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Question Number 164019 by ajfour last updated on 13/Jan/22

x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0

\Rightarrow x^{2} + \frac{d}{x^{2}} + a x + \frac{c}{x} + b = 0

s a y \frac{1}{x} = t \Rightarrow

x^{2} + a x + b + {d t}^{2} + c t = 0

l e t x^{2} + a x + p = 0

& {d t}^{2} + c t + q = 0

p + q = b

x = - \frac{a}{2} \pm \sqrt{\frac{a^{2}}{4} - p}

t = - \frac{c}{2 d} \pm \sqrt{\frac{c^{2}}{4 d^{2}} - \frac{q}{d}}

t x = 1 \Rightarrow

\frac{a c}{4 d} \mp \frac{a}{2} \sqrt{\frac{c^{2}}{4 d^{2}} - \frac{q}{d}} \mp \frac{c}{2 d} \sqrt{\frac{a^{2}}{4} - p}

+ \sqrt{\frac{a^{2}}{4} - p} \sqrt{\frac{c^{2}}{4 d^{2}} - \frac{q}{d}} = 1

l e t p = \frac{a^{2}}{4} \Rightarrow

\frac{a c}{4 d} \mp \frac{a}{2} \sqrt{\frac{c^{2}}{4 d^{2}} - \frac{b}{d} + \frac{a^{2}}{4 d}} = 1

\Rightarrow \frac{c^{2}}{4 d^{2}} - \frac{b}{d} + \frac{a^{2}}{4 d} = \frac{4}{a^{2}} + \frac{c^{2}}{4 d^{2}} - \frac{2 c}{a d}

\Rightarrow 4 a^{2} b - a^{4} + 16 d - 8 a c = 0

i f x = s + h

s^{4} + 4 h s (s^{2} + h^{2}) + 6 h^{2} s^{2} + h^{4}

+ a {s^{3} + 3 h s (s + h) + h^{2}}

+ b {s^{2} + 2 h s + h^{2}} + c (s + h) + d = 0

A = 4 h + a

B = 6 h^{2} + 3 a h + b

C = 4 h^{3} + 3 {a h}^{2} + 2 b h + c

D = h^{4} + {a h}^{3} + {b h}^{2} + c h + d

\Rightarrow

{(4 h + a)}^{2} (8 h^{2} + 4 a h + 4 b - a^{2})

+ 16 (h^{4} + {a h}^{3} + {b h}^{2} + c h + d)

= 8 (4 h + a) (4 h^{3} + 3 {a h}^{2} + 2 b h + c)

\dots . .