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

Answer 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~ :)

Answer 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?