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# Weird Question

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Let $$f(x)=(x^2+6x+9)^{50}-4x+3$$, and let $$r_1,r_2,\ldots,r_{100}$$be the roots of $$f(x).$$

Compute $$(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}.$$

-------Thanks!

Apr 1, 2020

### 6+0 Answers

#1
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I see you are on the Vieta's Formulas Week of Intermediate Algebra right now. This hint has probably already been given, but do you notice anything interesting about the first term of the function. Namely, does it remind you of a certain (a+b)^2?

P.S. Ask in the AoPS forum! Their advice will be way better than mine.

This is a very, very interesting problem, and I don't want to spoil the answer for you. Probably one of my favorites so far.

The reason I say it is because the solution is very simple and elegant. Also, this is one of the few writing problems that may be possible by yourself.

Apr 1, 2020
edited by Impasta  Apr 1, 2020
edited by Impasta  Apr 1, 2020
edited by Impasta  Apr 1, 2020
#2
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Are you still there, or did you find the answer somewhere else?

Impasta  Apr 1, 2020
#3
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Wow, to think I typed all that up for nothing :(

Impasta  Apr 1, 2020
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Umm... I got the (a+b)^2, its (x+3)^2, how would I continue Impasta. Or someone else?

madyl  Apr 1, 2020
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LOL. I don't think it was for nothing though...

I liked it very much!

Apr 1, 2020
#6
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Let $$f(x)=(x^2+6x+9)^{50}-4x+3$$, and let $$r_1,\ r_2,\ \ldots,\ r_{100}$$ be the roots of $$f(x)$$.
Compute $$(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$$.

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array}$$

$$\mathbf{\text{vieta:}}$$
For any polynomial equation
$$0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0$$
with the solutions $$r_1\dots r_n$$, the relatively simple formulas for $$a_0$$ and $$a_{n-1}$$ are:
$$a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k$$

$$\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array}$$

Apr 1, 2020
edited by heureka  Apr 1, 2020