SOLUTION: Let {{{f(x) = x^2-3x}}}. For what values of x is {{{f(f(x)) = f(x)}}}? Enter all the solutions, separated by commas.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Let {{{f(x) = x^2-3x}}}. For what values of x is {{{f(f(x)) = f(x)}}}? Enter all the solutions, separated by commas.      Log On


   

Question 1156960: Let f%28x%29+=+x%5E2-3x. For what values of x is f%28f%28x%29%29+=+f%28x%29? Enter all the solutions, separated by commas.
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39629) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%5E2-3x%29%5E2-3%28x%5E2-3x%29=x%5E2-3x
.
.
x%28x%5E3-6x%5E2%2B5x%2B12%29=0
.
.
x%28x%2B1%29%28x-3%29%28x-4%29=0

Values which should work are -1, 0, 3, 4.

Answer by greenestamps(13208) About Me  (Show Source):
You can put this solution on YOUR website!


The response from the other tutor makes solving the problem harder by multiplying polynomials when not necessary.

x%5E2-3x+=+%28x%5E2-3x%29%5E2-3%28x%5E2-3x%29

Treat the x%5E2-3x as a unit (a new "variable"):

0+=+%28x%5E2-3x%29%5E2-4%28x%5E2-3x%29

Factor out one occurrence of the "variable" from each term:

0+=+%28x%5E2-3x%29%28%28x%5E2-3x%29-4%29
0+=+%28x%5E2-3x%29%28x%5E2-3x-4%29

Now solve by factoring each quadratic.

0+=+%28x%29%28x-3%29%28x-4%29%28x%2B1%29

ANSWERS: x=0; x=3; x=4; x=-1