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Question Number 171846 by Mikenice last updated on 21/Jun/22

solve:  x^9 −2x^5 −3x=0

$${solve}: \\ $$$${x}^{\mathrm{9}} −\mathrm{2}{x}^{\mathrm{5}} −\mathrm{3}{x}=\mathrm{0} \\ $$

Commented by kaivan.ahmadi last updated on 21/Jun/22

x(x^8 −2x^4 −3)=0⇒{_(x^8 −2x^4 −3=0) ^(x=0)   y=x^4 ⇒y^2 −2y−3=0⇒  (y−3)(y+1)=0⇒{_(y+1=0⇒y=−1) ^(y−3=0⇒y=3)   if y=3⇒x^4 =3⇒x=(3)^(1/4)   if y=−1⇒x^4 =−1 there is no real answer

$${x}\left({x}^{\mathrm{8}} −\mathrm{2}{x}^{\mathrm{4}} −\mathrm{3}\right)=\mathrm{0}\Rightarrow\left\{_{{x}^{\mathrm{8}} −\mathrm{2}{x}^{\mathrm{4}} −\mathrm{3}=\mathrm{0}} ^{{x}=\mathrm{0}} \right. \\ $$$${y}={x}^{\mathrm{4}} \Rightarrow{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{3}=\mathrm{0}\Rightarrow \\ $$$$\left({y}−\mathrm{3}\right)\left({y}+\mathrm{1}\right)=\mathrm{0}\Rightarrow\left\{_{{y}+\mathrm{1}=\mathrm{0}\Rightarrow{y}=−\mathrm{1}} ^{{y}−\mathrm{3}=\mathrm{0}\Rightarrow{y}=\mathrm{3}} \right. \\ $$$${if}\:{y}=\mathrm{3}\Rightarrow{x}^{\mathrm{4}} =\mathrm{3}\Rightarrow{x}=\sqrt[{\mathrm{4}}]{\mathrm{3}} \\ $$$${if}\:{y}=−\mathrm{1}\Rightarrow{x}^{\mathrm{4}} =−\mathrm{1}\:{there}\:{is}\:{no}\:{real}\:{answer} \\ $$

Commented by Mikenice last updated on 23/Jun/22

thanks sir

$${thanks}\:{sir} \\ $$

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