There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. what are the integers?
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Let the first be x
second be x + 1
third be x + 2
x² + ( x + 1 ) ( x + 2 ) = 154
x² + x² + 2x + x + 2 = 154
2x² + 3x + 2 - 154 = 0
2x² + 3x - 152 = 0
2x² + 19x - 16x - 152 = 0
x ( 2x + 19 ) - 8 ( 2x + 19 ) = 0
( x - 8 ) ( 2x + 19 ) = 0
* ( x - 8 ) = 0
x = 8
* ( 2x + 19 ) = 0
x = -19/2 ( it is neglected because it is asked in the que that the no. are positive integer )
So,
° x = 8
° x + 1
8 + 1 = 9
° x + 2
8 + 2 = 10
The three Consecutive numbers are 8 , 9 and 10 !!
second be x + 1
third be x + 2
x² + ( x + 1 ) ( x + 2 ) = 154
x² + x² + 2x + x + 2 = 154
2x² + 3x + 2 - 154 = 0
2x² + 3x - 152 = 0
2x² + 19x - 16x - 152 = 0
x ( 2x + 19 ) - 8 ( 2x + 19 ) = 0
( x - 8 ) ( 2x + 19 ) = 0
* ( x - 8 ) = 0
x = 8
* ( 2x + 19 ) = 0
x = -19/2 ( it is neglected because it is asked in the que that the no. are positive integer )
So,
° x = 8
° x + 1
8 + 1 = 9
° x + 2
8 + 2 = 10
The three Consecutive numbers are 8 , 9 and 10 !!
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