$\large f(x)= x^{150} + x^{2} -7x + 1$

Let $x_{1}$ , $x_{2}$ , $x_{3}$ $\ldots$ $x_{150}$ be the roots of the above equation. Define

$\large S = \huge \sum_{i=1}^{150}\dfrac{1}{(1+x_{i})^{2}}$

Find $|S|$ .

The answer is 1982.39.

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Since you tagged this problem with algebra, what would be an algebraic approach?

Bingo!! Same method! Amazing problem and congrats on 150 followers!

Aditya Kumar
- 5 years, 9 months ago

Good. It took me a few seconds only to crack the idea of this problem as I had previously posted a similar problem: An Intriguing Polynomial Problem

Also Shivamani had added a similar problem being inspired my my problem: 400 Followers Problem

I've also linked your problem with mine as a Sister Problem! :D

Satyajit Mohanty
- 5 years, 9 months ago

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but it's little different than yours. :)

Surya Prakash
- 5 years, 9 months ago

I think even transformation of the polynomial helps...

Abhishek Bakshi
- 5 years, 9 months ago

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Yaah! Nice idea +1. but it makes the solution little lengthy

Surya Prakash
- 5 years, 9 months ago

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that's true but its a pass for someone who doesn't want to use or doesn't know calculus..

Abhishek Bakshi
- 5 years, 9 months ago

U really back?? No studies?? :P

Aditya Kumar
- 5 years, 9 months ago

Vieta is all you need.

Billy Sugiarto
- 5 years, 9 months ago

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An algebraic approach would be using the transformations.

Surya Prakash
- 5 years, 9 months ago

Awesome!Brilliant!

Adarsh Kumar
- 5 years, 7 months ago

i solved it using algebra and nice problem!!

Dev Sharma
- 5 years, 5 months ago

This is knowledge.

Lu Chee Ket
- 5 years, 8 months ago

My failure in first attempt was caused by a careless mistake of 7 being taken as 1.

Lu Chee Ket
- 5 years, 8 months ago

i did same

Dev Sharma
- 5 years, 5 months ago

2 Helpful
0 Interesting
0 Brilliant
0 Confused

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Since, $x_{1}, x_{2}, \ldots x_{150}$ are the roots of $f(x)$ .

So, $f(x) = (x-x_{1})(x-x_{2}) \ldots (x-x_{150})$ .

Differentiating it w.r.t $x$ and dividing it with $f(x)$ . We get

$\dfrac{f'(x)}{f(x)} = \sum_{i=1}^{150} \dfrac{1}{x-x_{i}}$

Differentiating again w.r.t $x$ . We get,

$\dfrac{(f'(x))^{2} - f(x)f''(x)}{f^2 (x)} = - \sum_{i=1}^{150}\dfrac{1}{\left(x -x_{i}\right)^{2}}$

Taking $x = -1$ , we get, $\dfrac{(f'(-1))^{2} - f(-1)f''(-1)}{f^2 (-1)} = - \sum_{i=1}^{150}\dfrac{1}{\left(-1 -x_{i}\right)^{2}}$

But we have $f(-1) = 10$ , $f'(-1) = -159$ and $f''(-1) = 22352$ . Substituting and solving them we get,

$S = \sum_{i=1}^{150}\dfrac{1}{\left(1 +x_{i}\right)^{2}} = -1982.39$

So, $|S| = \boxed{1982.39}$ .