## ParthKohli one year ago Where could I possibly have gone wrong?!

1. ParthKohli

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:

2. ParthKohli

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 $$(x-1)(x+2)$$.

3. ParthKohli

@ganeshie8

4. ParthKohli

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

5. ganeshie8

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

6. ganeshie8

try @dan815

7. ganeshie8

@surjithayer

8. ParthKohli

@Miracrown

9. ParthKohli

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

10. ParthKohli

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

11. ParthKohli

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$

12. ParthKohli

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

13. ParthKohli

Thanks guys - I'm done.

14. dan815

good job :)

15. ganeshie8

Nice!