Question Number 214964 by cherokeesay last updated on 24/Dec/24
Answered by GDVilla last updated on 24/Dec/24
$$\mathrm{3}? \\ $$
Answered by A5T last updated on 24/Dec/24
Commented by A5T last updated on 24/Dec/24
$${x}={y};\:{y}^{\mathrm{2}} ={z}\left({z}+{z}\right)=\mathrm{2}{z}^{\mathrm{2}} \Rightarrow{z}^{\mathrm{2}} =\frac{{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{2}{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{2}{x}\left(\mathrm{2}{y}\right){cos}\theta \\ $$$$\Rightarrow\mathrm{4}{z}^{\mathrm{2}} =\mathrm{2}{y}^{\mathrm{2}} ={y}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} {cos}\theta\Rightarrow{cos}\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xycos}\theta=\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}^{\mathrm{2}} {cos}\left(\mathrm{180}−\theta\right) \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} =\frac{\mathrm{2}+\left(\mathrm{2}×\frac{\mathrm{3}}{\mathrm{4}}\right)}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}=\mathrm{14}\Rightarrow{x}=\sqrt{\mathrm{7}} \\ $$
Commented by cherokeesay last updated on 24/Dec/24
$${thank}\:{you}\:{sir}\:! \\ $$