Question

The sum of the squares of three consecutive even integers is 980. Determine the integers.

Answer 1

Answer:

  16, 18, 20

Step-by-step explanation:

We can estimate that the square of the middle integer will be about 1/3 of 980. Then the middle integer is about ...

  √(980/3) ≈ 18.09

The integers are 16, 18, 20.

_____

Check

  16^2 +18^2 +20^2 = 256 +324 +400 = 980

Answer 2

Answer:

16, 18, 20

Step-by-step explanation:

Let the three consecutive even integers be (x - 2), x, (x + 2)

$\therefore \: {(x -2 )}^{2} + {x}^{2} + {(x + 2)}^{2} = 980 \\ \therefore \: {x}^{2} - 4x + 4 + {x}^{2} + {x}^{2} + 4x + 4 = 980 \\ \therefore \: 3{x}^{2} + 8 = 980 \\ \therefore \: 3{x}^{2} = 980 - 8 \\ \therefore \: 3{x}^{2} = 972 \\ \\ \therefore \: {x}^{2} = \frac{972}{3} \\ \\ \therefore \: {x}^{2} = 324 \\ \therefore \: x = \pm \sqrt{324} \\ \therefore \: x = \pm \: 18 \\ \because \: x \: is \: even \: integer \\ \therefore \: x \neq - 18 \\ \therefore \: x = 18 \\ \\ x - 2 = 18 - 2 = 16 \\ x = 18 \\ x + 2 = 18 + 2 = 20$

Hence, three consecutive even integers are : 16, 18, 20.