SOLUTION: find two consecutive odd integers such that the square of the larger is 3 than 12 times the smaller x, x+2

Question 192149: find two consecutive odd integers such that the square of the larger is 3 than 12 times the smaller

x, x+2
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
let smaller +x
let larger = x+2
The problem only makes sense if " is three than" reads " is 3 smaller than"
to solve this problem
l^2 = 12 S-3
(x+2)^2 =12(x) -3
x^2 +4x+4 =12x -3
subt 12 x both sides
x^2 +4x -12x +4 =-3
now add 3 to both sides
x^2 +4x -12x +4 +3 =0
x^2 -8x +7 =0
factoring
(x-7) (x-1) =0
setting both =0
x-7=0, x=7 (smaller) & larger =9
or x-1=0, x=1 (smaller) & larger is 3
checking (l^2=12*s-3)
3^2=12(1)-3 ok
9^2 = 12(7)-3 or 81=84-3 ok
Therefore both sets are correct, 1 &3, and 7&9
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