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 represent the three consecutive integers algebraically.
Let's assume that the first integer is n. So, the three consecutive integers are n, n+1, and n+2.
According to the problem, the product of the first and second integers is two more than three times the third integer. We can write this as an equation:
n * (n+1) = 3 * (n+2) + 2
Now, let's solve this equation step by step to find the value of n.
Expanding the equation:
n^2 + n = 3n + 6 + 2
Combining like terms:
n^2 + n = 3n + 8
Bringing all terms to one side:
n^2 - 2n - 8 = 0
Now, we have a quadratic equation. We can factorize it, use the quadratic formula, or complete the square to solve for n. Let's factorize it:
(n - 4)(n + 2) = 0
Setting each factor to zero:
n - 4 = 0 or n + 2 = 0
Solving for n:
n = 4 or n = -2
So, the two possible values for n are 4 and -2.
If n = 4, then the consecutive integers are 4, 5, and 6.
If n = -2, then the consecutive integers are -2, -1, and 0.
Hence, the three consecutive integers that satisfy the given conditions are 4, 5, and 6 or -2, -1, and 0.