Question

Find two positive consecutive integers such that the square of the smaller integer is 11 more than the larger integer.

a) 4 and 5
b) 5 and 6
c) 6 and 7
d) 7 and 8

Answer 1

The two positive consecutive integers that satisfy the given condition are 4 and 5.

Step-by-step explanation:

To find two positive consecutive integers such that the square of the smaller integer is 11 more than the larger integer, we can represent the smaller integer as x and the larger integer as x + 1. According to the problem, the square of the smaller integer (x^2) is 11 more than the larger integer (x + 1). So, we can write the equation as x^2 = (x + 1) + 11. Solving this equation, we get x = 4. Therefore, the two positive consecutive integers are 4 and 5, which corresponds to option a) 4 and 5.