Find three consecutive intergers such that the product of the first and the second is two more than three times the third.
To solve this problem, let's first assume the three consecutive integers are n, n+1, and n+2.
According to the given information, the product of the first (n) and the second (n+1) is two more than three times the third (n+2). We can write this as an equation:
n * (n+1) = 3 * (n+2) + 2
Expanding the equation:
n^2 + n = 3n + 6 + 2
Simplifying further:
n^2 + n = 3n + 8
Now we can bring all terms to one side of the equation:
n^2 + n - 3n - 8 = 0
Combining similar terms:
n^2 - 2n - 8 = 0
To solve this quadratic equation, we can use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / 2a
In this equation, a = 1, b = -2, and c = -8.
n = (-(-2) ± √((-2)^2 - 4*1*(-8))) / (2*1)
Simplifying further:
n = (2 ± √(4 + 32)) / 2
n = (2 ± √36) / 2
n = (2 ± 6) / 2
We have two possible solutions:
1. When n = (2 + 6) / 2 = 8 / 2 = 4, the three consecutive integers are 4, 5, and 6.
2. When n = (2 - 6) / 2 = -4 / 2 = -2, the three consecutive integers are -2, -1, and 0.
So, the two sets of three consecutive integers that satisfy the given condition are (4, 5, 6) and (-2, -1, 0).