t-x-c-2-t-2-x-4-c-2-4-find-x-or-t-Given-0-lt-c-lt-2-3-3-

Question Number 141076 by ajfour last updated on 15/May/21

t + x = \frac{c}{2}

t^{2} + x^{4} = \frac{c^{2}}{4}

f i n d x o r t . G i v e n 0 < c < \frac{2}{3 \sqrt{3}}

Answered by Rasheed.Sindhi last updated on 15/May/21

t - \frac{c}{2} = - x

t^{2} - \frac{c^{2}}{4} = - x^{4} \Rightarrow (t - \frac{c}{2}) (t + \frac{c}{2}) = - x^{4}

\Rightarrow (- x) (t + \frac{c}{2}) + x^{4} = 0

\Rightarrow x (- t - \frac{c}{2} + x^{3}) = 0

\Rightarrow x = 0 ∣ - t - \frac{c}{2} + x^{3} = 0

x = 0 \Rightarrow t + 0 = \frac{c}{2} \Rightarrow t = \frac{c}{2}

(x, t) = (0, \frac{c}{2})

- t - \frac{c}{2} + x^{3} = 0 \dots \dots \dots . . (i)

t - \frac{c}{2} + x = 0 \dots \dots \dots \dots . (i i)

(i) + (i i)

x^{3} + x - c = 0 (★)

(i) \Rightarrow x^{3} = t + \frac{c}{2} \dots \dots . . (i i i)

(i i) \Rightarrow x^{3} = {(- t + \frac{c}{2})}^{3} \dots . (i v)

t + \frac{c}{2} = {(- t + \frac{c}{2})}^{3}

t + \frac{c}{2} = - t^{3} + \frac{c^{3}}{8} + 3 (- t) (\frac{c}{2}) (- t + \frac{c}{2})

t + \frac{c}{2} = - t^{3} + \frac{c^{3}}{8} + \frac{3}{2} {c t}^{2} - \frac{3}{2} c t

t^{3} - \frac{3}{2} {c t}^{2} + \frac{3}{2} c t - \frac{c^{3}}{8} + \frac{c}{2} = 0 (★)

(★) D {o n}^{'} t k n o w h o w t o s o l v e c u b i c

e q u a t i o n .

(x, t) = (0, \frac{c}{2})

\dots \dots

Commented by mr W last updated on 15/May/21

x^{3} + x - c = 0 (★)

h a s a l w a y s o n e r e a l r o o t :

x = \sqrt[3]{\sqrt{\frac{1}{27} + \frac{c^{2}}{4}} + \frac{c}{2}} - \sqrt[3]{\sqrt{\frac{1}{27} + \frac{c^{2}}{4}} - \frac{c}{4}}

Commented by Rasheed.Sindhi last updated on 15/May/21

T h α n k s m r W s i r!