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

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Question 189366: what is the smallest of three positive consecutive odd integers if the product of the second and third intergers is 63??
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the 3 consecutive odd integers
a, a+%2B+2, and a+%2B+4
given:
%28a+%2B+2%29%28a+%2B+4%29+=+63
a%5E2+%2B+6a+%2B+8+=+63
a%5E2+%2B+6a+=+55
I'll use complete the square method:
a%5E2+%2B+6a++%2B+%286%2F2%29%5E2+=+55+%2B+%286%2F2%29%5E2
a%5E2+%2B+6a++%2B+9+=+55+%2B+9
%28a+%2B+3%29%5E2+=+64
Take the square root of both sides
a+%2B+3+=+8
a+=+5
and
a+%2B+3+=+-8
a+=+-11 (I can't use a negative result)
a+=+5
a+%2B+2+=+7
a+%2B+4+=+9
(notice that 7%2A9+=+63)
The smallest number is 5