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Question Number 164155 by HongKing last updated on 14/Jan/22
if the root of equation x^9 - 10x + 1 = 0 are x_1 and x_2 x_1 x_2 = 1 find: x_1 + x_2 = ?
$$\mathrm{if}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{9}} \:-\:\mathrm{10x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\:\mathrm{are}\:\:\mathrm{x}_{\mathrm{1}} \:\:\mathrm{and}\:\:\mathrm{x}_{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\mathrm{find}:\:\:\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{x}_{\mathrm{2}} \:=\:? \\ $$
Commented by MJS_new last updated on 14/Jan/22
there′s no pair of roots of the given equation with x_1 x_2 =1
$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\mathrm{with}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\mathrm{1} \\ $$
Commented by mr W last updated on 15/Jan/22
you are right sir. but, is there an easy way to see this?
$${you}\:{are}\:{right}\:{sir}. \\ $$$${but},\:{is}\:{there}\:{an}\:{easy}\:{way}\:{to}\:{see}\:{this}? \\ $$
Answered by mr W last updated on 15/Jan/22
assume there exist such two roots. it means there exists a α, such that α and (1/α) are the roots of the eqn. α≠0, α≠1. α^9 −10α+1=0 (1/α^9 )−((10)/α)+1=0 ⇒1−10α^8 +α^9 =0 ⇒10α^8 −10α=0 ⇒10α^8 −10α=0 ⇒α^7 =1 ⇒α=e^((2kπi)/7) ...(I) ⇒α^2 −10α+1=0 ⇒α=5±2(√6) ...(II) (I) and (II) are contradiction. that means x^9 −10x+1=0 can not have two roots x_1 and x_2 with x_1 x_2 =1.
$${assume}\:{there}\:{exist}\:{such}\:{two}\:{roots}. \\ $$$${it}\:{means}\:{there}\:{exists}\:{a}\:\alpha,\:{such}\:{that} \\ $$$$\alpha\:{and}\:\frac{\mathrm{1}}{\alpha}\:{are}\:{the}\:{roots}\:{of}\:{the}\:{eqn}. \\ $$$$\alpha\neq\mathrm{0},\:\alpha\neq\mathrm{1}. \\ $$$$\alpha^{\mathrm{9}} −\mathrm{10}\alpha+\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{9}} }−\frac{\mathrm{10}}{\alpha}+\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{1}−\mathrm{10}\alpha^{\mathrm{8}} +\alpha^{\mathrm{9}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{10}\alpha^{\mathrm{8}} −\mathrm{10}\alpha=\mathrm{0} \\ $$$$\Rightarrow\mathrm{10}\alpha^{\mathrm{8}} −\mathrm{10}\alpha=\mathrm{0}\:\Rightarrow\alpha^{\mathrm{7}} =\mathrm{1}\:\Rightarrow\alpha={e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{7}}} \:\:\:…\left({I}\right) \\ $$$$\Rightarrow\alpha^{\mathrm{2}} −\mathrm{10}\alpha+\mathrm{1}=\mathrm{0}\:\Rightarrow\alpha=\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}}\:\:\:…\left({II}\right) \\ $$$$\left({I}\right)\:{and}\:\left({II}\right)\:{are}\:{contradiction}. \\ $$$${that}\:{means}\:{x}^{\mathrm{9}} −\mathrm{10}{x}+\mathrm{1}=\mathrm{0}\:{can}\:{not} \\ $$$${have}\:{two}\:{roots}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{with}\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\mathrm{1}. \\ $$
Commented by HongKing last updated on 15/Jan/22
thank you very much my dear Sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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