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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:x1215x3+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=1t415t+14=0,t=x3t4t14t+14=0(t4t)(14t14)=0t(t31)14(t1)=0t(t1)(t2+t+1)14(t1)=0(t1)(t3+t2+t14)=0(t1)[(t34t)+(t2+5t14)]=0(t1)[t(t24)+(t2)(t+7)]=0(t1)[t(t2)(t+2)+(t2)(t+7)]=0(t1)(t2)(t2+3t+7)=0t=1,t=2,t=3+i192,t=3i192x=1,x=23forxR

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=x3t415t+14=0t4t14t+14=0t(t31)14(t1)=0t(t1)(t2+t+1)14(t1)=0(t1)(t3+t2+t14)=023+22+214=0t3+t2+t14=(t2)(t2+3t+7)t415t+14=(t1)(t2)(t2+3t+7)=0t{1,2,3±94.72}Rt{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

x1215x3+14=0(x63x3+2)(x6+3x3+7)=0(x1)(x2+x+1)(x32)(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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