1-Show-that-the-polynomial-P-x-x-n-ax-b-n-2-a-b-R-2-has-at-most-3-real-distinct-roots-2-Show-that-the-equation-x-4-4ax-b-0-has-no-more-than-2-real-distinct-roots-

Question Number 105340 by Ar Brandon last updated on 27/Jul/20

1 ∖ Show that the polynomial, P (x) = x^{n} + ax + b, n ⩾ 2

(a, b) \in R^{2} has at most 3 real distinct roots .

2 ∖ Show that the equation x^{4} + 4 ax + b = 0 has no

more than 2 real distinct roots .

Answered by LifeOnMars last updated on 28/Jul/20

P^{'} (x) = 0

{n x}^{n - 1} + a = 0

(1) n = 2 k \Rightarrow 2 {k x}^{2 k - 1} + a = 0 \Leftrightarrow x = - \sqrt[2 k - 1]{\frac{a}{2 k}} \Rightarrow

P (x) h a s 1 e x t r e m e

(2) n = 2 k + 1 \Rightarrow (2 k + 1) x^{2 k} + a = 0 \Leftrightarrow x = \pm \sqrt[2 k]{- \frac{a}{2 k + 1}} \Rightarrow

P (x) h a s {\begin{cases} 2 e x t r e m e s; a < 0 \\ 1 e x t r e m e; a = 0 \\ n o e x t r e m e; a > 0 \end{cases}

\Rightarrow

P (x) h a s a t m o s t 2 e x t r e m e s t h u s 3

d i s t i n c t z e r o s

t h e r e s t s h o u l d b e e a s y

Commented by LifeOnMars last updated on 28/Jul/20

b t w I h a t e t h i s f o r u m y o u c a n n o t

s o l v e t h e e a s i e s t p r o b l e m s

Commented by abdomsup last updated on 28/Jul/20

s e a r c h a n o t h e r w h a t d o y o u t h i n k

y o u a r e \dots ?

Commented by Rasheed.Sindhi last updated on 28/Jul/20

I f y o u h a t e t h e f o r u m, w h y d o

y o u c o m e t o t h e f o r u m a g a i n &

a g a i n b y c h a n g i n g y o u r {i d}^{'} s ? ? ?