Question Number 25738 by yesaditya22@gmail.com last updated on 13/Dec/17
Answered by mrW1 last updated on 13/Dec/17
$$\frac{{a}+{b}}{\sqrt{{ab}}}=\mathrm{6} \\ $$$$\frac{\frac{{a}}{{b}}+\mathrm{1}}{\sqrt{\frac{{a}}{{b}}}}=\mathrm{6} \\ $$$${let}\:{t}^{\mathrm{2}} =\frac{{a}}{{b}}\geqslant\mathrm{0} \\ $$$$\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}}=\mathrm{6} \\ $$$${t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{4}}}{\mathrm{2}}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{{a}}{{b}}={t}^{\mathrm{2}} =\mathrm{6}{t}−\mathrm{1}=\mathrm{18}\pm\mathrm{12}\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{17}\pm\mathrm{12}\sqrt{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 14/Dec/17
$$\mathrm{Nice}\:\mathrm{Sir}! \\ $$