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Question Number 39026 by maxmathsup by imad last updated on 01/Jul/18

find the roots of 8x^3 −4x−1 =0

$${find}\:{the}\:{roots}\:{of}\:\:\mathrm{8}{x}^{\mathrm{3}} \:−\mathrm{4}{x}−\mathrm{1}\:=\mathrm{0} \\ $$

Answered by behi83417@gmail.com last updated on 02/Jul/18

x_1 =−(1/2) 8x^3 −4x−1=(2x+1)(4x^2 +ax+b) x=0⇒−1=(1)(b)⇒b=−1 x=1⇒3=(3)(4+a−1)⇒a=−2 ⇒4x^2 −2x−1=0⇒x=((2±(√(4+16)))/8)=(1/4)(1±(√5)) ⇒x=−(1/2),(1/4)(1±(√5)) .■

$${x}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{1}=\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{2}} +{ax}+{b}\right) \\ $$$${x}=\mathrm{0}\Rightarrow−\mathrm{1}=\left(\mathrm{1}\right)\left({b}\right)\Rightarrow{b}=−\mathrm{1} \\ $$$${x}=\mathrm{1}\Rightarrow\mathrm{3}=\left(\mathrm{3}\right)\left(\mathrm{4}+{a}−\mathrm{1}\right)\Rightarrow{a}=−\mathrm{2} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{16}}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}\pm\sqrt{\mathrm{5}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}\pm\sqrt{\mathrm{5}}\right)\:.\blacksquare \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jul/18

8x^3 −4x−(2−1)=0 8x^3 −4x−2+1=0 8x^3 +1−2(2x+1)=0 (2x+1)(4x^2 −2x.1+1^2 )−2(2x+1)=0 (2x+1)(4x^2 −2x+1−2)=0 (2x+1)(4x^2 −2x−1)=0 x=−(1/2) 4x^2 −2x−1=0 x=((2±(√(4+16)) )/8) =((2±2(√5) )/8)=((1±(√5) )/4)

$$\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}−\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{2}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{8}{x}^{\mathrm{3}} +\mathrm{1}−\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}.\mathrm{1}+\mathrm{1}^{\mathrm{2}} \right)−\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{16}}\:}{\mathrm{8}} \\ $$$$=\frac{\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{8}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}\:}{\mathrm{4}} \\ $$

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