Question

Find four consecutive multiples of three that the product of the first and third miners minus eight times the fourth number is negative 48

Answer 1

The four consecutive multiples of three are 6, 9, 12, and 15

Explanation:

let the first multiple of 3 = n

second multiple of 3 = (n + 3)

third multiple of 3 =(n + 6)

fourth multiple of 3 =(n + 9)

product of the first and the third = n(n + 6) = n² + 6n

eight times the fourth number = 8(n + 9) = 8n + 72

The difference between the two = n² + 6n - (8n + 72) = -48

n² + 6n - 8n - 72 = -48

n² - 2n = -48 + 72

n² - 2n = 24

n² - 2n - 24 = 0

n² - 2n - 24 = 0

n² + 4n - 6n - 24 = 0

n(n + 4) - 6(n + 4) = 0

(n - 6)(n + 4) = 0

n = 6 or - 4

n should be a positive number, thus n = 6

The second multiple of 3 = (n + 3) = 6 + 3 = 9

The third multiple of 3 =(n + 6) = 6 + 6 = 12

The fourth multiple of 3 = (n + 9) = 6 + 9 = 15