SOLUTION: the sum of the square of the two consecutive odd intergers is equal to 290 find the two intergers

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Question 165493: the sum of the square of the two consecutive odd intergers is equal to 290 find the two intergers
Found 2 solutions by Fombitz, jojo14344:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let n be the first integer, (n+2) would be the next consecutive odd integer.
n%5E2%2B%28n%2B2%29%5E2=290
n%5E2%2B%28n%5E2%2B4n%2B4%29=290
2n%5E2%2B4n-286=0
n%5E2%2B2n-143=0
You can factor this quadratic equation.
%28n%2B13%29%28n-11%29=0
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First solution:
n%2B13=0
n=-13
The two integers are then -13 and -11.
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Second solution:
n-11=0
n=11
The two integers are then 11 and 13.

Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!

Two consecutive ODD Integers ----system%28x%2Cx%2B2%29
Condition:
x%5E2%2B%28x%2B2%29%5E2=290, EQN 1
Continuing,
x%5E2%2Bx%5E2%2B4x%2B4=290
2x%5E2%2B4x%2B4-290 ---> 2x%5E2%2B4x-286=0, divide whole eqn by 2

x%5E2%2B2x-143-0
Remember ---system%28a=1%2Cb=2%2Cc=-143%29
By PYTH.THEOREM:
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x=%28-2%2B-sqrt%282%5E2-4%281%29%28-143%29%29%29%2F%282%2A1%29
x=%28-2%2B-sqrt%284%2B572%29%29%2F2
x=%28-2%2B-sqrt%28576%29%29%2F2 ----> x=%28-2%2B-24%29%2F2
2 VALUES:
x=%28-2%2B24%29%2F2=22%2F2
highlight%28x=11%29
x=%28-2-24%29%2F2=-26%2F2
x=-13
USE HIGHLIGHTED ONE:
x=11 -------------> 1st odd intgr.
x%2B2=11%2B2=13 ------> 2nd odd intgr.
Check via EQN 1,
11%5E2%2B13%5E2=290
121%2B169=290
290=290
Thank you,
Jojo