Question

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

Answer 1

Let's call the smaller integer "x" and the larger integer "x+1".

According to the problem statement, we can set up the following equation:

(x+1)^2 = 9x + 19

Expanding the left side of the equation, we get:

x^2 + 2x + 1 = 9x + 19

Bringing all the terms to one side of the equation:

x^2 - 7x - 18 = 0

Factoring the quadratic equation:

(x - 9)(x + 2) = 0

So, x could be either 9 or -2. Since we are looking for positive integers, we can discard the negative value of x.

Therefore, the smaller integer is 9 and the larger integer is 10.

To check:

10^2 = 100 = 9(9) + 19 = 82.

The equation is true, and we have found our two consecutive positive integers: 9 and 10.