Question

Find two consecutive positive even integers such that the square of the first decreased by 64 equals 3 times the second.

Answer 1

Answer:

10 and 12

Step-by-step explanation:

let the consecutive even integers be n and n + 2 , then

n² - 64 = 3(n + 2) ← distribute parenthesis

n² - 64 = 3n + 6 ( subtract 3n + 6 from both sides )

n² - 3n - 70 = 0 ← in standard form

(n - 10)(n + 7) = 0 ← in factored form

Equate each factor to zero and solve for n

n - 10 = 0 ⇒ n = 10

n + 7 = 0 ⇒ n = - 7

Since n must be a positive even integer then n = 10 and n + 2 = 10 + 2 = 12

The 2 numbers are 10 and 12