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Si:
P(x) = (a+b) x2+ (b–c)x + c–3
Por lo tanto:
c - 3 = 0 => c = 3
Además:
b - c = 0
b - (3) = 0
b = 3
Y:
a + b = 0
a + (3) = 0
a = -3
Entonces: a = -3, b = 3 y c = 3
Reemplazando los valores:
R = (b - a ) / c
R = (3 - (-3) ) / 3
R = (3 + 3 ) / 3
R = 6 / 3
R = 2