Solve-x-4-5x-3-4x-2-7x-1-0-

Question Number 64758 by Tawa1 last updated on 21/Jul/19

Solve : x^{4} + {5 x}^{3} - {4 x}^{2} + 7 x - 1 = 0

Answered by ajfour last updated on 21/Jul/19

x = t - \frac{5}{4}

s a y w e t h e n g e t

t^{4} - {a t}^{2} + b t - c = 0

\Rightarrow (t^{2} - p t + q) (t^{2} + p t + r) = 0

\Rightarrow t^{4} + (r + q - p^{2}) t^{2} + (- p r + p q) t + q r = 0

\Rightarrow p^{2} - (q + r) = a

p (q - r) = b

q r = - c

\Rightarrow {(p^{2} - a)}^{2} - \frac{b^{2}}{p^{2}} = - 4 c

l e t p^{2} = s

\Rightarrow s^{3} - 2 {a s}^{2} + (a^{2} + 4 c) s - b^{2} = 0

l e t s = k + \frac{2 a}{3}

\Rightarrow k^{3} + \frac{4 a^{2} k}{3} + \frac{8 a^{3}}{27} - \frac{8 a^{2} k}{3} - \frac{8 a^{3}}{9}

+ (a^{2} + 4 c) k + \frac{2 a (a^{2} + 4 c)}{3} - b^{2} = 0

s a y

k^{3} + P k + Q = 0

k = {[- \frac{Q}{2} + \sqrt{\frac{Q^{2}}{4} + \frac{P^{3}}{27}}]}^{1 / 3}

+ {[- \frac{Q}{2} - \sqrt{\frac{Q^{2}}{4} + \frac{P^{3}}{27}}]}^{1 / 3}

P = 4 c - \frac{a^{2}}{3};

Q = \frac{2 a^{3}}{27} + \frac{8 a c}{3} - b^{2}

Commented by Tawa1 last updated on 21/Jul/19

Wow, waiting for the rest sir

Answered by MJS last updated on 21/Jul/19

no “ {beautiful}^{″} solution

x_{1} \approx - 5.88645

x_{2} \approx . 153681

x_{3, 4} \approx . 366384 \pm . 985486 i

Commented by ajfour last updated on 21/Jul/19

T h a n k s s i r, i h a d m a d e b l u n d e r,

h a d p r a c t i c e d q u i n t i c e x c e s s i v e l y

b e f o r e, t h a t s w h y . .

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