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2-2x-1-1-3-x-3-1-How-much-the-x-is-

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Question Number 167508 by Bagus1003 last updated on 18/Mar/22
2((2x+1))^(1/3) =x^3 −1 How much the x is?
22x+13=x3−1Howmuchthexis?
Answered by alephzero last updated on 18/Mar/22
2((2x+1))^(1/3) = x^3 −1 ⇒ ((2x+1))^(1/3) = ((x^3 −1)/2) ⇒ 2x+1 = ((x^9 −3x^6 +3x^3 −1)/8) ⇒ 2x = ((x^9 −3x^6 +3x^3 −2)/8) ⇒ x = ((x^9 −3x^6 +3x^3 −2)/(16)) ⇒ ((x^9 −3x^6 +3x^3 −8x−2)/(16)) = 0 ⇒ x^9 −3x^6 +3x^3 −8x−2 = 0 x_1 ≈ −0.9592 x_2 ≈ −0.2564 x_3 ≈ 1.4962
22x+13=x3−1⇒2x+13=x3−12⇒2x+1=x9−3x6+3x3−18⇒2x=x9−3x6+3x3−28⇒x=x9−3x6+3x3−216⇒x9−3x6+3x3−8x−216=0⇒x9−3x6+3x3−8x−2=0x1≈−0.9592x2≈−0.2564x3≈1.4962
Answered by Jamshidbek last updated on 18/Mar/22
Solution. 2x+1=t ⇒ 2x=t−1 2((2x+1))^(1/3) =x^3 −1 ⇒ 16((2x+1))^(1/3) =(2x)^3 −8 ⇒ 16t=(t−1)^3 −8 ⇒ t^3 −3t^2 +3t−1−8=16t t^3 −3t^2 −13t−9=0 t^3 +t^2 −4t^2 −4t−9t−9=0 t^2 (t+1)−4t(t+1)−9(t+1)=0 (t+1)(t^2 −4t−9)=0 t_1 =−1 t_(2,3) =((4±(√(52)))/2)=2±(√(13)) x_1 =−1 x_(2,3) =((2±(√(13))−1)/2)=((1±(√(13)))/2) Telegram:@math_undergraduate
Solution.2x+1=t⇒2x=t−122x+13=x3−1⇒162x+13=(2x)3−8⇒16t=(t−1)3−8⇒t3−3t2+3t−1−8=16tt3−3t2−13t−9=0t3+t2−4t2−4t−9t−9=0t2(t+1)−4t(t+1)−9(t+1)=0(t+1)(t2−4t−9)=0t1=−1t2,3=4±522=2±13x1=−1x2,3=2±13−12=1±132Telegram:@math_undergraduate
Commented by mr W last updated on 18/Mar/22
x_1 =−1 x_(2,3) =((1±(√5))/2)
x1=−1x2,3=1±52

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