Question

Find two consecutive odd integers such that their product is 23 more than 5 times their sum.

Answer 1

If x is the smaller one, and is odd, the next consecutive odd integer is x+2.

$x(x+2) = 23 + 5(x+ (x+2))$

$x^2 + 2x = 23 +5(2x+2)$

$x^2 + 2x = 23 + 10x + 10$

$x^2 -8x - 33 = 0$

$(x-11)(x+3) = 0$

$x = 11\textrm{ or }x=-3$

Check:

11(13) = 143

23 + 5(11 + 13) = 23 + 120 = 143, check

-3(-1) = 3

23 + 5(-3 + -1) = 3, check

Answer: 11,13 or -3,-1