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Question-90162




Question Number 90162 by I want to learn more last updated on 21/Apr/20
Answered by maths mind last updated on 21/Apr/20
x=y−1  ⇔(y−2)^5 +(y+2)^5 =242y  ⇔2y^5 +80y^3 +160y=242y  ⇒2y(y^4 −40y^2 −41)=0  easy now
$${x}={y}−\mathrm{1} \\ $$$$\Leftrightarrow\left({y}−\mathrm{2}\right)^{\mathrm{5}} +\left({y}+\mathrm{2}\right)^{\mathrm{5}} =\mathrm{242}{y} \\ $$$$\Leftrightarrow\mathrm{2}{y}^{\mathrm{5}} +\mathrm{80}{y}^{\mathrm{3}} +\mathrm{160}{y}=\mathrm{242}{y} \\ $$$$\Rightarrow\mathrm{2}{y}\left({y}^{\mathrm{4}} −\mathrm{40}{y}^{\mathrm{2}} −\mathrm{41}\right)=\mathrm{0} \\ $$$${easy}\:{now} \\ $$
Commented by I want to learn more last updated on 21/Apr/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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