Finding the Value of \( \mathbf{x^2 + \frac{1}{x^2}} \)

We are given the value of \( \mathbf{x} \) as \( 4 + \sqrt{15} \). The question asks us to calculate the specific expression \( \rm \left( {{x^2}\, + \,\frac{1}{{{x^2}}}} \right) \). We can solve this by first finding \( \mathbf{x^2} \) and \( \frac{1}{x^2} \) separately, and then adding them together. A more efficient method involves calculating \( x + \frac{1}{x} \) first.

Method 1: Calculating \( \mathbf{x^2} \) and \( \mathbf{\frac{1}{x^2}} \) Separately

Calculating \( \mathbf{x^2} \)

Given \( x = 4 + \sqrt{15} \), we square it:

$$ x^2 = (4 + \sqrt{15})^2 $$

Using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \):

$$ x^2 = 4^2 + 2(4)(\sqrt{15}) + (\sqrt{15})^2 $$ $$ x^2 = 16 + 8\sqrt{15} + 15 $$ $$ x^2 = 31 + 8\sqrt{15} $$

Calculating \( \mathbf{\frac{1}{x}} \)

Now, let's find the reciprocal of \( \mathbf{x} \):

$$ \frac{1}{x} = \frac{1}{4 + \sqrt{15}} $$

To simplify, we rationalize the denominator by multiplying the numerator and denominator by the conjugate, \( 4 - \sqrt{15} \):

$$ \frac{1}{x} = \frac{1}{4 + \sqrt{15}} \times \frac{4 - \sqrt{15}}{4 - \sqrt{15}} $$ $$ \frac{1}{x} = \frac{4 - \sqrt{15}}{4^2 - (\sqrt{15})^2} $$ $$ \frac{1}{x} = \frac{4 - \sqrt{15}}{16 - 15} $$ $$ \frac{1}{x} = \frac{4 - \sqrt{15}}{1} = 4 - \sqrt{15} $$

Calculating \( \mathbf{\frac{1}{x^2}} \)

Now we square the value of \( \frac{1}{x} \):

$$ \frac{1}{x^2} = \left(\frac{1}{x}\right)^2 = (4 - \sqrt{15})^2 $$

Using the formula \( (a-b)^2 = a^2 - 2ab + b^2 \):

$$ \frac{1}{x^2} = 4^2 - 2(4)(\sqrt{15}) + (\sqrt{15})^2 $$ $$ \frac{1}{x^2} = 16 - 8\sqrt{15} + 15 $$ $$ \frac{1}{x^2} = 31 - 8\sqrt{15} $$

Calculating the Final Expression

Finally, we add \( \mathbf{x^2} \) and \( \frac{1}{x^2} \):

$$ \rm \left( {{x^2}\, + \,\frac{1}{{{x^2}}}} \right) = (31 + 8\sqrt{15}) + (31 - 8\sqrt{15}) $$ $$ \rm \left( {{x^2}\, + \,\frac{1}{{{x^2}}}} \right) = 31 + 31 + 8\sqrt{15} - 8\sqrt{15} $$ $$ \rm \left( {{x^2}\, + \,\frac{1}{{{x^2}}}} \right) = 62 $$

Method 2: Using the Value of \( \mathbf{x + \frac{1}{x}} \)

Calculating \( \mathbf{x + \frac{1}{x}} \)

First, find the sum of \( \mathbf{x} \) and \( \frac{1}{x} \):

$$ x = 4 + \sqrt{15} $$ $$ \frac{1}{x} = 4 - \sqrt{15} $$ $$ x + \frac{1}{x} = (4 + \sqrt{15}) + (4 - \sqrt{15}) $$ $$ x + \frac{1}{x} = 4 + 4 + \sqrt{15} - \sqrt{15} $$ $$ x + \frac{1}{x} = 8 $$

Relating \( \mathbf{x + \frac{1}{x}} \) to \( \mathbf{x^2 + \frac{1}{x^2}} \)

We know the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Let \( a=x \) and \( b=\frac{1}{x} \).

$$ \left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 $$ $$ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} $$

Rearranging this formula to find \( x^2 + \frac{1}{x^2} \):

$$ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 $$

Final Calculation

Substitute the value of \( x + \frac{1}{x} = 8 \) into the rearranged formula:

$$ x^2 + \frac{1}{x^2} = (8)^2 - 2 $$ $$ x^2 + \frac{1}{x^2} = 64 - 2 $$ $$ x^2 + \frac{1}{x^2} = 62 $$

Conclusion

Both methods show that the value of the expression \( \rm \left( {{x^2}\, + \,\frac{1}{{{x^2}}}} \right) \) when \( x = 4 + \sqrt{15} \) is 62.