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

Answer:

9 and 10

Step-by-step explanation:

To start, let's have what we know:

a and b are two consecutive positive integers

a = first integer

b = second integer, 1 greater than variable a

b = a + 1

b^2 = 19 + 9a

To solve this, we would only have to substitute values in

b^2 = 19 + 9a

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

a^2 + 2a +1     = 19 + 9a

       -9a -19    -19 - 9a

a^2 -7a - 18 = 0

a^2 + 2a - 9a - 18

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

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

So solving this, we would get a = 9 or a = -2. Since this is supposed to be a positive integer, a will be 9.

Solving for b,

b = a + 1

  = 9 + 1

  = 10

Let me know in the comments if you have any questions! If you could mark this answer as brainliest, I would very much appreciate it!