The only real root of x^3 - 1999x^2+x-1999=0 is 1999. What is the only real root of (x-1)^3-1999(x-1)^2 +(x-1)-1999=0?
A. 1
B. 1998
C. 1999
D. 2000
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If 1999 is a root of
x^3 - 1999x^2+x-1999=0
then a root of
(x-1)^3-1999(x-1)^2 +(x-1)-1999=0
is a value of x so that
x-1 = 1999
In other words, x=2000
To find the only real root of the equation (x-1)^3 - 1999(x-1)^2 + (x-1) - 1999 = 0, we can use a simple algebraic transformation.
Step 1: Simplify the equation.
Expand (x-1)^3 and (x-1)^2:
(x^3 - 3x^2 + 3x - 1) - 1999(x^2 - 2x + 1) + (x - 1) - 1999 = 0
Simplify further:
x^3 - 3x^2 + 3x - 1 - 1999x^2 + 3998x - 1999 + x - 1 - 1999 = 0
Combine like terms:
x^3 - (3x^2 + 1999x^2) + (3x + 3998x + x) - (1 + 1999 + 1 + 1999) = 0
Simplify again:
x^3 - 2002x^2 + 4002x - 4000 = 0
Step 2: Apply the given information.
Based on the given information, we know that the only real root of the equation x^3 - 1999x^2 + x - 1999 = 0 is 1999. So, we need to determine the value of x that corresponds to this root.
Since we have transformed the original equation (x-1)^3 - 1999(x-1)^2 + (x-1) - 1999 = 0 to x^3 - 2002x^2 + 4002x - 4000 = 0, we can conclude that:
x = 1999 + 1 = 2000
Therefore, the only real root of the equation (x-1)^3 - 1999(x-1)^2 + (x-1) - 1999 = 0 is x = 2000.
Hence, the correct answer is D) 2000.