Answers

2014-08-11T20:40:05+05:30
Let the number be=X and (X+1)

according to the sum, 
   x(x+1) = 3192
   x(square) + x -3192=0
   x(square) +57x - 56x - 3192 =0
   x(x + 57) - 56(x + 57) =0
  (x-56)(x + 57) =0
  
value of x= 56, -57
 the required number, that is the smallest number is-56.


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haven't i typed the same ? except for the square, i do not know how to do that..
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i'm sorry , you've not seen the answer properly.. this is how , you ought to solve..
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2014-08-11T20:49:59+05:30
The sum is as follows

let the two numbers be x and x+1
so product is x(x+1)
given x(x+1)= 3192
x²+x-3192=0
x²+57x-56-x-3192=0
x(x+57)-56(x+57)=0
(x-56)(x+57)=0
the numbers are 56 and -57
since it is two positive integers -57 is not valid
so the numbers are 56 and 56+1
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