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 such that the product of the first and the second is two more than three times the third, we can represent the three consecutive integers in terms of variables. Let's assume the first integer is x, then the second integer will be x+1, and the third integer will be x+2.
Now, we can set up the equation based on the given condition. The product of the first and the second integer is (x)(x+1), and three times the third integer is 3(x+2). According to the question, the product of the first and the second is two more than three times the third, so we can write the equation as:
(x)(x+1) = 3(x+2) + 2
Now, we can solve this equation to find the value of x, which will give us the three consecutive integers. Expanding the equation, we get:
x^2 + x = 3x + 6 + 2
Simplifying further:
x^2 + x = 3x + 8
Rearranging the terms:
x^2 + x - 3x - 8 = 0
Combining like terms:
x^2 - 2x - 8 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Let's use factoring in this case.
Factoring the quadratic equation:
(x - 4)(x + 2) = 0
Now, setting each factor to zero and solving for x:
x - 4 = 0 or x + 2 = 0
Solving each equation:
x = 4 or x = -2
So, we have two possible values for x.
If x = 4, then the three consecutive integers are 4, 5, and 6.
If x = -2, then the three consecutive integers are -2, -1, and 0.
Thus, we have two sets of three consecutive integers that satisfy the given condition: {4, 5, 6} and {-2, -1, 0}.