Vamos calcular os valores f(1), f(2) e f(3), temos:
f(x) = x² + bx + c
f(1) = -1
f(1) = 1^2 + b.1 +c → f(1) = 1 + b + c → 1 + b + c = -1
b + c = -1-1 → b + c = -2
f(2) = 2^2 + b.2 +c → f(2) = 4 + 2b + c
f(3) = 3² + b3 + c → f(3) = 9 + 3b + c
Substituindo os valores encontrados, temos:
f(2) - f(3) = 1
4 + 2b + c - 9 + 3b + c = 1
5b + 2c = 1 + 5 → 5b + 2c = 6
Resolvendo o sistema, temos: ( método da substituição )
b + c = -2 → b = -2-c
5b + 2c = 6
5.(-2-c) + 2c = 6
-10-5c + 2c = 6
-10-(-5+2)c = 6
-3c = 6 + 10
C = - 16/3
b = -2-(-16/3)
b = -6+16/3
b = 10/3
f(x) = x² + 10/3x -16/3
f(x) = 0
x² + 10/3x -16/3 = 0
∆ = b^2 - 4.a.c
∆ = (10/3)^2 - 4.1.(-16/3)
∆ = 100/9 + 64/3 = 100+192/9 = 292/9
√∆ = √292/√9 = 2√73/3
x = -10/3±2√73/3/2.1
x1 = -10/3 - 2√73/3/2 = -10 -2√73/6 ( : 2)
x2 = -10/3 + 2√73/3/2 = -10+2√73/6 ( : 2)
x1 = -10 -2√73/6 ( : 2) = -5-√73/3 = -4,51467
x2 = -10+2√73/6 ( : 2) = -5+√73/3 = 1,18133
Resp.: Alternativa d. Aproximadamente -5.
Espero ter ajudado!
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Vamos calcular os valores f(1), f(2) e f(3), temos:
f(x) = x² + bx + c
f(1) = -1
f(1) = 1^2 + b.1 +c → f(1) = 1 + b + c → 1 + b + c = -1
b + c = -1-1 → b + c = -2
f(2) = 2^2 + b.2 +c → f(2) = 4 + 2b + c
f(3) = 3² + b3 + c → f(3) = 9 + 3b + c
Substituindo os valores encontrados, temos:
f(2) - f(3) = 1
4 + 2b + c - 9 + 3b + c = 1
5b + 2c = 1 + 5 → 5b + 2c = 6
Resolvendo o sistema, temos: ( método da substituição )
b + c = -2 → b = -2-c
5b + 2c = 6
5.(-2-c) + 2c = 6
-10-5c + 2c = 6
-10-(-5+2)c = 6
-3c = 6 + 10
C = - 16/3
b = -2-(-16/3)
b = -6+16/3
b = 10/3
f(x) = x² + 10/3x -16/3
f(x) = 0
x² + 10/3x -16/3 = 0
∆ = b^2 - 4.a.c
∆ = (10/3)^2 - 4.1.(-16/3)
∆ = 100/9 + 64/3 = 100+192/9 = 292/9
√∆ = √292/√9 = 2√73/3
x = -10/3±2√73/3/2.1
x1 = -10/3 - 2√73/3/2 = -10 -2√73/6 ( : 2)
x2 = -10/3 + 2√73/3/2 = -10+2√73/6 ( : 2)
x1 = -10 -2√73/6 ( : 2) = -5-√73/3 = -4,51467
x2 = -10+2√73/6 ( : 2) = -5+√73/3 = 1,18133
Resp.: Alternativa d. Aproximadamente -5.
Espero ter ajudado!