n^2 + (n+2)^2 = 164
2n^2 + 4n - 160 = 0
n^2 + 2n - 80 = 0
(n+10)(n-8) = 0
...
The sum of the squares of two consecutive positive even integers is one hundred sixty-four. Find the two integers.
2 answers
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)
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)
1 answer