The product of two consecutive positive integers is 11 more than their sum. What the integers? Please help.
x(x + 1) = x + (x + 1) + 11
Your move.
x^2 + x = 2x + 12
x^2 -x -12 = 0
(x-4)(x+3) = 0
The only positive solution for x is
x = 4. That is the smaller of the two numbers.
To solve this problem, let's assume that the two consecutive positive integers are x and x+1.
According to the given information, the product of these two consecutive positive integers is 11 more than their sum. Mathematically, this can be written as:
x(x+1) = (x + (x+1)) + 11
Now, let's simplify the equation:
x(x+1) = 2x + 12
Expanding the left side:
x^2 + x = 2x + 12
Moving all the terms to one side:
x^2 + x - 2x - 12 = 0
Combining like terms:
x^2 - x - 12 = 0
Now, we need to factorize this quadratic equation. The factors are:
(x - 4)(x + 3) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 3 = 0
Solving for x:
x = 4 or x = -3
Since we're looking for positive integers, we discard the negative value. Therefore, x = 4.
The consecutive positive integers are x = 4 and x+1 = 4+1 = 5.
So, the two integers are 4 and 5.