Posted by **LaShawn** on Monday, February 4, 2013 at 1:23am.

The sum of the squares of two consecutive positive even integers is one hundred sixty-four. Find the two integers.

- Algebra -
**Steve**, Monday, February 4, 2013 at 5:24am
n^2 + (n+2)^2 = 164

2n^2 + 4n - 160 = 0

n^2 + 2n - 80 = 0

(n+10)(n-8) = 0

...

- Algebra -
**HELP HELP HELP MNHS**, Monday, February 4, 2013 at 5:31am
x 1st + even integer

x+2 2nd + even integer

x^2+(x+2)^2= 164

x^2+x^2+4x+4-164=0

1/2(2x^2+4x-160=0)

x^2+2x-80=0

(x+10)(x-8)=0

x=-10 & x=8, disregard -10 since we are looking for positive integers...

so, x=8 (1st even)

x+2= 10 (2nd even)

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