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Find all real numbers x such that f(x) = f(f(x)) + 16x + 5, where f(x) = x^2 - 3x + 1.

 Jan 18, 2024
 #1
avatar+128772 
+1

f(x)  = f(f(x)) + 16x + 5

 

x^2  - 3x + 1 =   [ x^2 - 3x + 1]^2 -3[ x^2 - 3x + 1] + 1 + 16x + 5 

 

x^2 - 3x + 1 =  x^4 - 6 x^3 + 11 x^2 - 6 x + 1 -3x^2 + 9x - 3 + 1 + 16x + 5

 

x^4 -6x^3 + 7x^2+ 22x -3  =  0

 

By graph, the solutions  for  x are

 

x ≈ -1.2305 

 

x ≈  -0.14377

 

 

cool cool cool

 Jan 18, 2024

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