Let's denote the number by $x$. According to the problem, the sum of the number and its reciprocal is given as:

$x + \frac{1}{x} = \frac{25}{12}$

To solve for $x$, we multiply both sides of the equation by $x$ to eliminate the fraction:

$x^2 + 1 = \frac{25}{12}x$

Now, let's multiply the entire equation by 12 to get rid of the denominator:

$12x^2 + 12 = 25x$

Next, rearrange the equation to standard quadratic form:

$12x^2 - 25x + 12 = 0$

Now, we can solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 12$, $b = -25$, and $c = 12$. Plugging in these values:

$x = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(12)(12)}}{2(12)}$

$x = \frac{25 \pm \sqrt{625 - 576}}{24}$

$x = \frac{25 \pm \sqrt{49}}{24}$

$x = \frac{25 \pm 7}{24}$

This gives us two possible solutions:

1. $x = \frac{25 + 7}{24} = \frac{32}{24} = \frac{4}{3}$
2. $x = \frac{25 - 7}{24} = \frac{18}{24} = \frac{3}{4}$

So, the number can be either $\frac{4}{3}$ or $\frac{3}{4}$.

Would you like more details on this solution or have any questions?

Here are 5 related questions:

1. How do you derive the quadratic formula?
2. What are some examples of equations where the sum of a number and its reciprocal is an integer?
3. Can you verify the solutions by substituting them back into the original equation?
4. How does multiplying both sides of an equation by the same value affect the solution?
5. What are the properties of quadratic equations and their solutions?

Tip: When solving quadratic equations, always check for possible simplifications before applying the quadratic formula.