Question Number 33766 by .none. last updated on 23/Apr/18
![a−(1/a)=4 (√(a^2 +(1/a^2 )))−(√((a+(1/a))^2 ))−(√((a−2)^2 ))=(√x)−(√y) what is [x+y]?](https://www.tinkutara.com/question/Q33766.png)
$${a}−\frac{\mathrm{1}}{{a}}=\mathrm{4} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}−\sqrt{\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} }−\sqrt{\left({a}−\mathrm{2}\right)^{\mathrm{2}} }=\sqrt{{x}}−\sqrt{{y}} \\ $$$${what}\:{is}\:\left[{x}+{y}\right]? \\ $$
Answered by MJS last updated on 23/Apr/18
$${a}=\mathrm{2}\pm\sqrt{\mathrm{5}} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\sqrt{\left({a}−\mathrm{2}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$$\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{5}}−\sqrt{\mathrm{5}}=\sqrt{{x}}−\sqrt{{y}} \\ $$$$\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{5}}=\sqrt{{x}}−\sqrt{{y}} \\ $$$$\sqrt{\mathrm{18}}−\sqrt{\mathrm{45}}=\sqrt{{x}}−\sqrt{{y}} \\ $$$${x}=\mathrm{18} \\ $$$${y}=\mathrm{45} \\ $$$${x}+{y}=\mathrm{63} \\ $$