The question. Find two consecutive even integers whose sum of their squares is 244? Please help
2007-06-13 08:59:31 · 9 answers · asked by Ashley C 1 in Science & Mathematics ➔ Mathematics
We need to set up the following equation to solve for the first number of the two:
x^2 + (x + 2)^2 = 244, where
x = "the first number"
x + 2 = "the second number"
So, we just solve:
x^2 + (x + 2)^2 = 244
x^2 + (x^2 + 4x + 4) = 244
2x^2 + 4x + 4 = 244
2x^2 + 4x - 240 = 0
x^2 + 2x - 120 = 0
(x + 12)(x - 10) = 0
x = 10, x = -12.
So, we have two values for "the first number," so we have two sets of solutions:
10 and 12
-12 and -10
2007-06-13 09:02:06 · answer #1 · answered by C-Wryte 3 · 0⤊ 0⤋
Let the two consecutive even integers be 2n and (2n+2) for some n greater than or equal to 1. Then you are seeking to find the n that satisfies: (2n)^2 + (2n+2)^2 = 244. Simplifying this leads to 4n^2 + 4(n+1)^2 = 244 or n^2 + (n+1)^2 = 61. You can then solve for n as follows:
n^2 + (n^2 + 2n +1) = 61 implying 2n^2 +2n - 60 = 0 or n^2 + n - 30 = 0. This can be factored via trial and error or by using the quadratic equation as (n + 6)(n - 5) = 0. Since we are looking for positive answers, n = 5. This means that the two consecutive even integers are 10 and 12 (check to see that this right).
[Note that -12 and -10 also work and are consecutive (negative) even intergers. This is the answer you get when n = -6.]
Math Rules!
2007-06-13 09:10:45 · answer #2 · answered by Math Chick 4 · 0⤊ 0⤋
The integers are represented by x and x+2
x^2+(x+2)^2=244 which simplifies to
x^2-2x-120=0 which has roots of 10 and -12
There are two answers 10 and 12 or -12 and -10
2007-06-13 09:07:01 · answer #3 · answered by Anonymous · 0⤊ 0⤋
10 & 12
2007-06-13 09:04:26 · answer #4 · answered by obsolete professor 4 · 0⤊ 0⤋
2x = 1st integer
2x+2 = nex integer
(2x)^2 + (2x+2)^2 = 244
4x^2 + 4x^2 +8x + 4 = 244
8x^2 +8x -240 = 0
x^2 + x -30 = 0
x+6)(x-5) = 0
x = 5 and x = -6
2x = 10 = 1st integer
2x+2 = 12 = 2nd integer
2x = -12
2x+2 = -10
Abo are a 2nd set of answers
2007-06-13 09:08:56 · answer #5 · answered by ironduke8159 7 · 0⤊ 0⤋
10 and 12
10^2 = 100
12^2 = 144
100+144 = 244
2007-06-13 09:04:08 · answer #6 · answered by Bryan W 1 · 1⤊ 0⤋
x^2 +(x+2)^2 =244
x^2 +x^2 +4 +4x=244
2x^2 +4x -240=0
x^2 +2x -120=0
(x-10)(x+12)=0
so x=10
so the two nos are 10 and 12.
2007-06-13 09:02:49 · answer #7 · answered by sweet n simple 5 · 0⤊ 1⤋
n^2 + (n+2)^2 = 244
2n^2 + 4n +4 = 244
2n^2 + 4n -240 = 0
n^2 + 2n - 120 = 0
(n+12)(n-10) = 0
n = -12 or n = 10
So you have two answers:
(-12 and -10) or (10 and 12)
2007-06-13 09:07:03 · answer #8 · answered by MathProf 4 · 0⤊ 0⤋
why are you doing homework... ITS SUMMER!!!!!!!!!
2007-06-13 09:03:31 · answer #9 · answered by hoopla 2 · 0⤊ 3⤋