find two consecutive positive integers such that their product added to four times the smaller integer equals 36
let the two numbers be x and x+1
4x + x(x+1) = 36
x^2 + 5x - 36 = 0
(x+9)(x-4) = 0
x = 4 or x = -9, but x>0
x = 4
the two numbers are 4 and 5
Let's solve this problem step by step.
Let's suppose the smaller integer is 'x'. According to the problem, the larger integer will be 'x + 1' since it's mentioned that they are consecutive positive integers.
Given that their product added to four times the smaller integer equals 36, we can create an equation:
x(x + 1) + 4x = 36
Now, let's solve this equation:
x^2 + x + 4x = 36
Combine like terms:
x^2 + 5x = 36
Rearrange the equation to bring all terms to one side:
x^2 + 5x - 36 = 0
Now we can solve this quadratic equation by factoring, completing the square, or by using the quadratic formula. Let's solve it by factoring:
(x - 4)(x + 9) = 0
To find the possible values of x, we set each factor equal to zero:
x - 4 = 0 or x + 9 = 0
Solving for x, we get:
x = 4 or x = -9
Since we are looking for positive integers, we ignore the solution x = -9.
Therefore, the smaller integer is x = 4. The larger integer is x + 1 = 5.
So, the two consecutive positive integers are 4 and 5.