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Question Number 162622 by HongKing last updated on 30/Dec/21

Solve for real numbers: x^(12) - 15x^3 + 14 = 0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{x}^{\mathrm{12}} \:-\:\mathrm{15x}^{\mathrm{3}} \:+\:\mathrm{14}\:=\:\mathrm{0} \\ $$

Answered by Ar Brandon last updated on 30/Dec/21

x=1 t^4 −15t+14=0 , t=x^3 t^4 −t−14t+14=0 (t^4 −t)−(14t−14)=0 t(t^3 −1)−14(t−1)=0 t(t−1)(t^2 +t+1)−14(t−1)=0 (t−1)(t^3 +t^2 +t−14)=0 (t−1)[(t^3 −4t)+(t^2 +5t−14)]=0 (t−1)[t(t^2 −4)+(t−2)(t+7)]=0 (t−1)[t(t−2)(t+2)+(t−2)(t+7)]=0 (t−1)(t−2)(t^2 +3t+7)=0 t=1, t=2, t=((−3+i(√(19)))/2), t=((−3−i(√(19)))/2) x=1, x=(2)^(1/3) for x∈R

$${x}=\mathrm{1} \\ $$$$ \\ $$$${t}^{\mathrm{4}} −\mathrm{15}{t}+\mathrm{14}=\mathrm{0}\:,\:{t}={x}^{\mathrm{3}} \\ $$$${t}^{\mathrm{4}} −{t}−\mathrm{14}{t}+\mathrm{14}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{4}} −{t}\right)−\left(\mathrm{14}{t}−\mathrm{14}\right)=\mathrm{0} \\ $$$${t}\left({t}^{\mathrm{3}} −\mathrm{1}\right)−\mathrm{14}\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)−\mathrm{14}\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}−\mathrm{14}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left[\left({t}^{\mathrm{3}} −\mathrm{4}{t}\right)+\left({t}^{\mathrm{2}} +\mathrm{5}{t}−\mathrm{14}\right)\right]=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left[{t}\left({t}^{\mathrm{2}} −\mathrm{4}\right)+\left({t}−\mathrm{2}\right)\left({t}+\mathrm{7}\right)\right]=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left[{t}\left({t}−\mathrm{2}\right)\left({t}+\mathrm{2}\right)+\left({t}−\mathrm{2}\right)\left({t}+\mathrm{7}\right)\right]=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{7}\right)=\mathrm{0} \\ $$$${t}=\mathrm{1},\:{t}=\mathrm{2},\:{t}=\frac{−\mathrm{3}+{i}\sqrt{\mathrm{19}}}{\mathrm{2}},\:{t}=\frac{−\mathrm{3}−{i}\sqrt{\mathrm{19}}}{\mathrm{2}} \\ $$$${x}=\mathrm{1},\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$

Commented by HongKing last updated on 31/Dec/21

cool my dear Sir thank you

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

Answered by aleks041103 last updated on 30/Dec/21

t=x^3 t^4 −15t+14=0 t^4 −t−14t+14=0 t(t^3 −1)−14(t−1)=0 t(t−1)(t^2 +t+1)−14(t−1)=0 (t−1)(t^3 +t^2 +t−14)=0 2^3 +2^2 +2−14=0 ⇒t^3 +t^2 +t−14=(t−2)(t^2 +3t+7) ⇒t^4 −15t+14=(t−1)(t−2)(t^2 +3t+7)=0 ⇒t∈{1,2,((−3±(√(9−4.7)))/2)}∩R ⇒t∈{1,2} ⇒x=1;(2)^(1/3)

$${t}={x}^{\mathrm{3}} \\ $$$${t}^{\mathrm{4}} −\mathrm{15}{t}+\mathrm{14}=\mathrm{0} \\ $$$${t}^{\mathrm{4}} −{t}−\mathrm{14}{t}+\mathrm{14}=\mathrm{0} \\ $$$${t}\left({t}^{\mathrm{3}} −\mathrm{1}\right)−\mathrm{14}\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)−\mathrm{14}\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}−\mathrm{14}\right)=\mathrm{0} \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}−\mathrm{14}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +{t}−\mathrm{14}=\left({t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{7}\right) \\ $$$$\Rightarrow{t}^{\mathrm{4}} −\mathrm{15}{t}+\mathrm{14}=\left({t}−\mathrm{1}\right)\left({t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}\in\left\{\mathrm{1},\mathrm{2},\frac{−\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}.\mathrm{7}}}{\mathrm{2}}\right\}\cap\mathbb{R} \\ $$$$\Rightarrow{t}\in\left\{\mathrm{1},\mathrm{2}\right\} \\ $$$$\Rightarrow{x}=\mathrm{1};\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$ \\ $$

Commented by HongKing last updated on 31/Dec/21

thank you my dear Sir cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

Answered by MJS_new last updated on 31/Dec/21

x^(12) −15x^3 +14=0 (x^6 −3x^3 +2)(x^6 +3x^3 +7)=0 (x−1)(x^2 +x+1)(x^3 −2)(x^6 +3x^3 +7)=0 x_1 =1 x_2 =2^(1/3) no other real solutions

$${x}^{\mathrm{12}} −\mathrm{15}{x}^{\mathrm{3}} +\mathrm{14}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{6}} −\mathrm{3}{x}^{\mathrm{3}} +\mathrm{2}\right)\left({x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{7}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left({x}^{\mathrm{3}} −\mathrm{2}\right)\left({x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{7}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\mathrm{2}^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solutions} \\ $$

Commented by HongKing last updated on 31/Dec/21

cool my dear Sir thank you

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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