2 = (-2 - 12)/2 = -7 --> x2 = 2y2 = -14
pronto
x+y² = 35
x-2y = 0
um menos o outro
y² + 2y - 35 = 0
y' = -7
y = 5
x-2.(5) = 0
x - 10 = 0
x' = 10
x-2.(-7) = 0
x'' = -14