Question: Directions: Use Polya’s four-step problem-solving strategy to solve this problem:
• Find two consecutive positive integers whose product is 240.
1. **Understand the Problem**
We need to find two consecutive positive integers, say $n$ and $n+1$, such that their product is 240.
2. **Devise a Plan**
Let the first integer be $n$. Then the next consecutive integer is $n+1$. The product is given by:
$$n(n+1) = 240$$
We will solve this quadratic equation to find $n$.
3. **Carry out the Plan**
Start with the equation:
$$n(n+1) = 240$$
Expanding:
$$n^2 + n = 240$$
Rewrite as a standard quadratic equation:
$$n^2 + n - 240 = 0$$
Use the quadratic formula:
$$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a=1$, $b=1$, and $c=-240$.
Calculate the discriminant:
$$\Delta = b^2 - 4ac = 1^2 - 4 \times 1 \times (-240) = 1 + 960 = 961$$
Calculate the square root:
$$\sqrt{961} = 31$$
Substitute back:
$$n = \frac{-1 \pm 31}{2}$$
Two possible solutions:
$$n = \frac{-1 + 31}{2} = \frac{30}{2} = 15$$
$$n = \frac{-1 - 31}{2} = \frac{-32}{2} = -16$$
Since we want positive integers, we take $n=15$.
Therefore, the two consecutive positive integers are $15$ and $16$.
4. **Look Back**
Check the product:
$$15 \times 16 = 240$$
This confirms our solution is correct.
**Final answer:** The two consecutive positive integers are $15$ and $16$.