Question

product of two consecutive even integers is 168, find integers.

Answer 1

Let first integer be x , then devon will be x+2
A/Q x(x+2)=168
x^2 +2x=168
x^2+2x-168=0
x^2+14x-12x-168=0
×(x+14) -12(x+14)=0
so, x+14=0
and x-12=0
so x=. -14
and x=. 12

Answer 2

Let the two consecutive even integers be (2x) , (2x+2).
According to question,
=> (2x)(2x+2) = 168
=> 4x(x+1) = 168
=> x(x+1) = 168/4
=> x^2 + x = 42
=> x^2 + x - 42 = 0
=> x^2 + 7x - 6x - 42 = 0
=> (x+7)(x-6) = 0
=> x = -7 , 6
So, when x = -7, then
integers are 2x and 2x+2 i.e 2(-7) and 2(-7)+2
i.e -14 , -12.
and when x= 6 then,
integers are 2x and 2x+2 i.e 2(6) and 2(6)+2
i.e 12 and 14.