how-is-the-solution-of-this-qution-x-x-1-x-2-x-3-1-when-determinant-x-50-determinant-

Question Number 175715 by zaheen last updated on 05/Sep/22

$${how}\:{is}\:{the}\:{solution}\:{of}\:{this}\:{qution} \\ $$$$ \\ $$$$\sqrt{\left({x}\right)\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1}} \\ $$$${when}\:\:\:\:\:\begin{array}{|c|}{{x}=\mathrm{50}}\\\hline\end{array}\begin{array}{|c|c|}\\\\\hline\end{array} \\ $$

Commented by Frix last updated on 05/Sep/22

(x+n)(x+n+1)(x+n+2)(x+n+3)+1= =(x^2 +(2n+3)x+n^2 +3n+1)^2

$$\left({x}+{n}\right)\left({x}+{n}+\mathrm{1}\right)\left({x}+{n}+\mathrm{2}\right)\left({x}+{n}+\mathrm{3}\right)+\mathrm{1}= \\ $$$$=\left({x}^{\mathrm{2}} +\left(\mathrm{2}{n}+\mathrm{3}\right){x}+{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Answered by ajfour last updated on 05/Sep/22

let x=t−(3/2) ⇒ x+3=t+(3/2) f(t)=(√((t^2 −(9/4))(t^2 −(1/4))+1)) let t^2 −(5/4)=s f(s)=(√(s^2 −1+1))=s =t^2 −(5/4)=(x+(3/2))^2 −(5/4) f(50)=((103^2 −5)/4)=2651

$${let}\:\:{x}={t}−\frac{\mathrm{3}}{\mathrm{2}}\:\:\Rightarrow\:\:{x}+\mathrm{3}={t}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${f}\left({t}\right)=\sqrt{\left({t}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{1}} \\ $$$${let}\:\:\:\:{t}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}={s} \\ $$$${f}\left({s}\right)=\sqrt{{s}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}={s} \\ $$$$\:\:\:\:\:\:\:\:\:={t}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}=\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${f}\left(\mathrm{50}\right)=\frac{\mathrm{103}^{\mathrm{2}} −\mathrm{5}}{\mathrm{4}}=\mathrm{2651} \\ $$

Commented by Tawa11 last updated on 05/Sep/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by behi834171 last updated on 05/Sep/22

x(x+3)=x^2 +3x (x+1)(x+2)=x^2 +3x+2 x^2 +3x=t ⇒f(t)=(√((t)(t+2)+1))=(√(t^2 +2t+1))=t+1 ⇒f(x)=x^2 +3x+1 ⇒f(x)=50^2 +3×50+1=2651 . ■

$${x}\left({x}+\mathrm{3}\right)={x}^{\mathrm{2}} +\mathrm{3}{x} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)={x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}={t} \\ $$$$\Rightarrow{f}\left({t}\right)=\sqrt{\left({t}\right)\left({t}+\mathrm{2}\right)+\mathrm{1}}=\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}}=\boldsymbol{{t}}+\mathrm{1} \\ $$$$\Rightarrow\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}+\mathrm{1} \\ $$$$\Rightarrow\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\mathrm{50}^{\mathrm{2}} +\mathrm{3}×\mathrm{50}+\mathrm{1}=\mathrm{2651}\:.\:\blacksquare \\ $$

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