Question 405935
"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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