the product of two consecutive positive even integers is 14 more than their sum .set up an equation that can be used to find the two numbers and solve it
Let's call the two consecutive positive even integers x and (x + 2), where x is the smaller integer.
The product of the two integers is given by x(x + 2).
The sum of the two integers is given by x + (x + 2).
According to the problem statement, the product of the two integers is 14 more than their sum:
x(x + 2) = (x + (x + 2)) + 14
To solve this equation, we can first simplify it:
x(x + 2) = 2x + 16
Expanding the left side of the equation:
x^2 + 2x = 2x + 16
Now, let's get all the terms on one side of the equation:
x^2 + 2x - 2x - 16 = 0
Simplifying further:
x^2 - 16 = 0
Now, we can factor the quadratic equation:
(x - 4)(x + 4) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 4 = 0
Solving for x in each case:
For x - 4 = 0:
x = 4
For x + 4 = 0:
x = -4
Since we are looking for positive even integers, x = -4 is not a valid solution.
Therefore, the smaller integer is x = 4, and the larger integer is (x + 2) = 6.
So the two consecutive positive even integers are 4 and 6.
To solve this problem, let's start by breaking down the information given into equations.
Let's assume that the first positive even integer is x. Since the next even integer is consecutive, we can represent it as x + 2.
The product of two consecutive positive even integers is given by:
x * (x + 2)
The sum of two consecutive positive even integers is given by:
x + (x + 2)
According to the problem, the product of the two integers is 14 more than their sum. We can write this as an equation:
x * (x + 2) = (x + (x + 2)) + 14
Now, we can simplify the equation and solve for x:
x^2 + 2x = 2x + 4 + 14
x^2 + 2x = 2x + 18
x^2 + 2x - 2x - 18 = 0
x^2 - 18 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula.
Factoring:
(x - 3)(x + 3) = 0
Setting each factor equal to zero gives two possibilities:
x - 3 = 0 --> x = 3
x + 3 = 0 --> x = -3 (However, since we are considering positive even integers, this solution is not valid.)
So, the first positive even integer (x) is 3. The next even integer (x + 2) is 3 + 2 = 5.
Therefore, the two positive even integers are 3 and 5.
(2k)(2k+2) = 2k + 2k+2 + 14
I used 2k, because we need to be sure that the numbers are even. We might have used
x(x+2) = x + x+2 + 14
but it might turn out that x is odd.