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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.
$$\mathrm{1}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polynomial},\:\mathrm{P}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} +\mathrm{ax}+\mathrm{b},\:\mathrm{n}\geqslant\mathrm{2} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \:\mathrm{has}\:\mathrm{at}\:\mathrm{most}\:\mathrm{3}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{4}} +\mathrm{4ax}+\mathrm{b}=\mathrm{0}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{2}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$
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
$${P}'\left({x}\right)=\mathrm{0} \\ $$$${nx}^{{n}−\mathrm{1}} +{a}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{n}=\mathrm{2}{k}\:\Rightarrow\:\mathrm{2}{kx}^{\mathrm{2}{k}−\mathrm{1}} +{a}=\mathrm{0}\:\Leftrightarrow\:{x}=−\sqrt[{\mathrm{2}{k}−\mathrm{1}}]{\frac{{a}}{\mathrm{2}{k}}}\Rightarrow \\ $$$${P}\left({x}\right)\:{has}\:\mathrm{1}\:{extreme} \\ $$$$\left(\mathrm{2}\right)\:{n}=\mathrm{2}{k}+\mathrm{1}\:\Rightarrow\:\left(\mathrm{2}{k}+\mathrm{1}\right){x}^{\mathrm{2}{k}} +{a}=\mathrm{0}\:\Leftrightarrow\:{x}=\pm\sqrt[{\mathrm{2}{k}}]{−\frac{{a}}{\mathrm{2}{k}+\mathrm{1}}}\Rightarrow \\ $$$${P}\left({x}\right)\:{has}\:\begin{cases}{\mathrm{2}\:{extremes};\:{a}<\mathrm{0}}\\{\mathrm{1}\:{extreme};\:{a}=\mathrm{0}}\\{{no}\:{extreme};\:{a}>\mathrm{0}}\end{cases} \\ $$$$\Rightarrow \\ $$$${P}\left({x}\right)\:{has}\:{at}\:{most}\:\mathrm{2}\:{extremes}\:{thus}\:\mathrm{3} \\ $$$${distinct}\:{zeros} \\ $$$${the}\:{rest}\:{should}\:{be}\:{easy} \\ $$
Commented by LifeOnMars last updated on 28/Jul/20
btw I hate this forum you cannot solve the easiest problems
$$\boldsymbol{{btw}}\:\boldsymbol{{I}}\:\boldsymbol{{hate}}\:\boldsymbol{{this}}\:\boldsymbol{{forum}}\:\boldsymbol{{you}}\:\boldsymbol{{cannot}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{easiest}}\:\boldsymbol{{problems}} \\ $$
Commented by abdomsup last updated on 28/Jul/20
search another what do you think you are...?
$${search}\:{another}\:{what}\:{do}\:{you}\:{think} \\ $$$${you}\:{are}…? \\ $$
Commented by Rasheed.Sindhi last updated on 28/Jul/20
If you hate the forum, why do you come to the forum again & again by changing your id′s???
$${If}\:\:{you}\:{hate}\:{the}\:{forum},\:{why}\:{do} \\ $$$${you}\:{come}\:{to}\:{the}\:{forum}\:{again}\:\& \\ $$$${again}\:{by}\:{changing}\:{your}\:{id}'{s}??? \\ $$

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