Sum of Squared Roots

Since $f(x)$ is monic and equal to $x^9 - 3x^8 + 17x$ for $x \in \{1, 2, ... , 9, 10\}$ , we know:

$f(x) - x^9 + 3x^8 - 17x = (x-1)(x-2)...(x-10)$

We can see then that both sides of the equation are a 10th-degree polynomial, therefore:

$f(x) = (x-1)(x-2)...(x-10) + x^9 - 3x^8 +17x$

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Since we now know $f(x)$ , we can use vieta to find the sum of its squared roots:

$r_1^2 + r_2^2 + ... + r_{10}^2 =(r_1 + r_2 + ... + r_{10})^2 - 2(r_1r_2 + r_1r_3 + ... + r_9r_{10})$

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We can see in order to find the sum of the squared roots, we need to know the coefficient of the $x^9$ term and the $x^8$ term.

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$x^9$ term of $(x-1)(x-2)...(x-10)$ :

$\displaystyle\sum_{i = 1}^{10} -i = -55$

Adding the lone $x^9$ :

$-55x^9 + x^9 = -54x^9$

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$x^8$ term of $(x-1)(x-2)...(x-10)$ :

$\hspace{10pt}1(2 + 3 + ... + 10)$

$+$ $2(3 + 4 + ... + 10)$

$+$ $3(4 + 5 + ... + 10)$

$\hspace{40pt} \vdots$

$+$ $9(10)$

$= \displaystyle\sum_{i = 1}^{10} i(55 - \frac{i^2 + i}{2})$

$= \displaystyle\sum_{i = 1}^{10} 55i - \frac{1}{2}i^2 - \frac{1}{2}i^3$

$= \frac{(55)(10)(11)}{2} - \frac{(10)(11)(21)}{12} - \frac{(10)^2(11)^2}{8}$

$= 3025 - 192.5 - 1512.5$

$= 1320$

Adding the $-3x^8$ :

$1320x^8 - 3x^8 = 1317x^8$

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Therefore,

$\displaystyle\sum_{i = 1}^{10} r_i^2 = (-54)^2 - 2\cdot1317 = \fbox{282}$

We note that

$\begin{aligned} f(x) - g(x) & = (x-1)(x-2)(x-3) \cdots (x-10) \\ f(x) - x^9 + 3x^8 - 17x & = x^{10} - 55x^9 + 1320x^8 - \cdots + \prod_{k=1}^{10} k \\ \implies f(x) & = x^{10} - \blue{54}x^9 + \red{1317} x^8 - \cdots + \prod_{k=1}^{10} k \end{aligned}$

By Vieta's formula , $\displaystyle \sum_{i=1}^{10} r_i = \blue{54}$ and $\displaystyle \sum_{i \ne j}^{10} r_ir_j = \red{1317}$ and by Newton's sums or identities ,

$\sum_{i=1}^{10} r_i^2 = \left(\sum_{i=1}^{10} r_i \right)^2 - 2 \sum_{i \ne j}^{10} r_i r_j = 54^2 - 2\times 1317 = \boxed{282}$