<PRE><B>Question 36391</B>
find two consecutive odd integers such that twice the square of the first plus the square of the second totals ninety-nine.


Let the 2 odd integers be x, x+2
equation is 

2x^2 + (x + 2)^2 = 99
2x^2 + x^2 + 4x + 4 = 99
3x^2 + 4x + 4 = 99
3x^2 + 4x - 95 = 0
3x^2 + 19x - 15x - 95  = 0
x(3x + 19) - 5 (3x + 19) = 0
(x - 5)(3x + 19) = 0

x = 5, -19/3


Since x is a integer x = 5
and the other integer is 7


Verify
2x^2 + (x + 2)^2 = 99
2(5)^2 + (7)^2 = 99
2(25) + 49 = 99
50 + 49 = 99
99 = 99
</PRE>