Write and solve an equation to find three consecutive even integers such that the product of the first and the second integers are 8 more than the third integer.

Answer 1

FS= 8+T F= first number, S=second, T= third consecutive even integer

S=F+2

T=F+4

F(F+2) = 8 +F+4

F^2 +2F = F+12

F^2 +F -12 = 0

(F+4)(F-3) = 0

F =3 or -4

only -4 is even

S=-2

T = 0

the 3 consecutive even numbers are -4, -2 and 0

(-4)(-2) = 8 + 0