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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
Solveforrealnumbers:x12−15x3+14=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=1t4−15t+14=0,t=x3t4−t−14t+14=0(t4−t)−(14t−14)=0t(t3−1)−14(t−1)=0t(t−1)(t2+t+1)−14(t−1)=0(t−1)(t3+t2+t−14)=0(t−1)[(t3−4t)+(t2+5t−14)]=0(t−1)[t(t2−4)+(t−2)(t+7)]=0(t−1)[t(t−2)(t+2)+(t−2)(t+7)]=0(t−1)(t−2)(t2+3t+7)=0t=1,t=2,t=−3+i192,t=−3−i192x=1,x=23forx∈R
Commented by HongKing last updated on 31/Dec/21
cool my dear Sir thank you
coolmydearSirthankyou
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=x3t4−15t+14=0t4−t−14t+14=0t(t3−1)−14(t−1)=0t(t−1)(t2+t+1)−14(t−1)=0(t−1)(t3+t2+t−14)=023+22+2−14=0⇒t3+t2+t−14=(t−2)(t2+3t+7)⇒t4−15t+14=(t−1)(t−2)(t2+3t+7)=0⇒t∈{1,2,−3±9−4.72}∩R⇒t∈{1,2}⇒x=1;23
Commented by HongKing last updated on 31/Dec/21
thank you my dear Sir cool
thankyoumydearSircool
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
x12−15x3+14=0(x6−3x3+2)(x6+3x3+7)=0(x−1)(x2+x+1)(x3−2)(x6+3x3+7)=0x1=1x2=21/3nootherrealsolutions
Commented by HongKing last updated on 31/Dec/21
cool my dear Sir thank you
coolmydearSirthankyou

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