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.


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Question Number 105340 by Ar Brandon last updated on 27/Jul/20
$1\Show that the polynomial, P(x)=x^n +ax+b, n≥2 (a,b)∈R^2 has at most 3 real distinct roots. 2\Show that the equation x^4 +4ax+b=0 has no more than 2 real distinct roots.$
$1 ∖ Show that the polynomial, P (x) = x^{n} + ax + b, n ⩾ 2$ $(a, b) \in R^{2} has at most 3 real distinct roots .$ $2 ∖ Show that the equation x^{4} + 4 ax + b = 0 has no$ $more than 2 real distinct roots .$
	Answered by LifeOnMars last updated on 28/Jul/20
	$P′(x)=0 nx^(n−1) +a=0 (1) n=2k ⇒ 2kx^(2k−1) +a=0 ⇔ x=−((a/(2k)))^(1/(2k−1)) ⇒ P(x) has 1 extreme (2) n=2k+1 ⇒ (2k+1)x^(2k) +a=0 ⇔ x=±((−(a/(2k+1))))^(1/(2k)) ⇒ P(x) has { ((2 extremes; a<0)),((1 extreme; a=0)),((no extreme; a>0)) :} ⇒ P(x) has at most 2 extremes thus 3 distinct zeros the rest should be easy$
	$P^{'} (x) = 0$ ${n x}^{n - 1} + a = 0$ $(1) n = 2 k \Rightarrow 2 {k x}^{2 k - 1} + a = 0 \Leftrightarrow x = - \sqrt[2 k - 1]{\frac{a}{2 k}} \Rightarrow$ $P (x) h a s 1 e x t r e m e$ $(2) n = 2 k + 1 \Rightarrow (2 k + 1) x^{2 k} + a = 0 \Leftrightarrow x = \pm \sqrt[2 k]{- \frac{a}{2 k + 1}} \Rightarrow$ $P (x) h a s {\begin{cases} 2 e x t r e m e s; a < 0 \\ 1 e x t r e m e; a = 0 \\ n o e x t r e m e; a > 0 \end{cases}$ $\Rightarrow$ $P (x) h a s a t m o s t 2 e x t r e m e s t h u s 3$ $d i s t i n c t z e r o s$ $t h e r e s t s h o u l d b e e a s y$
	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

	$s e a r c h a n o t h e r w h a t d o y o u t h i n k$ $y o u a r e . . . ?$
	Commented by Rasheed.Sindhi last updated on 28/Jul/20

	$I f y o u h a t e t h e f o r u m, w h y d o$ $y o u c o m e t o t h e f o r u m a g a i n &$ $a g a i n b y c h a n g i n g y o u r {i d}^{'} s ? ? ?$
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