The product of the smallest and largest of three consecutive integers is three times the middle integer. Find the integers?

Answer 1

x-1, x, x+1 are 3 consecutive integers

(x-1)(x+1) = x^2 -1 = 3x

x^2 -3x -1 = 0

x^3 - 3x + 9/4 = 1 + 9/4 = 13/4

(x-9/2)^2 = 13/4

x-9/2 = + or - sqr(13/4)

x = 9/2 + or - (1/2)sqr13 = 4.5 + or - (1/2)(3.6) = 4.5 + or - 1.8 = 6.3 or 2.7

3 consecutive numbers, each 1 more than the other are 1.7, 2.7 and 3.7 or 5.3, 6.3 and 7.3 approximately

closest integers are 2, 3, and 4 or 5, 6 and 7

2x4 = 3x3 almost

8=9

5x7 = 3x6

35 isn't close to 18

try 3 integers as x, x+1 and x+2 where x= 1st integer

x(x+2) = 3(x+1)

x^2 +2x = 3x +3

x^2 -x -3 = 0

again it doesn't factor, so there are no 3 consecutive integers where the 1st x 3rd = 3 x 2nd

x^2 - x + 1/4 = 3 + 1/4

(x-1/2)^2 = 13/4

x = 1/2 + or - (1/2)sqr13 = .5 + or - 1.8 = -1.3 or 2.3,

3 consecutive numbers, each one more than the preceding are 2.3, 3.3 and 4.3

about 2, 3 & 4

2x4 = close to 3x3

8 is about 9

or -1.3, -0.3 and 1.3

but (-1.3)(1.3) = -1.69 which is closer to 3(0.3) = 0.9 than 8 is to 9

but closest integers are -1, 0 and 1, where (-1)(1) = -1 which is within 1 of 3(0) = 0

Best answer is there are no 3 consecutive integers satisfying 1st x 3rd = 3 x 2nd

but if you had to pick 3 numbers, best ones are -1.3, -0.3 and 1.3

best integers are either 2,3 and 4, or -1,0 and 1