For my intermediate algebra class
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I guess the product should be 156. If p is one page number the other is p+1 and p(p+1)=156.

p²+p-156=0=(p+13)(p-12) so p=12 and p+1=13. The book is open at pages 12 and 13.

by Top Rated User (1.0m points)
I think that the product of the two facing pages of the book is only 156.

If we work with the 156156 we dont get integer numbers as a solution to the problem, as it should be.

So if the P is for the one page the next facing page is P + 1 then we will have P(P+1) = 156

so P2 + P -  156 = 0  we get P = 12 (the second negative solution is rejected) the next page will be P+1 = 13

chck; 12 x 13 = 156
by Level 5 User (13.1k points)

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