f ( x ) = x 1 5 0 + x 2 − 7 x + 1
Let x 1 , x 2 , x 3 … x 1 5 0 be the roots of the above equation. Define
S = i = 1 ∑ 1 5 0 ( 1 + x i ) 2 1
Find ∣ S ∣ .
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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!
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
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but it's little different than yours. :)
I think even transformation of the polynomial helps...
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Yaah! Nice idea +1. but it makes the solution little lengthy
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that's true but its a pass for someone who doesn't want to use or doesn't know calculus..
U really back?? No studies?? :P
Vieta is all you need.
Since you tagged this problem with algebra (as opposed to calculus), what would be an algebraic approach?
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An algebraic approach would be using the transformations.
Awesome!Brilliant!
i solved it using algebra and nice problem!!
This is knowledge.
My failure in first attempt was caused by a careless mistake of 7 being taken as 1.
i did same
let t=1/1+x =>x=(1-t)/t sub x=(1-t)/t in f(x)=o we get, (1-t)^{150}+t^{148}*(1-t)^{2}-7t^{149}(1-t)+t^{150}=0 we get binomial expansion given the pic.
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Since, x 1 , x 2 , … x 1 5 0 are the roots of f ( x ) .
So, f ( x ) = ( x − x 1 ) ( x − x 2 ) … ( x − x 1 5 0 ) .
Differentiating it w.r.t x and dividing it with f ( x ) . We get
f ( x ) f ′ ( x ) = i = 1 ∑ 1 5 0 x − x i 1
Differentiating again w.r.t x . We get,
f 2 ( x ) ( f ′ ( x ) ) 2 − f ( x ) f ′ ′ ( x ) = − i = 1 ∑ 1 5 0 ( x − x i ) 2 1
Taking x = − 1 , we get, f 2 ( − 1 ) ( f ′ ( − 1 ) ) 2 − f ( − 1 ) f ′ ′ ( − 1 ) = − i = 1 ∑ 1 5 0 ( − 1 − x i ) 2 1
But we have f ( − 1 ) = 1 0 , f ′ ( − 1 ) = − 1 5 9 and f ′ ′ ( − 1 ) = 2 2 3 5 2 . Substituting and solving them we get,
S = i = 1 ∑ 1 5 0 ( 1 + x i ) 2 1 = − 1 9 8 2 . 3 9
So, ∣ S ∣ = 1 9 8 2 . 3 9 .