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Question Number 33200 by pieroo last updated on 12/Apr/18

Solve 2^(3n+2) −7×2^(2n+2) −31×2^n −8=0, n∈R. I need some help with this

$$\mathrm{Solve}\:\mathrm{2}^{\mathrm{3n}+\mathrm{2}} \:−\mathrm{7}×\mathrm{2}^{\mathrm{2n}+\mathrm{2}} \:−\mathrm{31}×\mathrm{2}^{\mathrm{n}} \:−\mathrm{8}=\mathrm{0},\:\mathrm{n}\in\boldsymbol{\mathrm{R}}. \\ $$$$\mathrm{I}\:\mathrm{need}\:\mathrm{some}\:\mathrm{help}\:\mathrm{with}\:\mathrm{this} \\ $$

Answered by MJS last updated on 12/Apr/18

2^(3n+2) =4×8^n =4×(2^n )^3 −7×2^(2n+2) =−28×4^n =−28×(2^n )^2 x=2^n 4x^3 −28x^2 −31x−8=0 x_1 x_2 x_3 =−8 ⇒ try ±1; ±2; ±4; ±8 ⇒ x_1 =8 ((4x^3 −28x^2 −31x−8)/(x−8))=4x^2 +4x+1 4x^2 +4x+1=0 x^2 +x+(1/4)=0 (x+(1/2))^2 =0 ⇒ x_2 =x_3 =−(1/2) x=8 ⇒ 2^n =8 ⇒ n=3 x=−(1/2) ⇒ 2^n =−(1/2) ⇒ no solution in R

$$\mathrm{2}^{\mathrm{3}{n}+\mathrm{2}} =\mathrm{4}×\mathrm{8}^{{n}} =\mathrm{4}×\left(\mathrm{2}^{{n}} \right)^{\mathrm{3}} \\ $$$$−\mathrm{7}×\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} =−\mathrm{28}×\mathrm{4}^{{n}} =−\mathrm{28}×\left(\mathrm{2}^{{n}} \right)^{\mathrm{2}} \\ $$$${x}=\mathrm{2}^{{n}} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} −\mathrm{28}{x}^{\mathrm{2}} −\mathrm{31}{x}−\mathrm{8}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} =−\mathrm{8}\:\Rightarrow\:\mathrm{try}\:\pm\mathrm{1};\:\pm\mathrm{2};\:\pm\mathrm{4};\:\pm\mathrm{8} \\ $$$$\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{8} \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{3}} −\mathrm{28}{x}^{\mathrm{2}} −\mathrm{31}{x}−\mathrm{8}}{{x}−\mathrm{8}}=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{2}} ={x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\mathrm{8}\:\Rightarrow\:\mathrm{2}^{{n}} =\mathrm{8}\:\Rightarrow\:{n}=\mathrm{3} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{2}^{{n}} =−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{R} \\ $$

Commented by pieroo last updated on 13/Apr/18

thanks very much boss

$$\mathrm{thanks}\:\mathrm{very}\:\mathrm{much}\:\mathrm{boss} \\ $$

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