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find-the-roots-of-8x-3-4x-1-0-




Question Number 39026 by maxmathsup by imad last updated on 01/Jul/18
find the roots of  8x^3  −4x−1 =0
findtherootsof8x34x1=0
Answered by behi83417@gmail.com last updated on 02/Jul/18
x_1 =−(1/2)  8x^3 −4x−1=(2x+1)(4x^2 +ax+b)  x=0⇒−1=(1)(b)⇒b=−1  x=1⇒3=(3)(4+a−1)⇒a=−2  ⇒4x^2 −2x−1=0⇒x=((2±(√(4+16)))/8)=(1/4)(1±(√5))  ⇒x=−(1/2),(1/4)(1±(√5)) .■
x1=128x34x1=(2x+1)(4x2+ax+b)x=01=(1)(b)b=1x=13=(3)(4+a1)a=24x22x1=0x=2±4+168=14(1±5)x=12,14(1±5).◼
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jul/18
8x^3 −4x−(2−1)=0  8x^3 −4x−2+1=0  8x^3 +1−2(2x+1)=0  (2x+1)(4x^2 −2x.1+1^2 )−2(2x+1)=0  (2x+1)(4x^2 −2x+1−2)=0  (2x+1)(4x^2 −2x−1)=0  x=−(1/2)  4x^2 −2x−1=0  x=((2±(√(4+16)) )/8)  =((2±2(√5) )/8)=((1±(√5) )/4)
8x34x(21)=08x34x2+1=08x3+12(2x+1)=0(2x+1)(4x22x.1+12)2(2x+1)=0(2x+1)(4x22x+12)=0(2x+1)(4x22x1)=0x=124x22x1=0x=2±4+168=2±258=1±54

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