please help me to show that the equation X^n +aX+c=0 can not have more than 3 reals solutions


Question and Answers Forum

All Questions Topic List
Permutation and Combination Questions
Previous in All Question Next in All Question
Previous in Permutation and Combination Next in Permutation and Combination
Question Number 106815 by pticantor last updated on 07/Aug/20

$p l e a s e h e l p m e t o s h o w$ $t h a t t h e e q u a t i o n$ $X^{n} + a X + c = 0 c a n n o t h a v e$ $m o r e t h a n 3 r e a l s s o l u t i o n s$
	Answered by 1549442205PVT last updated on 07/Aug/20

	$f (X) = X^{n} + aX + c \Rightarrow f^{'} (X) = {nX}^{n - 1} + a (1)$ $i) The n odd \Rightarrow n = 2 k + 1$ $(1) \Leftrightarrow (2 k + 1) X^{2 k} = - a (2)$ $+ If a > 0 then (2) has no roots$ $\Rightarrow f (X) has one unique root since f (X)$ $is increasing function$ $+ If a = 0 \Rightarrow f (X) has only root x =^{2 k + 1} \sqrt{- c}$ $+ If a < 0 then (2) has two roots$ $X = \pm^{2 k} \sqrt{\frac{- a}{2 k + 1}}$ $\Rightarrow f (X) has no more three roots$ $ii) The n even \Rightarrow n = 2 k . Then$ $(1) \Leftrightarrow {2 kX}^{2 k - 1} = - a \Leftrightarrow X =^{2 k - 1} \sqrt{\frac{- a}{2 k}}$ $is the unique root$ $\Rightarrow f (X) has no more two roots$ $Thus, in all cases we see that f (X) has$ $no more three solutions (q . e . d)$ $Note : Here we need to understand that$ $the roots of the eqn . X^{n} + aX + c = 0 that$ $we are mentioning be real roots$ $because if not as we were known$ $a n arbitrary polynomial of degree n$ $being always has n roots in C$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum