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## Answers

**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

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

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