a positive integer squared plus 8 times its consecutive integer is equal to 41. find the integers

Answer 1

Devin got it right.

But also check the answer

3^2 + 8(4) = 9+32 = 41. It works. Bingo.

Answer 2

1. Let n = the positive integer and Let n+1 = The consecutive positive integer.
2. Set up your equation:

n²+8(n+1) = 41

3) Expand and get = 0 (since it's a quadratic equation)

n²+ 8n - 33 = 0

4) Factor and solve.

(n+11)(n-3) = 0

n = -11 n = 3

5) Get rid of the negative solution since it was to be a positive integer.

Therefore, the integers would be:

n = 3 and n+1 = 4.