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Question Number 208681 by Noorzai last updated on 20/Jun/24

Answered by Rasheed.Sindhi last updated on 21/Jun/24

$$\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=a+b\sqrt{3}(>0)\left(let\right)$$$$Wherea,b\in \mathbb{Z}$$$$7+4\sqrt{3}=a^2+3b^2+2ab\sqrt{3}$$$$a^2+3b^2=7\mathrm{\&}2ab=4\Rightarrow b=\frac{2}{a}$$$$a^2+3{\left(\frac{2}{a}\right)}^2=7$$$$a^2+\frac{12}{a^2}=7$$$$a^4-7a^2+12=0$$$$(a^2-3)(a^2-4)=0$$$$a^2=4\Rightarrow a=\pm 2$$$$b=\frac{2}{\pm 2}=\pm 1$$$$a+b\sqrt{3}=2+\sqrt{3},-2-\sqrt{3}(<0)$$$$\therefore a+b\sqrt{3}=2+\sqrt{3}$$

Answered by a.lgnaoui last updated on 20/Jun/24

$$=\sqrt{7+4\sqrt{3}}$$$$=\sqrt{(4+3)+4\sqrt{3}}$$$$=\sqrt{{(2+\sqrt{3})}^2}$$$$=2+\sqrt{3}$$

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