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

$$howisthesolutionofthisqution$$$$$$$$\sqrt{\left(x\right)(x+1)(x+2)(x+3)+1}$$$$when\overline{)\begin{array}{c}x=50\end{array}}\overline{)\begin{array}{c}\\ \end{array}}$$

Commented by Frix last updated on 05/Sep/22

$$(x+n)(x+n+1)(x+n+2)(x+n+3)+1=$$$$={(x^2+(2n+3)x+n^2+3n+1)}^2$$

Answered by ajfour last updated on 05/Sep/22

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

Commented by Tawa11 last updated on 05/Sep/22

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

Answered by behi834171 last updated on 05/Sep/22

$$x(x+3)=x^2+3x$$$$(x+1)(x+2)=x^2+3x+2$$$$x^2+3x=t$$$$\Rightarrow f\left(t\right)=\sqrt{\left(t\right)(t+2)+1}=\sqrt{t^2+2t+1}=\mathit{t}+1$$$$\Rightarrow \mathit{f}\left(\mathit{x}\right)={\mathit{x}}^2+3\mathit{x}+1$$$$\Rightarrow \mathit{f}\left(\mathit{x}\right)={50}^2+3\times 50+1=2651.◼$$

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