Question 33038: Can you please solve this for me? I have to solve this using factoring:
The sum of the squares of two consecutive negative even integers is 340. Find the integers.
Found 3 solutions by xcentaur, ikleyn, greenestamps:
Answer by xcentaur(357)   (Show Source):
You can put this solution on YOUR website! sum of the squares of two consecutive negative even integers is 340
let the first number be x
it must be negative and even,so it becomes '-2x'
Then the second number becomes '-2x+2'
So we get,
(-2x)^2+(-2x+2)^2=340
4x^2+(4x^2+4-8x)=340
8x^2-8x+4-340=0
8x^2-8x-316=0
Solve for x. then '-2x' and '-2x+2' give you the required numbers
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
.
Can you please solve this for me? I have to solve this using factoring:
The sum of the squares of two consecutive negative even integers is 340. Find the integers.
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I will solve it in as simple way as I can.
We are looking for two consecutive even integer numbers n and (n+2).
I will start from the central integer number 'm' between n and (n+2), so that
n = m-1, n+2 = m+1.
Then my equation is
(m-1)^2 + (m+1)^2 = 340,
(m^2 - 2m + 1) + (m^2 + 2m + 1) = 340,
2m^2 + 2 = 340,
2m^2 = 340 - 2 = 338,
m^2 = 338/2 = 169,
m = +/-  = +/- 13.
We are looking for two consecutive negative numbers, so these numbers are -14 and - 12. ANSWER
CHECK. (-14)^2 + (-12)^2 = 196 + 144 = 340. ! Precisely correct !
Notice that the other tutor reduced the problem to solution of a quadratic equation, but left the solution to you.
I solved the problem completely in a simplest way, practically mentally
to the end, without solving a quadratic equation.
Answer by greenestamps(13326)  (Show Source):
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