Find three consecutive integers such that the product of the first and third integer is 9 greater than 3 times the second

Answer 1

Let the 3 numbers be: x-1, x, and x+1

(x-1)(x+1)>9+3x

x^2-1>9+3x

x^2-3x -10 > 0

(x-5)(x+2)>0

x-5>0 or x+2<0

x>5 or x<-2

(negative infinity,-2) U (5,infinity) is the range of values of x

6, 7, 8 has 6x8 > 9+3(7)

48 > 30

7, 8, 9 has 7x9 > 9+3(8)

63 > 33

any three consecutive integers with the 2nd greater than 5 will satisfy the inequality

5, 6, 7 has 5x7 > 9+3(6)

35 > 27

but

4,5,6 has 4x6 = 9+3(5)

24 = 24

so 4,5,6 won't work

-5,-4,--3 has

(-5)(-3) > 9+ 3(-4)

15 > -3

but

-3,-2,-1 has

(-3)(-1) = 9+3(-2)

3=3

the 2nd integer has to be less than -2 or greater than 5