x^4 +bx^2 +cx+d=0 let cx=m+px^2 +x^4 ⇒ 2x^4 +(b+p)x^2 +m+d=0 x^2 =−(((b+p)/4))±(√((((b+p)/4))^2 −(((m+d)/2)))) (p−b)x^2 −2cx+m−d=0 x=(c/(p−b))±(√(((c/(p−b)))^2 −(((m−d)/(p−b))))) ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −8(p−b)^2 (m+d))/(16))) +((2c^2 )/(b−p))∓(√((4c^4 −4c^2 (p−b)(m−d))/((p−b)^2 ))) +m−d=0 Just if m=d ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −16d(p−b))/(16))) +((4c^2 )/(b−p))=0 ⇒ (((b^2 −p^2 )/4)+((4c^2 )/(b−p)))^2 +d(p−b)=(((b^2 −p^2 )^2 )/(16)) ⇒ ((16c^4 )/((b−p)^2 ))+(d+2c^2 )(b+p)=0 let b−p=z z^2 (2b−z)=−((16c^4 )/(2c^2 +d)) z^3 −2bz^2 −k=0 (k>0) if b=−1 , d=0, c→−c z^3 +2z^2 −8c^2 =0 let z=((2c)/t) 8c^3 +8c^2 t−8c^2 t^3 =0 ⇒ t^3 −t=c back to square one.


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Question Number 156404 by ajfour last updated on 10/Oct/21

$x^{4} + {bx}^{2} + cx + d = 0$ $let cx = m + {px}^{2} + x^{4}$ $\Rightarrow {2 x}^{4} + (b + p) x^{2} + m + d = 0$ $x^{2} = - (\frac{b + p}{4}) \pm \sqrt{{(\frac{b + p}{4})}^{2} - (\frac{m + d}{2})}$ $(p - b) x^{2} - 2 cx + m - d = 0$ $x = \frac{c}{p - b} \pm \sqrt{{(\frac{c}{p - b})}^{2} - (\frac{m - d}{p - b})}$ $\Rightarrow$ $\frac{(b^{2} - p^{2})}{4} \pm \sqrt{\frac{{(b^{2} - p^{2})}^{2} - 8 {(p - b)}^{2} (m + d)}{16}}$ $+ \frac{{2 c}^{2}}{b - p} \mp \sqrt{\frac{{4 c}^{4} - {4 c}^{2} (p - b) (m - d)}{{(p - b)}^{2}}}$ $+ m - d = 0$ $J u s t i f m = d \Rightarrow$ $\frac{(b^{2} - p^{2})}{4} \pm \sqrt{\frac{{(b^{2} - p^{2})}^{2} - 16 d (p - b)}{16}}$ $+ \frac{{4 c}^{2}}{b - p} = 0$ $\Rightarrow$ ${(\frac{b^{2} - p^{2}}{4} + \frac{{4 c}^{2}}{b - p})}^{2} + d (p - b) = \frac{{(b^{2} - p^{2})}^{2}}{16}$ $\Rightarrow \frac{{16 c}^{4}}{{(b - p)}^{2}} + (d + {2 c}^{2}) (b + p) = 0$ $let b - p = z$ $z^{2} (2 b - z) = - \frac{{16 c}^{4}}{{2 c}^{2} + d}$ $z^{3} - {2 bz}^{2} - k = 0 (k > 0)$ $if b = - 1, d = 0, c \to - c$ $z^{3} + {2 z}^{2} - {8 c}^{2} = 0$ $let z = \frac{2 c}{t}$ ${8 c}^{3} + {8 c}^{2} t - {8 c}^{2} t^{3} = 0$ $\Rightarrow t^{3} - t = c$ $back to square one .$
Commented by Tawa11 last updated on 10/Oct/21

$Sir, I thought you got one formula that work for power 4 already .$ $You want to get another one ?$
Commented by ajfour last updated on 11/Oct/21

$I havent yet derived a very$ $genersl one that doesnt need$ $to adopt imaginary numbers$ $and inverse trigonometry .$
Commented by Tawa11 last updated on 11/Oct/21

$Ohh . Altight sir . God will help you more .$
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