if the product of two consecutive negative multiples of 4 is 2112. what are the two numbers?
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Answered by
2
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
-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
Answered by
2
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
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
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