Question Number 105124 by ajfour last updated on 26/Jul/20

x^4 +ax^3 +bx^2 +cx+d=0  Find x.

Answered by OlafThorendsen last updated on 26/Jul/20

x = X−(b/4)  the term in x^3  disappears  and we have :  X^4 = pX^2 +qX+r  now it′s easy.

Commented byajfour last updated on 26/Jul/20

thanks sir, i think you mean  x=X−(a/4)  .

Commented byOlafThorendsen last updated on 26/Jul/20

yes sorry, x = X−(a/4) of course sir.

Answered by ajfour last updated on 26/Jul/20

let  (x^2 +px+q)(x^2 +rx+m)=0  ⇒   p+r=a          q+m+pr=b          mp+qr = c             qm = d  Let   mp+(b−m−pr)r=c  ⇒     m = ((c−r(b−pr))/(p−r))  And    (b−pr−q)p+qr=c  ⇒      q=((c−p(b−pr))/(r−p))  ⇒   [((c−r(b−pr))/(p−r))][((c−p(b−pr))/(r−p))]=d  ⇒   c^2 −ac(b−pr)+pr(b−pr)^2                   +d(a^2 −4pr)=0  let   pr=t   ⇒  t^3 −2bt^2 +(b^2 +ac−4d)t+(c^2 −abc+a^2 d)      = 0  let   t=z+s  ⇒  (z^3 +3sz^2 +3s^2 z+s^3 )−2b(z^2 +2sz+s^2 )  +(b^2 +ac−4d)(z+s)+(c^2 −abc+a^2 d)=0  ⇒  z^3 +(3s−2b)z^2 +(3s^2 −4bs+b^2 +ac−4d)z  +[s^3 −2bs^2 +s(b^2 +ac−4d)+c^2 −abc+a^2 d]     = 0  If   3s=2b  ⇒  z^3 +(((4b^2 )/3)−((8b^2 )/3)+b^2 +ac−4d)z      +(((8b^3 )/(27))−((8b^3 )/9)+((2b^3 )/3)+((2abc)/3)−((8bd)/3)                   +c^2 −abc+a^2 d)=0  ⇒  z^3 +(ac−4d−(b^2 /3))z+(((2b^3 )/(27))−((abc)/3)−((8bd)/3)                                                   +c^2 +a^2 d) = 0  finding z by Cardano′s formula      t=pr= z+((2b)/3)  and  as  p+r=a  ⇒   p ,r = (a/2)±(√((a^2 /4)−z−((2b)/3)))         q= (d/m) = (d/([((c−r(b−pr))/(p−r))]))  Now,     x^2 +px+q = 0   ....