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Question Number 114965 by joki last updated on 22/Sep/20

find all irational numbers x such that x 2+20x+20 and x^(3 ) −2020x+1 both is a rasional number.

$${find}\:{all}\:{irational}\:{numbers}\:{x}\:{such}\:{that}\:{x} \\ $$$$\mathrm{2}+\mathrm{20}{x}+\mathrm{20}\:{and}\:{x}^{\mathrm{3}\:} −\mathrm{2020}{x}+\mathrm{1}\:\:\:{both}\:{is}\:{a} \\ $$$${rasional}\:{number}. \\ $$

Answered by MJS_new last updated on 22/Sep/20

(1) x^2 +20x+20∈Q (2) x^3 −2020x+1∈Q x=a±(√b); a, b ∈Q; b∉Q (1) ±2(a+10)(√b)+a^2 +20a+b+20∈Q ⇔ 2(a+10)(√b)∈Q (2) ±(3a^2 +b−2020)(√b)+a^3 +3ab−2020a+1∈Q ⇔ (3a^2 +b−2020)(√b)∈Q now we have (1) 2(a+10)(√b)∈Q (2) (3a^2 +b−2020)(√b)∈Q ⇒ either b=0 but then x is not irrational or a+10=0∧3a^2 +b−2020=0 ⇒ a=−10∧b=1720 ⇒ x=−10±2(√(430))

$$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}} +\mathrm{20}{x}+\mathrm{20}\in\mathbb{Q} \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} −\mathrm{2020}{x}+\mathrm{1}\in\mathbb{Q} \\ $$$$ \\ $$$${x}={a}\pm\sqrt{{b}};\:{a},\:{b}\:\in\mathbb{Q};\:{b}\notin\mathbb{Q} \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$$\pm\mathrm{2}\left({a}+\mathrm{10}\right)\sqrt{{b}}+{a}^{\mathrm{2}} +\mathrm{20}{a}+{b}+\mathrm{20}\in\mathbb{Q} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{2}\left({a}+\mathrm{10}\right)\sqrt{{b}}\in\mathbb{Q} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\pm\left(\mathrm{3}{a}^{\mathrm{2}} +{b}−\mathrm{2020}\right)\sqrt{{b}}+{a}^{\mathrm{3}} +\mathrm{3}{ab}−\mathrm{2020}{a}+\mathrm{1}\in\mathbb{Q} \\ $$$$\Leftrightarrow \\ $$$$\left(\mathrm{3}{a}^{\mathrm{2}} +{b}−\mathrm{2020}\right)\sqrt{{b}}\in\mathbb{Q} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\left({a}+\mathrm{10}\right)\sqrt{{b}}\in\mathbb{Q} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{3}{a}^{\mathrm{2}} +{b}−\mathrm{2020}\right)\sqrt{{b}}\in\mathbb{Q} \\ $$$$\Rightarrow \\ $$$$\mathrm{either}\:{b}=\mathrm{0}\:\mathrm{but}\:\mathrm{then}\:{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{irrational} \\ $$$$\mathrm{or}\:{a}+\mathrm{10}=\mathrm{0}\wedge\mathrm{3}{a}^{\mathrm{2}} +{b}−\mathrm{2020}=\mathrm{0} \\ $$$$\Rightarrow\:{a}=−\mathrm{10}\wedge{b}=\mathrm{1720} \\ $$$$\Rightarrow\:{x}=−\mathrm{10}\pm\mathrm{2}\sqrt{\mathrm{430}} \\ $$

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