Question Number 185315 by Shrinava last updated on 20/Jan/23
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}!\:=\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:+\:\mathrm{20} \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/23
$${f}\left({x}\right)=\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:+\:\mathrm{20}−{x}! \\ $$$${x}>\mathrm{4}\Rightarrow{f}\left({x}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{19}\:× \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{17}\:× \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{16}\:× \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{14}\:× \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{0}\:\:\checkmark\Rightarrow{x}=\mathrm{4} \\ $$$${f}\left({x}>\mathrm{4}\right)<\mathrm{0}\:× \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jan/23
$${x}!={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{20} \\ $$$${x}\neq\mathrm{0},\:{So}\:{dividing}\:\:{by}\:{x}\:{to}\:{both}\:{sides}\: \\ $$$$\left({x}−\mathrm{1}\right)!={x}−\mathrm{3}+\frac{\mathrm{20}}{{x}}\Rightarrow{x}\mid\mathrm{20} \\ $$$${possible}\:{values}\:{of}\:{x} \\ $$$$\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{10},\mathrm{20} \\ $$$${Only}\:{x}=\mathrm{4}\:{satisfies}\:{the}\:{equation}. \\ $$