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-

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 .

Facebook Tweet Pin