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Question Number 133981 by bemath last updated on 26/Feb/21

$$\mathrm{If}x=5+2\sqrt{6}\mathrm{then}\frac{x-1}{\sqrt{x}}=?$$

Answered by Dwaipayan Shikari last updated on 26/Feb/21

$$x=5+2\sqrt{6}={(\sqrt{3}+\sqrt{2})}^2$$$$\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}$$

Commented by malwan last updated on 26/Feb/21

$$\frac{2\sqrt{6}}{2}=\sqrt{6}=\sqrt{2}\times \sqrt{3}$$$$suchthat{\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2=5$$$$thankyousirshikari$$$$andthismethodcanbeused$$$$tofindthesquarerootof$$$$complexnumbers$$$$verysimple$$

Commented by Dwaipayan Shikari last updated on 26/Feb/21

$$Generalway$$$$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$$$$a+\sqrt{b}=x+y+2\sqrt{xy}$$$$a=x+yb=4xy$$$$x-y=\sqrt{a^2-b}$$$$x=\frac{a+\sqrt{a^2-b}}{2}y=\frac{a-\sqrt{a^2-b}}{2}$$$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$$$\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+\sqrt{25-24}}{2}}+\sqrt{\frac{5-\sqrt{25-24}}{2}}$$$$=\sqrt{3}+\sqrt{2}$$

Answered by malwan last updated on 26/Feb/21

$$x=5+2\sqrt{6}$$$$\Rightarrow \frac{x-1}{\sqrt{x}}=\frac{4+2\sqrt{6}}{\sqrt{5+2\sqrt{6}}}\times \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5+2\sqrt{6}}}$$$$=\frac{2(2+\sqrt{6})\sqrt{5+2\sqrt{6}}}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}$$$$=\frac{2(10+\sqrt{6}-12)\sqrt{5+2\sqrt{6}}}{25-4\times 6}$$$$=\frac{2(\sqrt{6}-2)\sqrt{5+2\sqrt{6}}}{1}$$$$=2\sqrt{(6-4\sqrt{6}+4)(5+2\sqrt{6})}$$$$=2\sqrt{2(5-2\sqrt{6})(5+2\sqrt{6})}$$$$=2\sqrt{2(25-24)}=2\sqrt{2\times 1}=2\sqrt{2}$$

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