Question Number 216477 by Jubr last updated on 08/Feb/25
Answered by A5T last updated on 08/Feb/25
$$\left.\mathrm{2}\right)\:\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\mathrm{u}\:;\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}=\mathrm{v} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{2}\:;\:\mathrm{uv}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}}=\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}} =\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} +\mathrm{2uv}=\mathrm{6} \\ $$$$\Rightarrow\mathrm{u}+\mathrm{v}=\sqrt{\mathrm{6}} \\ $$
Commented by Jubr last updated on 08/Feb/25
$${Thanks}\:{sir}. \\ $$$${I}\:{appreciate}\:{sir}. \\ $$
Answered by mr W last updated on 10/Feb/25
$$\left.\mathrm{2}\right) \\ $$$${say}\:{x}=\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}+\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{1}−\sqrt{−\mathrm{3}}+\mathrm{1}+\sqrt{−\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\mathrm{2}+\mathrm{2}×\mathrm{2}=\mathrm{6} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{6}} \\ $$
Commented by Jubr last updated on 10/Feb/25
$${Thanks}\:{sir}. \\ $$$${I}\:{appreciate}\:{sir} \\ $$