Given

$\rm 2^x + \frac{1}{2^x} = 4 \frac{1}{4}$

Let us consider $\rm a = 2^x$. Substituting the values of a in the above equations gives us

$\rm a + \frac{1}{a} = 4 \frac{1}{4}$

$\rm or, a + \frac{1}{a} = \frac{4 \cdot 4 + 1}{4}$

$\rm or, a + \frac{1}{a} = \frac{17}{4}$

Multiplying both sides of the equation by a, we get,

$\rm or, a \cdot a + \frac{1}{a} \cdot a = 4 \frac{17}{4} \cdot a$

$\rm or, a^2 + 1 = \frac{17a}{4}$

Multiplying both sides of the equation by 4, we get,

$\rm or, a^2 \cdot 4 + 1 \cdot 4 = \frac{17a}{4} \cdot 4$

$\rm or, 4 a^2 + 4 = 17a$

Subtracting 17a from both sides of the equation, we get,

$\rm or, 4a^2 + 4 - 17a = 17a - 17a$

$\rm or, 4a^2 - 17a + 4 = 0$

The above equation is quadratic in a. We use the factorization method to solve it.

$\rm or, 4a^2 - (16 + 1)a + 4 = 0$

$\rm or, 4a^2 - 16a - a + 4 = 0$

$\rm or, 4a (a - 4) - 1 (a - 4) = 0$

$\rm or, (4a - 1)(a -4) = 0$

Either

$\rm 4a - 1 = 0$

$\rm or, 4a = 1$

$\rm or, a = \frac{1}{4}$

Substituting the value, $\rm a = 2^x$, we get,

$\rm or, 2^x = \frac{1}{2^2}$

$\rm or, 2^x = 2^{-2}$

$\rm \therefore x = -2$

Or

$\rm a - 4 = 0$

$\rm or, a = 4$

Substituting the value, $\rm a = 2^x$, we get,

$\rm or, 2^x = 2^2$

$\rm \therefore x = 2$

Hence, the required values of x = {-2, 2}.