The difference between two positive integers is $6$ and their product is $40$. What is the sum of the integers?
Let the two positive integers be $x$ and $y$. We are given that $x-y=6$ and $xy=40$. Substituting $y = x-6$ into the second equation, you get $x(x-6) = 40$, which simplifies to $x^2 - 6x - 40 = 0$. This quadratic factors as $(x-10)(x+4)= 0$, so the possibilities for $x$ are 10 and -4. Since $x$ must be positive, we have $x = 10$, and so $y = x-6 = 4$. Thus, $x+y = \boxed{14}$.
To solve this problem, let's call the two positive integers $x$ and $y$, where $x > y$.
According to the problem, the difference between the two integers is $6$, so we can represent this as an equation:
$x - y = 6 \quad \text{(Equation 1)}$
The product of the two integers is $40$, so we can also represent this as an equation:
$x \cdot y = 40 \quad \text{(Equation 2)}$
We can now solve this system of equations to find the values of $x$ and $y$.
First, let's solve Equation 1 for $x$:
$x = y + 6$
Substituting this value of $x$ into Equation 2, we get:
$(y + 6) \cdot y = 40$
Expanding this equation, we get:
$y^2 + 6y = 40$
Rearranging this equation by moving everything to one side, we get:
$y^2 + 6y - 40 = 0$
Now we can solve for $y$ by factoring or by using the quadratic formula. To make it simpler in this case, we can factor the quadratic equation:
$(y - 4)(y + 10) = 0$
Setting each factor equal to zero, we get:
$y - 4 = 0 \quad \text{or} \quad y + 10 = 0$
Solving each equation for $y$, we find two possible solutions:
$y = 4 \quad \text{or} \quad y = -10$
Since we are given that the integers are positive, we can discard the solution $y = -10$. Therefore, $y = 4$.
Now we can substitute this value of $y$ back into Equation 1 to solve for $x$:
$x - 4 = 6$
Adding $4$ to both sides, we find:
$x = 10$
Therefore, the two positive integers are $10$ and $4$.
The sum of the integers is $10 + 4 = 14$.