Find-x-if-x-4-a-x-0-lt-a-lt-2-9-

Question Number 187233 by ajfour last updated on 15/Feb/23

F i n d x i f : x^{4} + a = x \forall (0 < a < \frac{2}{9})

Commented by MJS_new last updated on 15/Feb/23

maybe off topic but it could be of interest

(x - \frac{p^{2}}{2} - \frac{\sqrt{2 - p^{6}}}{2 p}) (x - \frac{p^{2}}{2} + \frac{\sqrt{2 - p^{6}}}{2 p}) (x + \frac{p^{2}}{2} - \frac{\sqrt{2 + p^{6}}}{2 p} i) (x + \frac{p^{2}}{2} + \frac{\sqrt{2 + p^{6}}}{2 p} i) = 0

(x^{2} - p^{2} x - \frac{1 - p^{6}}{2 p^{2}}) (x^{2} + p^{2} x + \frac{1 + p^{6}}{2 p^{2}}) = 0

x^{4} - x - \frac{1 - p^{12}}{4 p^{4}} = 0

Answered by ajfour last updated on 15/Feb/23

(x^{2} + p x + h) (x^{2} - p x + k) = 0

p (k - h) = - 1

h + k = p^{2}

h k = a

p^{4} - \frac{1}{p^{2}} = 4 a

p^{6} - 4 {a p}^{2} - 1 = 0

D = \frac{1}{4} - \frac{64 a^{3}}{27}

D > 0

\Rightarrow a^{3} < \frac{27}{256}

o r 0 < a < \frac{2}{9} < \frac{3}{4} {(\frac{1}{4})}^{1 / 3}

h e n c e p^{2} i s u n i q u e a n d > 0

2 k = p^{2} - \frac{1}{p}

2 h = p^{2} + \frac{1}{p}

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

= - \frac{p}{2} \pm \sqrt{- \frac{1}{2 p} - \frac{p^{2}}{4}}

h e n c e p > 0 g i v e s x \in C

w h i l e p < 0 m i g h t y i e l d x \in R

i f p^{3} > - 2

h e n c e x c o u l d t u r n r e a l o n l y i f

- 2^{1 / 3} ⩽ p < 0

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