Question

Find three consecutive positive even integers such that three times the square of the least integer plus seven times the square of the middle integer minus five times the square of the third integer is 328.

Answer 1

Let the three consecutive positive even integers be: x, x + 2, x + 4.

Given,

Three times the square of least integer plus seven times the square of middle integer minus five time the square of the third integer is 328.

⇒ 3x² + 7(x + 2)² − 5(x + 4)² = 328

⇒ 3x² + 7(x² + 4x + 4) − 5(x² + 8x + 16) = 328

⇒ 3x² + 7x² + 28x + 28 − 5x² − 40x − 80 = 328

⇒ 3x² + 7x² − 5x² + 28x − 40x − 80 + 28 = 328

⇒ 5x² − 12x − 52 = 328

⇒ 5x² − 12x − 52 − 328 = 0

⇒ 5x² − 12x − 380 = 0

⇒ 5x² − 50x + 38x − 380 = 0

⇒ 5x(x − 10) + 38(x − 10) = 0

⇒ (5x + 38)(x − 10) = 0

⇒ (5x + 38) = 0 or (x − 10) = 0

⇒ x = `−38/5`​ or x = 10

Since integers must be positive even numbers, x = 10.

Three consecutive positive even integers are:

x = 10

x + 2 = 10 + 2 = 12

x + 4 = 10 + 4 = 14

Hence, three consecutive positive even integers are 10, 12, 14.