If p = $$5 - 2\sqrt{6}$$, then find the value of $$p^2 + \frac{1}{p^2}$$​​

Correct option is C

​Given:

​$$p = 5 - 2\sqrt{6}$$ 

Now, we need to calculate the value of $$p^2 + \frac{1}{p^2}$$  

Formula Used: 

​$$(a+b)^2 = a^2 + b^2+ 2ab$$​​

Solution:

Since $$p = 5 - 2\sqrt{6}$$​  its conjugate is $$p = 5 + 2\sqrt{6}$$    Now, 

$$\frac{1}{p} = \frac{1}{5 - 2\sqrt{6}} \times \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} = \frac{5 + 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})}$$ 

​The denominator simplifies as:

​$$(5 - 2\sqrt{6})(5 + 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1$$ 

Thus, 

​$$\frac{1}{p} = 5 + 2\sqrt{6}$$ 

Now, 

$$p + \frac{1}{p} = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10$$ 

​$$\text{Using the identity: } p^2 + \frac{1}{p^2} = \left( p + \frac{1}{p} \right)^2 - 2$$ 

$$\text{Since } p + \frac{1}{p} = 10, \text{ So:}$$

$$p^2 + \frac{1}{p^2} = \left( 10 \right)^2 - 2$$ =  98  

​$$\text{Thus, the value of } p^2 + \frac{1}{p^2} \text{ is }\bf 98.$$​​

​​