the sum of the squares of two consecutive positive integers is 244, find the integers
umm what the heck, haha someone please help an explain!!
2007-10-23 02:41:43 · 4 answers · asked by sparklechic2010 5 in Science & Mathematics ➔ Mathematics
two consecutive positive even integers
sorry!!!!
2007-10-23 03:01:20 · update #1
I don't think there exist 2 CONSECUTIVE positive integers where the sum of their squares is 244.
10 squared + 11 squared = 100 + 121 = 221
11 squared + 12 squared = 121 + 144 = 265
Could you have perhaps meant two consecutive EVEN
integers?
10 squared + 12 squared = 100 + 144 = 244.
In that case let x be one even integer. Let x + 2 be the next even integer.
x^2 + (x + 2)^2 = 244
x^2 + (x^2 + 4x + 4) = 244
2x^2 + 4x + 4 = 244
2x^2 + 4x + -240 = 0
Use the quadratic formula to solve for x, and you would get 10.
2007-10-23 02:48:31 · answer #1 · answered by SoulDawg 4 UGA 6 · 1⤊ 0⤋
(n)^2 + (n+1)^2 = 244
n^2 + n^2 + n + 1 =244
2n^2 + n + 1 = 244
2n^2 + n = 243
(2n+1)n = 243
dosent work??? only with fractions
2007-10-23 09:45:25 · answer #2 · answered by Gengi 5 · 1⤊ 0⤋
something wrong with your question there are no two consecutive integer #
2007-10-23 09:58:03 · answer #3 · answered by xmashi 3 · 0⤊ 0⤋
x^2+(x+1)^2=244
x^2+x^2+2X+1=244
2x^2+2x-243=0
so we cant get integers. only fraction will come.
question s wrong
2007-10-23 09:51:13 · answer #4 · answered by sathyanarayanan k 2 · 0⤊ 0⤋