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Question Number 56991 by Tawa1 last updated on 28/Mar/19

$$\mathrm{x}!-{\mathrm{x}}^2=8,\mathrm{Find}\mathrm{x}$$$$$$

Commented by Joel578 last updated on 28/Mar/19

$$x=4$$

Commented by Tawa1 last updated on 28/Mar/19

$$\mathrm{Any}\mathrm{workings}\mathrm{please}$$

Commented by MJS last updated on 28/Mar/19

$$\mathrm{if}x\in \mathbb{N}\mathrm{we}\mathrm{can}\mathrm{only}\mathrm{try}$$$$\Rightarrow x=4$$$$\mathrm{if}x\in \mathbb{R}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{solve}\mathrm{this}:$$$$\mathrm{\Gamma }(x+1)-x^2=8$$$$\mathrm{where}\mathrm{\Gamma }\left(x\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{t}^{x-1}{e}^{-t}dt$$$$\mathrm{and}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{are}\mathrm{able}\mathrm{to}$$

Commented by maxmathsup by imad last updated on 28/Mar/19

$$sirtawa1youmustfirstdefinex!$$

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