The difference between two positive integers is 6 and their product is 112. What is the sum of the integers?
Let the integers be $x$ and $y$, with $x>y$. We have the equations \begin{align*}
x-y&=6\\
xy&=112
\end{align*}Squaring the first equation, we get  \[(x-y)^2=6^2\Rightarrow x^2-2xy+y^2=36\]Multiplying the second equation by four, we get $4xy = 4\cdot112=448$. Adding these last two equations, we have \[x^2-2xy+y^2+4xy=36+448 \Rightarrow (x+y)^2=484 \Rightarrow x+y = 22\]In the last step, we take the positive square root because both $x$ and $y$ are given to be positive. The sum of the two integers is $\boxed{22}$.
22
Level 2
Algebra
Problem: The difference between two positive integers is 6 and their product is 112. What is the sum of the integers?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the two positive integers be represented as x and y. According to the problem, we know two key pieces of information: the difference between the integers is 6, and their product is 112. 

From the difference, we can express one integer in terms of the other: 
x - y = 6, which can be rearranged to x = y + 6.

Next, we substitute this expression for x into the product equation:
x * y = 112 becomes (y + 6) * y = 112.

Expanding this gives us:
y^2 + 6y = 112.

To solve for y, we rearrange the equation:
y^2 + 6y - 112 = 0.

Now, we apply the quadratic formula, where a = 1, b = 6, and c = -112:
y = [-b ± sqrt(b^2 - 4ac)] / 2a.

Calculating the discriminant:
b^2 - 4ac = 6^2 - 4 * 1 * (-112) = 36 + 448 = 484.

Taking the square root of the discriminant:
sqrt(484) = 22.

Now substituting back into the quadratic formula:
y = [-6 ± 22] / 2.

This gives us two potential solutions:
y = (16) / 2 = 8 and y = (-28) / 2 = -14. 

Since we are looking for positive integers, we take y = 8. 

Now, substituting y back to find x:
x = y + 6 = 8 + 6 = 14.

Finally, we calculate the sum of the integers:
x + y = 14 + 8 = 22.

Thus, the final answer is 22.<|END_OF_LATENT|> The final answer is $22$.