Question

There are three consecutive positive integers such that the sum of their square of first and the product of the other two is 154.find the integer which is a multiple of 3

Answer 1

let three consecutive positive integers
= x,x+1,x+2
so, x²+(x+1)(x+2) = 154
= x²+x²+2x+x +2= 154
2x²+3x-152= 0
= 2x²+19x-16x-154=0
( x-8)(2x+19)
x= 8
so, x+1 = 9
x+2= 10 ans.
so, the multiple of 3 is 9
so, the integer is x+1ans.
please ,mark as brainlist.!!

Answer 2

HOPE KI ANS ACHA HOGA AR BHUT ACHE SE SMJ BHI AYA HOGA... SO i.e..WHY MARK IT IS AS BRAINLIST. .

let first consecutive positive integer is X
second is X+1
third is X+2
A.T.Q

${x}^{2} +(x + 1)(x + 2) = 154 \\ {x }^{2} + {x}^{2} + 3x + 2 = 154 \\ 2 {x}^{2} + 3x + 2 - 154 = 0 \\ 2 {x}^{2} + 3x - 152 = 0 \\ 2 {x}^{2} + 19x - 16x - 152 = 0 \\ x(2x + 19) - 8(2x + 19) = 0 \\ (2x + 19)(x - 8) = 0 \\ 2x + 19 = 0 \: \: or \: \: x - 8 = 0 \\ 2x = - 19 \: \: \: or \: \: \: x = 8 \\ x = \frac{ - 19}{2} \: \: \: or \: \: \: x = 8 \\ \\ so \: x \: equal \: to \: \: 8 \: is \: corect \\ and9 \: is \: a \: positive \: no \: which \: is \: divisible \: by \: 3$ is
9