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Question Number 45896 by Sanjarbek last updated on 18/Oct/18
solve for x  x^4 −4x+1=0
solveforxx44x+1=0
Commented by tanmay.chaudhury50@gmail.com last updated on 18/Oct/18
Commented by tanmay.chaudhury50@gmail.com last updated on 18/Oct/18
Answered by ajfour last updated on 19/Oct/18
let  x^4 −4x+1=0   ⇔ (x^2 +ax+b)(x^2 −ax+(1/b))=0    ⇔ x^4 +(b+(1/b)−a^2 )x^2 +a((1/b)−b)x+1=0  ⇒  (1/b)+b = a^2   ;   (1/b)−b = −(4/a)  adding:      a^2 −(4/a)= (2/b)     ...(i)  subtracting:   a^2 +(4/a) = 2b    ...(ii)  ⇒    b > 0    multiplying above two eqs.              a^4 −((16)/a^2 ) =4  let  s = a^2  , then         s^3 −4s−16 = 0  ⇒ s=a^2 = (8+(8/3)(√((26)/3)))^(1/3) +(8−(8/3)(√((26)/3)))^(1/3)      a^2 ≈ 3.04275941      a ≈ ±1.74435071      b = (a^2 /2)+(2/a)     with +ve a,  b_1 ≈ 2.66793813    with −ve a,  b_2 ≈ 0.374821282  both sets satisfy eqs. (i) & (ii)    For +ve a & b_1     x^2 +ax+b=0  provides      x=−(a/2)±(√((a^2 /4)−b))       no real roots; while   x^2 −ax+1/b =0   gives      x ≈ −1.49336 , −0.25099   For  −ve a & b_2     x^2 +ax+b = 0   gives      x ≈ 0.25099 , 1.49336    while x^2 −ax+1/b =0  gives     no real roots.  But   (d/dx)(x^4 −4x+1)= 4x^3 −4  ⇒  (dy/dx)=0  only at x=1  so one root is certainly > 1  ⇒ we need to reject the negative  set of roots;  the answer is indeed       x_1 ≈ 0.25099  &  x_2 ≈ 1.49336  .
letx44x+1=0(x2+ax+b)(x2ax+1b)=0x4+(b+1ba2)x2+a(1bb)x+1=01b+b=a2;1bb=4aadding:a24a=2b(i)subtracting:a2+4a=2b(ii)b>0multiplyingabovetwoeqs.a416a2=4lets=a2,thens34s16=0s=a2=(8+83263)1/3+(883263)1/3a23.04275941a±1.74435071b=a22+2awith+vea,b12.66793813withvea,b20.374821282bothsetssatisfyeqs.(i)&(ii)For+vea&b1x2+ax+b=0providesx=a2±a24bnorealroots;whilex2ax+1/b=0givesx1.49336,0.25099Forvea&b2x2+ax+b=0givesx0.25099,1.49336whilex2ax+1/b=0givesnorealroots.Butddx(x44x+1)=4x34dydx=0onlyatx=1soonerootiscertainly>1weneedtorejectthenegativesetofroots;theanswerisindeedx10.25099&x21.49336.
Commented by peter frank last updated on 18/Oct/18
second line sir clarify plz
secondlinesirclarifyplz
Commented by ajfour last updated on 18/Oct/18
coeff. of x^3  is 0 and constant  term is 1, and i assumed a degree 4  expression as a product of two  quadratic expressions..
coeff.ofx3is0andconstanttermis1,andiassumedadegree4expressionasaproductoftwoquadraticexpressions..
Commented by peter frank last updated on 20/Oct/18
and third line i dont get it
andthirdlineidontgetit
Commented by ajfour last updated on 20/Oct/18
second line expanded..
secondlineexpanded..
Commented by peter frank last updated on 20/Oct/18
thanks
thanks
Answered by mondodotto@gmail.com last updated on 18/Oct/18

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