Question

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

Answer 1

16, 18, 20

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) = (972)/(3) \\ \\ \therefore \: {x}^(2) = 324 \\ \therefore \: x = \pm √(324) \\ \therefore \: x = \pm \: 18 \\ \because \: x \: is \: even \: integer \\ \therefore \: x \\eq - 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.

Answer 2

16, 18, 20

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.

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Check

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