Question 1206861: given four consecutive odd numbers.when the square of the second number is subtrcted from the product of the first and last numbers,the answer is 22.find them.
Found 3 solutions by ankor@dixie-net.com, greenestamps, josgarithmetic:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! given four consecutive odd numbers.
n, n+2, n+4, n+6
when the square of the second number is subtracted from the product of the first and last numbers,the answer is 22.
n(n+6) - (n+2)^2 = 22
find them.
(n^2 + 6n) - (n^2 + 4n + 4) = 22
Combine like terms, removing bracket on the 2nd expression changes sign to -
n^2 - n^2 + 6n - 4n - 4 = 22
2n = 22 + 4
2n = 26
n = 13
The 4 numbers 13, 15, 17, 19
Check:
13*19 - 15^2 =
247 - 225 = 22
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
Looking ahead to see that the second of the four consecutive odd numbers is squared, I will guess that the algebra required to solve the problem might be easiest if I let x be the second number. So the four consecutive odd numbers are x-2, x, x+2, and x+4.
Then the problem says
The second number is x=15; the four numbers are then 13, 15, 17, and 19.
ANSWER: 13, 15, 17, and 19
CHECK: (13*19)-15^2=247-225=22
Answer by josgarithmetic(39630)  (Show Source):
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