The sum of two squares of TWO consecutive even integers is 340. Find the integers.

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  1. first we represent unknowns using variables,,
    let x = first number
    let x+2 = second number
    note that since they are consecutive but both even numbers, the difference between them is 2, that's why the second number is x+2
    now we set-up the equation,, since the sum of the squares (meaning both number are raised to 2) is equal to 340,
    (x)^2 + (x+2)^2 = 340
    x^2 + x^2 + 4x + 4 = 340
    combine similar terms and simplify:
    2x^2 + 4x + 4 - 340 = 0
    2x^2 + 4x - 336 = 0
    we can factor out 2 and cancel it:
    2(x^2 + 2x - 168) = 0
    x^2 + 2x - 168 = 0
    since this quadratic equation is factorable,
    (x + 14)(x - 12) = 0
    x = -14 and x = 12
    thus there are two pairs of answers:
    (i) x = -14 and x+2 = -12
    (ii) x = 12 and x+2 = 14

    hope this helps~ :)

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  2. So the average of the two squares is 340/2=170.
    Sqrt(170)=13 (approx.)
    What would you propose for the two consecutive even integers?

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