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Let f(x) = x^2 + 2x. Find all real numbers x such that f(x) = f(f(x)).

 Mar 3, 2021
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f(f(x))=(f(x))^2+2f(x)=x^4+4x^3+6x^2+4x

 

we want to find all numbers where

x^2+2x=x^4+4x^3+6x^2+4x

 or where

x^4+4x^3+5x^2+2x=0=x(x^3+4x^2+5x+2)

we can divide both sides by x, and get

x^3+4x^2+5x+2=0

we can then write

(x^3+3x^2+2x)+1(x^2+3x+2)=x(x^2+3x^2+2x)+1(x^2+3x+2)=(x+1)(x+2)(x+1)

and we know that 0 is also a solution, so all the real numbers possible are

$\boxed{-2,-1,0}$

 Mar 3, 2021

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