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

$$(x^2+x^3)=4$$$$findx.$$

Answered by ajfour last updated on 10/Apr/21

$$letx=\frac{4}{t}$$$$\frac{16}{t^2}+\frac{64}{t^3}=4$$$$\Rightarrow \frac{4t+16}{t^3}=1$$$$\Rightarrow t^3-4t-16=0$$$$If$$$$t^3+pt+q=0$$$$t={(-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}})}^{1/3}$$$$+{(-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}})}^{1/3}$$$$here=p=-4,q=-16$$$$sot={(8+\sqrt{64-\frac{64}{27}})}^{1/3}$$$$+{(8-\sqrt{64-\frac{64}{27}})}^{1/3}$$$$t={(8+\frac{8\sqrt{26}}{3\sqrt{3}})}^{1/3}+{(8-\frac{8\sqrt{26}}{3\sqrt{3}})}^{1/3}$$$$x=\frac{4}{t}$$$$x=\frac{4}{{(8+\frac{8\sqrt{26}}{3\sqrt{3}})}^{1/3}+{(8-\frac{8\sqrt{26}}{3\sqrt{3}})}^{1/3}}$$$$x=\frac{2}{{(1+\sqrt{\frac{26}{27}})}^{1/3}+{(1-\sqrt{\frac{26}{27}})}^{1/3}}$$$$x\approx 1.3146$$

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