Where could I possibly have gone wrong?!

Question

parthkohli · Answer

Consider all monic quadratic polynomials $f(x)$ satisfying the following relation:$$f(f(-1)) = f(f(0)) = 0$$Find the sum of $f(3)$ across all such polynomials. 

Here's my foolproof solution:

parthkohli · Answer

OK, so let $f(x) = x^2 + bx + c $ for a second. Now $f(0) = c \Rightarrow f(f(0)) = f(c) = 0$. This means that $c$ is a root of this quadratic! Now the other root must be $1$ so that the product of roots becomes $c$.

Therefore, so far, $f(x) = (x-1)(x-c)$.

We use the second relation to find $c$.

As you can clearly see, $f(-1)  = 2c + 2$ meaning that $f(2c + 2) = 0$. Thus, $2c+2$ is also a root of the polynomial - but $2c+2$ must either be $c$ or $1$. We consider both possibilities:

1. $2c + 2  = 1\Rightarrow c = -1/2$
2. $2c  + 2 = c\Rightarrow c = -2$

Thus, the two possible quadratics are $(x-1)(x+1/2)$ and $(x-1)(x+2)$.

parthkohli · Answer

@ganeshie8

parthkohli · Answer

The answer comes out to be $2\cdot 3.5 + 2\cdot 5 = \boxed{17}$. Which isn't the answer.

ganeshie8 · Answer

it looks very good to me! not able to find any mistakes

ganeshie8 · Answer

try @dan815

ganeshie8 · Answer

@surjithayer

parthkohli · Answer

@Miracrown

parthkohli · Answer

$$f(x) = x^2 + x$$also seems to satisfy this.

parthkohli · Answer

I didn't consider the case $c=0$ :)

parthkohli · Answer

If $c=0$, then the polynomial is in the form $f(x) = x^2 + bx$ so $f(0) = 0$ and so $f(f(-1))=0 \Rightarrow f(1-b) = 0 \Rightarrow (1-b)^2 + b(1 - b)  = 0 $.$$\Rightarrow b^2 - 2b + 1 + b - b^2 = 0$$$$\Rightarrow b = 1$$

parthkohli · Answer

$$f(3) = 9 + 3 = 12 $$so$$s = 29$$

parthkohli · Answer

Thanks guys - I'm done.

dan815 · Answer

good job :)

ganeshie8 · Answer

Nice!