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Question Number 157220 by MathSh last updated on 21/Oct/21

$$\mathrm{x}=\frac{1}{2}$$$$\mathrm{find}\sqrt{\mathrm{x}+1+\sqrt{{\mathrm{x}}^2+1}}=?$$

Answered by Rasheed.Sindhi last updated on 21/Oct/21

$$\mathrm{x}=\frac{1}{2};\sqrt{\mathrm{x}+1+\sqrt{{\mathrm{x}}^2+1}}=?$$$$\sqrt{\mathrm{x}+1+\sqrt{{\mathrm{x}}^2+1}}$$$$\sqrt{\frac{1}{2}+1+\sqrt{{\left(\frac{1}{2}\right)}^2+1}}$$$$\sqrt{\frac{3}{2}+\sqrt{\frac{1+4}{4}}}$$$$\sqrt{\frac{3}{2}+\frac{\sqrt{5}}{2}}=\sqrt{\frac{3+\sqrt{5}}{2}}=\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$$$$=\frac{\sqrt{6+2\sqrt{5}}}{2}$$$$▸\sqrt{6+2\sqrt{5}}=p+q\sqrt{5}\left(Say\right)$$$${(\sqrt{6+2\sqrt{5}}=p+q\sqrt{5})}^2;p,q\in \mathbb{Q}$$$$=6+2\sqrt{5}=p^2+2pq\sqrt{5}+5q^2$$$$p^2+5q^2=6\wedge 2pq=2$$$${\left(\frac{1}{q}\right)}^2+5q^2=6$$$$\frac{1}{q^2}+5q^2=6\Rightarrow 5q^4-6q^2+1=0$$$$(q-1)(q+1)\left(\underset{q\notin \mathbb{Q}}{\underbrace{5q^2-1}}\right)=0$$$$q=\pm 1\Rightarrow p=\frac{1}{q}=\frac{1}{\pm 1}=\pm 1$$$$\sqrt{6+2\sqrt{5}}=p+q\sqrt{5}=\pm 1\pm \sqrt{5}$$$$=1+\sqrt{5},-1-\sqrt{5}$$$$\mathrm{x}=\frac{\sqrt{6+2\sqrt{5}}}{2}=\frac{1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$$

Commented by MathSh last updated on 21/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{dear}\mathrm{Ser}$$

Commented by Tawa11 last updated on 21/Oct/21

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

Commented by Rasheed.Sindhi last updated on 22/Oct/21

$$Thanksmissforencouragingus!$$

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