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Question Number 185315 by Shrinava last updated on 20/Jan/23

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$$\mathrm{x}!={\mathrm{x}}^2-3\mathrm{x}+20$$

Answered by Rasheed.Sindhi last updated on 20/Jan/23

$$f\left(x\right)={\mathrm{x}}^2-3\mathrm{x}+20-x!$$$$x>4\Rightarrow f\left(x\right)<0$$$$f\left(0\right)=19\times$$$$f\left(1\right)=17\times$$$$f\left(2\right)=16\times$$$$f\left(3\right)=14\times$$$$f\left(4\right)=0✓\Rightarrow x=4$$$$f(x>4)<0\times$$

Answered by Rasheed.Sindhi last updated on 20/Jan/23

$$x!=x^2-3x+20$$$$x\ne 0,Sodividingbyxtobothsides$$$$(x-1)!=x-3+\frac{20}{x}\Rightarrow x\mid 20$$$$possiblevaluesofx$$$$1,2,4,5,10,20$$$$Onlyx=4satisfiestheequation.$$

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