Question

If $a=5+2\sqrt{6},\mathrm{then}a+\frac{1}{a}=\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{Given}:a=5+2\sqrt{6} \\ \mathrm{Now},\frac{1}{a}=\frac{1}{5+2\sqrt{6}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{RHS}\mathrm{by}\left(5-2\sqrt{6}\right),\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{1}{a}=\frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}} \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{25-24} \\ \Rightarrow \frac{1}{a}=\frac{5-2\sqrt{6}}{1} \\ \Rightarrow \frac{1}{a}=5-2\sqrt{6} \\ \mathrm{Thus}, \\ a+\frac{1}{a}=\left(5+2\sqrt{6}\right)+\left(5-2\sqrt{6}\right) \\ =5+2\sqrt{6}+5-2\sqrt{6} \\ =10 \end{gathered}$$


Hence, if $a=5+2\sqrt{6},\mathrm{then}a+\frac{1}{a}=\underline{10}.$