Find three consecutive intergers such that the product of the first and the second is two more than three times the third.
To find three consecutive integers that satisfy the given condition, let's assign variables to the integers.
Let's say the first integer is x, the second integer is (x+1) (consecutive to x), and the third integer is (x+2) (consecutive to x+1).
According to the given condition, the product of the first and the second is two more than three times the third, which can be written as:
x * (x+1) = 3 * (x+2) + 2
Now, let's solve this equation to find the value of x:
x^2 + x = 3x + 6 + 2
Simplifying the equation:
x^2 + x = 3x + 8
Rearranging and setting the equation equal to zero:
x^2 - 2x - 8 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Factoring here gives us:
(x - 4)(x + 2) = 0
So, the possible values of x are either x = 4 or x = -2.
If we substitute x = 4, we get:
(4)(4+1) = 3(4+2) + 2
20 = 18 + 2
The equation holds true, so the consecutive integers are 4, 5, and 6.
If we substitute x = -2, we get:
(-2)(-2 + 1) = 3(-2 + 2) + 2
2 = 2 + 2
The equation holds true in this case as well, so the consecutive integers can also be -2, -1, and 0.
Therefore, the two sets of consecutive integers that satisfy the given condition are 4, 5, 6, and -2, -1, 0.