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
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(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
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the numbers are 5,7,9
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}
