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

Correct option is C

Given:
p = 5 - 2 √6
Formula Used:
To find $$p^2 + \frac{1}{p^2}$$​, we can use the identity: $$p^2 + \frac{1}{p^2} = \left( p + \frac{1}{p} \right)^2 - 2 \\$$​​
Solution:
​$$\text{Given } p = 5 - 2 \sqrt{6}, \\\frac{1}{p} = \frac{1}{5 - 2 \sqrt{6}} \\\frac{1}{p} = \frac{ (5 + 2 \sqrt{6})}{(5 - 2 \sqrt{6})(5 + 2 \sqrt{6})} \\= \frac{5 + 2 \sqrt{6}}{25 - 24} = 5 + 2 \sqrt{6} \\p + \frac{1}{p} = (5 - 2 \sqrt{6}) + (5 + 2 \sqrt{6}) = 10 \\p^2 + \frac{1}{p^2} = \left( p + \frac{1}{p} \right)^2 - 2 \\p^2 + \frac{1}{p^2} = 10^2 - 2 = 100 - 2 = 98 \\\text{Answer:} \\p^2 + \frac{1}{p^2} = 98$$​​