SOLUTION: What is the answer for this problem? Find two consecutive even integers such that the sum of their squares is 52.

Question 180776: What is the answer for this problem?
Find two consecutive even integers such that the sum of their squares is 52.
Found 2 solutions by stanbon, nerdybill:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find two consecutive even integers such that the sum of their squares is 52.
-------------
1st: 2x
2nd: 2(x+1)
----------------
Equation:
(2x)^2 + (2(x+1))^2 = 52
4x^2 + 4(x+1)^2 = 52
x^2 + (x+1)^2 = 13
2x^2 + 2x + 1 = 13
2x^2 + 2x - 12 = 0
x^2 + x - 6 = 0
(x+3)(x-2) = 9
Positive solution:
x = 2
---------
1st: 2x = 4
2nd: 2x+2 = 6
---
Checking: 4^2 + 6^2 = 52
16 + 36 = 52
52 = 52
==================
Cheers,
Stan H.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Find two consecutive even integers such that the sum of their squares is 52.
.
Let x = 1st even integer
then
x+2 = 2nd even integer
.
x^2 + (x+2)^2 = 52
x^2 + x^2+4x+4 = 52
2x^2+4x+4 = 52
2x^2+4x-48 = 0
x^2+2x-24 = 0
(x+6)(x-4) = 0
.
x = {-6,4}
.
Solution: 4 and 6

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