find the smallest of three consecutive integers such that the product if the largest two integers is 626 more than the smallest integer

Answer 1

Let x be the smallest integer. Then the other two are x+1 and x+2.

(x+1)(x+2) = x + 626 = x²+3x+2 -> 0 = x²+2x-624

Using the quadratic formula

x = (-2 +/- sqrt(4-(4*(-624))))/2 = (-2 +/- sqrt(4-(-2496)))/2 = (-2 +/- sqrt(2500))/2 = (-2 +/- 50)/2 = -1 +/- 25

x = -26 or 24

-25 * -24 = 600 and -26 + 626 = 600

25 * 26 = 650 and 24 + 626 = 650

There are 2 sets of valid numbers: (-26,-25,-24) and (24,25,26)