Question

The number of Real distinct roots for the equation

7x^7 + 6x^6 - 3x^5 + 6x^4 +3x^2 + x - 1 = 0

Answer 1

Number of sign changes in f(x) + f(-x). I guess?
@shubhamsrg

Answer 2

descartes' only tells us possible no. of +ve and -ve roots, we have to find no. of real roots here.
that might not work.
and we don't add f(x) and f(-x) ?
we see the no. of alternate sign changes in each.

Answer 3

@electrokid @hartnn @satellite73

Answer 4

No i mean number of sign changes in f(x) + number of sign changes in f(-x). It wont work anyway. :/

Answer 5

using remainder theorem, see if \(\pm\Large{1\over7}\) is a root

Answer 6

No they arent @electrokid

Answer 7

trick there are "7" zeros in all, including complex let there be np positive real zeros and nn negative real zeros and 2nc complex roots (since they always occur in conjugate pairs). using Descartes rule of signs, we have
\[n_p=3-2k\\n_n=2-2l\\\;\\
n_p+n_n+2n_c=7\\3-2k+2-2l+2n_c=7\\
2n_c-2k-2l=2\\n_c-k-l=1\\k+l=n_c-1\\\;\\\quad\text{(number of real roots)}\\
n_r=n_p+n_n=5-2(k+l)=5-2(n_c-1)\]
now, \(n_c>0,k>0,l>0,n_p\ge1,n_n\ge0\)
possible values for \(n_r\)
\[
n_c=0\quad n_r=5\quad\text{not possible}\\
n_c=1\quad n_r=5\quad\text{possible, 3positive, 2 negative}\\
n_c=2\quad n_r=3\quad\text{possible, 1positive and 2 negative}\\
n_c=3\quad n_r=1\quad\text{possible, one positive}
\]

Answer 8

WoW Thats great thxxx...;)