Answers

2015-12-14T20:25:12+05:30
Let us take first number as x
Then the second number becomes x + 1
Acc. to question,
(x)(x+1) = 3192
x2 + x - 3192 = 0
x2 + 57x - 56x - 3192 = 0                                           (via factorisation)
x ( x + 57 ) - 56 ( x + 57 ) = 3192
(x - 56) ( x + 57)

so the smallest numbers are 56 and -56
since,
56 x 57=3192
-56 x -57 = 3192
1 5 1
please make correction in the last fourth line[ so the smallest no`s are 56 and -57] 
2015-12-14T20:45:50+05:30
Let the numbers be x and x+1
Given that the product of these two numbers = 3192
(x)(x+1) = 3192
⇒x² + x - 3192 = 0
⇒x² - 56x +57x - 3129 = 0
⇒x(x-56) + 57(x-56) = 0 
⇒(x-56) (x+57) = 0
⇒(x-56) = 0    or    (x+57) = 0 
⇒x = 56       or    x = -57
∴The numbers must be  x=56 , x+1=56+1=57
∵56×57= 3129                          
                                     OR
The numbers must be x = -57 , x+1= -57+1= -56
∵-57×-56=3129
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