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

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