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# Question-131214

Question Number 131214 by shaker last updated on 02/Feb/21
Answered by Ar Brandon last updated on 02/Feb/21
$$\mathrm{A}=\sqrt{\mathrm{3}\sqrt{\mathrm{5}\sqrt{\mathrm{3}\sqrt{\mathrm{5}…}}}}\:\Rightarrow\mathrm{A}=\sqrt{\mathrm{3}\sqrt{\mathrm{5A}}} \\$$$$\Rightarrow\mathrm{A}^{\mathrm{4}} =\mathrm{9}×\mathrm{5A}\:\Rightarrow\mathrm{A}^{\mathrm{4}} −\mathrm{45A}=\mathrm{0} \\$$$$\Rightarrow\mathrm{A}\left(\mathrm{A}^{\mathrm{3}} −\mathrm{45}\right)=\mathrm{A}\left(\mathrm{A}^{\mathrm{3}} −\mathrm{t}^{\mathrm{3}} \right)=\mathrm{0},\:\mathrm{t}^{\mathrm{3}} =\mathrm{45} \\$$$$\Rightarrow\mathrm{A}\left(\mathrm{A}−\mathrm{t}\right)\left(\mathrm{A}^{\mathrm{2}} +\mathrm{At}+\mathrm{t}^{\mathrm{2}} \right)=\mathrm{0} \\$$$$\Rightarrow\mathrm{A}=\mathrm{0},\:\mathrm{A}=\mathrm{t}\:,\:\mathrm{A}=\frac{−\mathrm{t}\pm\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}^{\mathrm{2}} }}{\mathrm{2}} \\$$$$\mathrm{A}\neq\mathrm{0}\:,\:\mathrm{A}\in\mathbb{R}\:\Rightarrow\:\mathrm{A}=\mathrm{t}=\sqrt[{\mathrm{3}}]{\mathrm{45}} \\$$
Commented by malwan last updated on 02/Feb/21
thank you so much sir
Commented by Ar Brandon last updated on 02/Feb/21
You're welcome, Sir