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- Feb 14, 2012
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The roots of $x^4-x^3-x^2-1=0$ are $p, q, r, s$. Find $f(p)+f(q)+f(r)+f(s)$, where $f(x)=x^6-x^5-x^3-x^2-x$.
Thanks for participating but I'm sorry, kaliprasad. Your answer is incorrect.X^4 = x^3+x^2 +1
So x^6 = x^5 + x^4 + x^2
So x^6 – x^5 – x^3 – x^2 – x =x^4-x^3 – x = x^2- x + 1
So f(x) = x^2
Sum p = 1 and sum pq = 0
Sum p^2 =( sum p)^2 – 2 sum pq = 1
sum p = 1
so sum p^2 - p + 1 = 1
Thanks anemone. My ans is incorrect. Here is the correct solutionThanks for participating but I'm sorry, kaliprasad. Your answer is incorrect.
Yeah, it's correct now! Well done, kali!Thanks anemone. My ans is incorrect. Here is the correct solution
We are given that(say the function is g)
g(x) = x^4-x^3 –x^2 – 1
As p,q,r,s are 4 roots we have
g(x) = 0 has solution p,q,r,s
Now x^4 – x^3 – x^2 – 1 = 0 ... (1) has solutions p,q,r,s
F(x) = x^6 – x^5 – x^3 – x^2 – x ..(2)
We need to reduce it t the lowest oder polynomial as possible
From (1) x^6 – x^5 = x^4 + x^2 ... (3)
From (2) and (3) F(x) = x^4 + x^2 – x^3 – x^2 – x = (x^3 + x^2 + 1) + x^2 – x^3 – x^2 – x
= x^2 – x + 1
So f(p) + f(q) + f(r) + f(s) = p^2 + q^2 + r^2 + s^2) – (p+q+r+s) + 4 ... (4)
Now as p q r s are roots of x^4 –x^3 – x^2 – x = 0
So using vietas formula
P + q + r + s = 1 ...(5) (
Pq + pr +ps + qr + qs + rs = - 1 ..(6)
Now p^2 + q^2 + r^2 + s^2 = (p+q+r+s)^2 – 2(pq+ Pq + pr +ps + qr + qs + rs) = 1 + 2 = 3 .... (7)
Using (5) and (7) in (4) we get
f(p) + f(q) + f(r) + f(s) = p^2 + q^2 + r^2 + s^2) – (p+q+r+s) + 4 = 3 – 1 + 4 = 6
I hope that solution is correct