# If the product of two consecutive negative multiples of 4 is 2112. what are the two numbers?

2
by sugu123

2015-12-24T13:07:01+05:30

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Let the two consequitive negative multiples be -4x and -4(x+1) where x is a positive integer.

-4x * -4(x+1) = 2112

=> 16x(x+1) = 2112

=> x(x+1) = 132

=> x^2 + x - 132 = 0

=> ( x + 12 ) ( x - 11 ) = 0

=> x = -12 or 11 , but since x is positive so x=11

so the multiples are : -4(11) = -44 & -4(12) = -48
2015-12-24T13:28:33+05:30

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Let the two consecutive negative numbers be 4x and 4(x+1)= 4x+4
Where x is less than 0
According to the question
(4x)(4x+4) = 2112
⇒16x² + 16 = 2112
⇒16x² + 16x - 2112 = 0
⇒16(x² + x - 132) = 0
⇒ x² +x - 132 = 0/16
⇒ x² +x -132 = 0
⇒x² +12x -11x -132 = 0
⇒x(x +12) - 11(x + 12) = 0
⇒(x+12)(x - 11) = 0
⇒ x +12 = 0             or          x - 11 =0
⇒x =-12                   or          x = 11
∵The x is less than 0, x=-12
The numbers are 4(-12) and 4(-12+1)
-48 and -44