Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 216477 by Jubr last updated on 08/Feb/25

Answered by A5T last updated on 08/Feb/25

$$2)\sqrt{1-\sqrt{-3}}=\mathrm{u};\sqrt{1+\sqrt{-3}}=\mathrm{v}$$$$\Rightarrow {\mathrm{u}}^2+{\mathrm{v}}^2=2;\mathrm{uv}=\sqrt{1^2-{\left(\sqrt{-3}\right)}^2}=\sqrt{4}=2$$$$\Rightarrow {(\mathrm{u}+\mathrm{v})}^2={\mathrm{u}}^2+{\mathrm{v}}^2+2\mathrm{uv}=6$$$$\Rightarrow \mathrm{u}+\mathrm{v}=\sqrt{6}$$

Commented by Jubr last updated on 08/Feb/25

$$Thankssir.$$$$Iappreciatesir.$$

Answered by mr W last updated on 10/Feb/25

$$2)$$$$sayx=\sqrt{1-\sqrt{-3}}+\sqrt{1+\sqrt{-3}}$$$$x^2=1-\sqrt{-3}+1+\sqrt{-3}+2\sqrt{1^2-{\left(\sqrt{-3}\right)}^2}$$$$=2+2\times 2=6$$$$\Rightarrow x=\sqrt{6}$$

Commented by Jubr last updated on 10/Feb/25

$$Thankssir.$$$$Iappreciatesir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com