Find four consecutive integers, such that the product of the first and second integer minus 2 is equal to 3 times the product of the third and fourth integer.

let the 4 consecutive intergers be

x-1, x , x+1 and x+2

(x-1)(x) -2 = 3(x+1)(x+2)
x^2 - x - 2 = 3x^2 + 9x + 6
2x^2 + 10x + 8 = 0
x^2 + 5x + 4 = 0
(x+1)(x+4) = 0
x = -1 or x = -4

case 1: the numbers are -2, -1, 0, and 1
or
case 2: the numbers are -5, -4, -3, and -2

Let's assume the first consecutive integer is "x". Since we are looking for four consecutive integers, the other three consecutive integers would be "x+1", "x+2", and "x+3".

According to the given condition, the product of the first and second integer minus 2 is equal to 3 times the product of the third and fourth integer:

(x) * (x+1) - 2 = 3 * (x+2) * (x+3)

Expanding the equation:

x^2 + x - 2 = 3(x^2 + 5x + 6)

Multiplying through:

x^2 + x - 2 = 3x^2 + 15x + 18

Rearranging the terms:

3x^2 + 14x + 20 = 0

To solve this quadratic equation, we can either factorize it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the equation 3x^2 + 14x + 20 = 0:

a = 3, b = 14, c = 20

x = (-14 ± √(14^2 - 4 * 3 * 20)) / (2 * 3)

x = (-14 ± √(196 - 240)) / 6

x = (-14 ± √(-44)) / 6

Since we have a square root of a negative number, there are no real solutions for x. Therefore, there are no consecutive integers that satisfy the given condition.

To solve this problem, let's break it down into steps:

Step 1: Set up the problem
Let's represent the consecutive integers as x, x+1, x+2, and x+3 (x being the first integer).

Step 2: Formulate the equation
According to the problem statement, the product of the first and second integer minus 2 is equal to 3 times the product of the third and fourth integer. In equation form, this can be written as:
(x)(x + 1) - 2 = 3((x + 2)(x + 3))

Step 3: Solve the equation
Let's solve the equation from Step 2 to find the value of x.

(x)(x + 1) - 2 = 3((x + 2)(x + 3))
x^2 + x - 2 = 3(x^2 + 5x + 6)
x^2 + x - 2 = 3x^2 + 15x + 18

Rearranging the equation and simplifying, we get:
2x^2 + 14x + 20 = 0

Step 4: Solve the quadratic equation
To solve the quadratic equation 2x^2 + 14x + 20 = 0, we can either use factoring or the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

For our equation, a = 2, b = 14, and c = 20.

x = (-14 ± √(14^2 - 4(2)(20)))/(2(2))
x = (-14 ± √(196 - 160))/(4)
x = (-14 ± √36)/(4)

Simplifying further, we have:
x = (-14 ± 6)/4
x = (-14 + 6)/4 or x = (-14 - 6)/4
x = -2/4 or -20/4
x = -1/2 or -5

Step 5: Find the consecutive integers
If x = -1/2, then the consecutive integers are:
-1/2, -1/2 + 1, -1/2 + 2, and -1/2 + 3
Which simplifies to:
-1/2, 1/2, 3/2, and 5/2

If x = -5, then the consecutive integers are:
-5, -5 + 1, -5 + 2, and -5 + 3
Which simplifies to:
-5, -4, -3, and -2

Therefore, the two sets of four consecutive integers that satisfy the given condition are:
-1/2, 1/2, 3/2, and 5/2
and
-5, -4, -3, and -2.

let the numbers be x,x+1,x+2,x+3

x(x+1)-2 = 3(x+2)(x+3)
x = -2 or -5

check:

-2(-1)-2 = 0 = 3(0)(1)
-5(-4)-2 = 18 = 3(-3)(-2)