Final answer:
To find the consecutive integers, we solve the equation x^2 + 2(x+1) = 37, which factors to (x + 7)(x - 5) = 0, giving us the pairs (-7, -6) and (5, 6).
Step-by-step explanation:
The student is asking for help in finding two consecutive integers where the square of the first plus two times the second equals 37. To solve this, let's denote the first integer as x. Because the integers are consecutive, the second integer will be x+1. The equation we need to solve is:
x^2 + 2(x+1) = 37
Expanding the equation, we get:
x^2 + 2x + 2 = 37
Subtracting 37 from both sides gives us:
x^2 + 2x - 35 = 0
Factoring this quadratic equation, we find that the factors are:
(x + 7)(x - 5) = 0
Setting each factor equal to zero gives us two possible values for x:
- x + 7 = 0 → x = -7
- x - 5 = 0 → x = 5
Therefore, the pairs of consecutive integers are: