Question Number 214247 by hardmath last updated on 02/Dec/24
$$\mathrm{If}\:\:\:\mathrm{2a}\:=\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{2a}^{\mathrm{2}} \:+\:\sqrt{\mathrm{a}}}{\mathrm{2a}}\:=\:? \\ $$
Answered by A5T last updated on 02/Dec/24
$$\mathrm{2}{a}=\mathrm{1}−\mathrm{2}\sqrt{{a}}\Rightarrow\begin{cases}{\sqrt{{a}}=\frac{\mathrm{2}{a}−\mathrm{1}}{−\mathrm{2}}}\\{\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{a}\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}=\mathrm{8}{a}}\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{2}{a}^{\mathrm{2}} +\sqrt{{a}}}{\mathrm{2}{a}}=\frac{\mathrm{2}{a}^{\mathrm{2}} +\frac{\mathrm{2}{a}−\mathrm{1}}{−\mathrm{2}}}{\mathrm{2}{a}}=\frac{−\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{1}}{−\mathrm{4}{a}} \\ $$$$=\frac{−\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{2}{a}}{−\mathrm{4}{a}}=\frac{−\mathrm{8}{a}+\mathrm{2}{a}}{−\mathrm{4}{a}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$