Question Number 189270 by Rupesh123 last updated on 14/Mar/23
Commented by maths_plus last updated on 14/Mar/23
$$\mathrm{good} \\ $$
Answered by HeferH last updated on 14/Mar/23
$$\:\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} =\:\mathrm{b}^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{x}^{\mathrm{2}} =\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:−\:\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2ab}\right) \\ $$$$\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{2ab} \\ $$
Commented by manxsol last updated on 14/Mar/23
$${excelent} \\ $$
Answered by manxsol last updated on 14/Mar/23
$${x}^{\mathrm{2}} ={b}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{b}.{bcos}\left(\mathrm{180}−\theta\right) \\ $$$${x}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} {cos}\theta \\ $$$${cos}\theta=\frac{{a}−{b}}{{b}} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \left(\mathrm{1}+\frac{{a}−{b}}{{b}}\right) \\ $$$${x}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \frac{{a}}{{b}} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}{ab} \\ $$