Understanding the Problem: Sum of a Number and its Reciprocal

The question provides a relationship between a number and its reciprocal. It states that when a number is added to its reciprocal, the sum is 6. We are asked to find the sum of the square of this number and the square of its reciprocal.

Let the unknown number be denoted by the variable $x$.

The reciprocal of the number $x$ is $\frac{1}{x}$.

According to the problem statement, the sum of the number and its reciprocal is 6. We can write this as an equation:

$\qquad x + \frac{1}{x} = 6$

We need to find the value of the sum of the square of the number and the square of its reciprocal. This means we need to calculate:

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

Calculating the Sum of Squares using Algebraic Identity

To find $x^2 + \frac{1}{x^2}$ from the given equation $x + \frac{1}{x} = 6$, we can use a common algebraic identity. The identity for squaring a sum is:

$\qquad (a+b)^2 = a^2 + 2ab + b^2$

Let's apply this identity to our given equation $x + \frac{1}{x} = 6$. We can square both sides of the equation:

$\qquad \left(x + \frac{1}{x}\right)^2 = 6^2$

Now, expand the left side using the identity where $a=x$ and $b=\frac{1}{x}$:

$\qquad x^2 + 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 = 36$

Simplify the middle term $2 \cdot x \cdot \frac{1}{x}$. Since $x \cdot \frac{1}{x} = 1$ (for $x \ne 0$), the term becomes $2 \cdot 1 = 2$.

$\qquad x^2 + 2 + \frac{1}{x^2} = 36$

We are looking for the value of $x^2 + \frac{1}{x^2}$. We can rearrange the equation to isolate this term:

$\qquad x^2 + \frac{1}{x^2} = 36 - 2$

$\qquad x^2 + \frac{1}{x^2} = 34$

Final Answer

The sum of the square of the number and the square of its reciprocal is 34.

This corresponds to Option 4.