SOLUTION: Find three consecutive odd numbers such that the square of the second number is 192 less than the square of the third number. WE ARE STUDYING POLYNOMIALS AND I NEED SOME HEL

Question 633892: Find three consecutive odd numbers such that the square of the second number is 192 less than the square of the third number.

WE ARE STUDYING POLYNOMIALS AND I NEED SOME HELP WITH THE WORD PROBLEMS PLEASE
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find three consecutive odd numbers such that the square of the second number is 192 less than the square of the third number.
------
1st: 2x-1
2nd: 2x+1
3rd: 2x+3
-------
Equation:
(2x+1)^2 = (2x+3)^2-192
---
4x^2+4x+1 = 4x^2+12x+9 - 192
---
-8x = -184
x = 23
---
1st: 2x-1 = 2*23-1 = 45
2nd: 47
3rd: 49
----------------
Cheers,
Stan H.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find three consecutive odd numbers such that the square of the second number
is 192 less than the square of the third number.
:
Let x = the 1st integer, then (x+2) and (x+4) are the next two odd integers
:
Write an equation for the statement:
"the square of the second number is 192 less than the square of the third
number."
(x+2)^2 = (x+4)^2 - 192
FOIL
x^2 + 4x + 4 = x^2 + 8x + 16 - 192
Combine the variables on the left
x^2 - x^2 + 4x - 8x = 16 - 192 - 4
:
-4x = -180
x =
x = +45 is the 1st odd number
then
47 and 49 are the 2nd and 3rd odd digits
:
:
Check
47^2 = 49^2 - 192
2209 = 2401 - 192

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