Question Number 76645 by naka3546 last updated on 29/Dec/19
Commented by mr W last updated on 29/Dec/19
$$\left(\mathrm{5}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{3}+{a}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\mathrm{16}{a}−\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{8}=\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{2}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{2}−{a}\right)^{\mathrm{2}} }−{a}=\mathrm{6}\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)−\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{8}=\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4} \\ $$$$\Rightarrow\frac{{a}}{{b}}=\frac{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}\right)}{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}}=\mathrm{2} \\ $$
Answered by benjo 1/2 santuyy last updated on 29/Dec/19
