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

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