SOLUTION: what is the smallest of three positive consecutive odd integers if the product of the second and third integers is 63?

Question 405935: what is the smallest of three positive consecutive odd integers if the product of the second and third integers is 63?
Found 2 solutions by mananth, algebrahouse.com:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let the consequtive odd integers be n , n+2, n+4
...
(n+2)(n+4)=63
n^2+6n+8=63
n^2+6n-55=0
n^2+11n-5n-55=0
n(n+11)-5(n+11)=0
(n+11)(n-5)=0
n=5 positive integer
...
the numbers are 5,7,9
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"what is the smallest of three positive consecutive odd integers if the product of the second and third integers is 63?"

x = first odd integer
x + 2 = second odd integer
x + 4 = 3rd odd integer

(x + 2)(x + 4) = 63 {product of second and third is 63}
x^2 + 6x + 8 = 63 {used foil method}
x^2 + 6x - 55 = 0 {subtracted 63 from both sides}
(x + 11)(x - 5) = 0 {factored into two binomials}
x + 11 = 0 or x - 5 = 0 {set each factor equal to 0}
x = 5 {first positive odd integer}
x + 2 = 7 {second positive odd integer}
x + 4 = 9 {third positive odd integer}

5 would be the smallest positive odd integer
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