Question Number 213201 by ajfour last updated on 01/Nov/24
Commented by ajfour last updated on 01/Nov/24
$${Find}\:{R} \\ $$
Answered by A5T last updated on 01/Nov/24
$$\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$
Commented by A5T last updated on 01/Nov/24
Commented by ajfour last updated on 01/Nov/24
$${Wow},\:{excellent}. \\ $$
Commented by ajfour last updated on 01/Nov/24
$$\left(\mathrm{2}\sqrt{{R}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{R} \\ $$$$\mathrm{2}{R}−\mathrm{4}\sqrt{\mathrm{3}}\sqrt{{R}}+\mathrm{2}=\mathrm{0} \\ $$$${R}−\mathrm{2}\sqrt{\mathrm{3}}\sqrt{{R}}+\mathrm{1}=\mathrm{0} \\ $$$$\sqrt{{R}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$${R}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$
Answered by cherokeesay last updated on 01/Nov/24