Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 158065 by zainaltanjung last updated on 30/Oct/21

Answered by Rasheed.Sindhi last updated on 30/Oct/21

$$\sqrt{2+2\sqrt{\frac{3}{4}}}$$$$=\sqrt{2+\sqrt{3}}=p+q\sqrt{3}$$$$=2+\sqrt{3}=p^2+3q^2+2pq\sqrt{3}$$$$p^2+3q^2=2\wedge 2pq=1$$$${\left(\frac{1}{2q}\right)}^2+3q^2-2=0$$$$12q^4-8q^2+1=0$$$$(2q^2-1)(6q^2-1)=0$$$$q=\pm \frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{6}}$$$$=\pm \frac{\sqrt{2}}{2},\pm \frac{\sqrt{6}}{6}$$$$\Rightarrow p=\frac{1}{2q}=\frac{1}{2(\pm \frac{\sqrt{2}}{2})},\frac{1}{2(\pm \frac{\sqrt{6}}{6})}$$$$p=\pm \frac{\sqrt{2}}{2},\pm \frac{3\sqrt{6}}{6}$$$$p+q\sqrt{3}=\pm \frac{\sqrt{2}}{2}\pm \frac{\sqrt{2}}{2}\left(\sqrt{3}\right)$$$$=\pm \frac{\sqrt{2}}{2}\pm \frac{\sqrt{6}}{2}=\pm \frac{\sqrt{2}+\sqrt{6}}{2}$$$$p+q\sqrt{3}=\pm \frac{3\sqrt{6}}{6}\pm \frac{\sqrt{6}}{6}\left(\sqrt{3}\right)=\pm \frac{3\sqrt{6}+3\sqrt{2}}{6}$$$$=\pm \frac{\sqrt{6}+\sqrt{2}}{2}\left(same\right)$$

Commented by zainaltanjung last updated on 31/Oct/21

$$\mathrm{Ok}.\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{brotber}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com