x-2-x-3-4-find-x-

Question Number 138135 by Deepak1298 last updated on 10/Apr/21

(x^{2} + x^{3}) = 4

f i n d x .

Answered by ajfour last updated on 10/Apr/21

l e t x = \frac{4}{t}

\frac{16}{t^{2}} + \frac{64}{t^{3}} = 4

\Rightarrow \frac{4 t + 16}{t^{3}} = 1

\Rightarrow t^{3} - 4 t - 16 = 0

I f

t^{3} + p t + q = 0

t = {(- \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}

h e r e = p = - 4, q = - 16

s o t = {(8 + \sqrt{64 - \frac{64}{27}})}^{1 / 3}

+ {(8 - \sqrt{64 - \frac{64}{27}})}^{1 / 3}

t = {(8 + \frac{8 \sqrt{26}}{3 \sqrt{3}})}^{1 / 3} + {(8 - \frac{8 \sqrt{26}}{3 \sqrt{3}})}^{1 / 3}

x = \frac{4}{t}

x = \frac{4}{{(8 + \frac{8 \sqrt{26}}{3 \sqrt{3}})}^{1 / 3} + {(8 - \frac{8 \sqrt{26}}{3 \sqrt{3}})}^{1 / 3}}

x = \frac{2}{{(1 + \sqrt{\frac{26}{27}})}^{1 / 3} + {(1 - \sqrt{\frac{26}{27}})}^{1 / 3}}

x \approx 1.3146

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