Solving for $p^2 + \frac{1}{p^2}$ with $p = 5 - 2\sqrt{6}$

We are given the value of $p$ and need to find $p^2 + \frac{1}{p^2}$.

Given: $p = 5 - 2\sqrt{6}$

Step 1: Find the reciprocal $\frac{1}{p}$

To find $\frac{1}{p}$, we take the reciprocal of $p$ and rationalize the denominator:

$ \frac{1}{p} = \frac{1}{5 - 2\sqrt{6}} $

Multiply the numerator and denominator by the conjugate of the denominator ($5 + 2\sqrt{6}$):

$ \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 - (2\sqrt{6})^2} $

$ \frac{1}{p} = \frac{5 + 2\sqrt{6}}{25 - (4 \times 6)} = \frac{5 + 2\sqrt{6}}{25 - 24} = \frac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6} $

Step 2: Use the algebraic identity

Let $p = a - b$ and $\frac{1}{p} = a + b$, where $a = 5$ and $b = 2\sqrt{6}$.

We need to find $p^2 + \frac{1}{p^2}$, which is $(a - b)^2 + (a + b)^2$.

Recall the identity: $(a - b)^2 + (a + b)^2 = (a^2 - 2ab + b^2) + (a^2 + 2ab + b^2) = 2a^2 + 2b^2$.

Step 3: Substitute values and calculate

Calculate $a^2$ and $b^2$:

$ a^2 = 5^2 = 25 $

$ b^2 = (2\sqrt{6})^2 = 2^2 \times (\sqrt{6})^2 = 4 \times 6 = 24 $

Now substitute these into the identity $2a^2 + 2b^2$:

$ p^2 + \frac{1}{p^2} = 2a^2 + 2b^2 = 2(25) + 2(24) $

$ p^2 + \frac{1}{p^2} = 50 + 48 = 98 $

Conclusion

The value of $p^2 + \frac{1}{p^2}$ is 98.