Question

If $$x=5-2\sqrt{6}$$, find the value of $$x+\dfrac{1}{x}$$.

Answer 1

$$x=5-2\sqrt{6}$$ [ Given ] ---- ( 1 )

Now,

$$\dfrac{1}{x}=\dfrac{1}{5-2\sqrt{6}}$$

$$\Rightarrow$$ $$\dfrac{1}{x}=\dfrac{1}{5-2\sqrt{6}}\times\dfrac{5+2\sqrt{6}}{5+2\sqrt{6}}$$ [ Rationalizing numerator and denominator ]

$$\Rightarrow$$ $$\dfrac{1}{x}=\dfrac{5+2\sqrt{6}}{(5)^2-(2\sqrt{6})^2}$$

$$\Rightarrow$$ $$\dfrac{1}{x}=\dfrac{5+2\sqrt{6}}{25-24}$$

$$\Rightarrow$$ $$\dfrac{1}{x}=5+2\sqrt{6}$$ ---- ( 2 )

Now,

$$\Rightarrow$$ $$x+\dfrac{1}{x}=5-2\sqrt{6}+5+2\sqrt{6}$$ [ From ( 1 ) and ( 2 ) ]

$$\Rightarrow$$ $$x+\dfrac{1}{x}=10$$