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Question Number 91660 by Zainal Arifin last updated on 02/May/20

Commented by Prithwish Sen 1 last updated on 02/May/20

$$\mathrm{LHS}$$$$\sqrt{\frac{10+2\sqrt{21}}{2}}+\sqrt{\frac{16+2\sqrt{55}}{2}}$$$$=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{5}+\sqrt{11}}{\sqrt{2}}\mathbf{considering}\mathbf{only}\mathbf{the}\mathbf{positive}\mathbf{value}$$$$=\frac{\sqrt{3}+\sqrt{11}+\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{3}+\sqrt{11}}{\sqrt{2}}+\frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}$$$$=\sqrt{\frac{(\sqrt{3}+\sqrt{{11)}^2}}{2}}+\sqrt{\frac{{(\sqrt{7}+\sqrt{5})}^2}{2}}$$$$=\sqrt{7+\sqrt{33}}+\sqrt{6+\sqrt{35}}\mathbf{RHS}\mathbf{proved}$$

Commented by jagoll last updated on 02/May/20

$$4)\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{...}}}}}=p,p>0$$$$\Rightarrow 1+p=p^2$$$$p^2-p-1=0\Rightarrow p=\frac{1+\sqrt{5}}{2}$$$$$$

Commented by jagoll last updated on 02/May/20

$$3)x^2=2x\Rightarrow x=2$$$$y^2-y-2=0\Rightarrow y=2$$$$x\times y=4$$

Commented by john santu last updated on 02/May/20

$$5)2x+4⩽-14\vee 2x+4⩾14$$$$x⩽-9\vee x⩾5$$

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