Question

Find three consecutive positive integers such that the sum of the square of the first integer and the product of the other 2 is 92

Answer 1

Let the first consecutive integer be x
second be ( x + 1 )
third be ( x + 2 )

• Given :- Sum of the square of the first integer and the product of the other two is 92

x² + ( x + 1 ) ( x + 2 ) = 92

x² + x² + 2x + x + 2 = 92

2x² + 3x = 92 - 2

2x² + 3x - 90 = 0

2x² + 15x - 12x - 90 = 0

x ( 2x + 15 ) - 6 ( 2x + 15 ) = 0

( x - 6 ) ( 2x + 15 ) = 0

• ( x - 6 ) = 0
x = 6

• 2x + 15 = 0

x = 15/2 ..... ( neglected bcoz it is given in the question that the integers are positive )

So,
x = 6

First integer is 6
Second is ( x + 1 ) i.e , 7
Third is ( x + 2 ) i.e, 8