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 ....