Question 113493

The product of two {{{consecutive}}}{{{ even}}} intergers is {{{360}}}. 

let first one be {{{x}}}, and second one be {{{x+2}}}

given:

{{{x(x+2) = 360}}}

{{{x^2 + 2x = 360}}}

{{{x^2 + 2x - 360 = 0}}}.....use square root formula

{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-2 +- sqrt (2^2 -4*1*(-360) )) / (2*1)}}}


{{{x[1,2]=(-2 +- sqrt (4 + 1440 ))/2}}}

{{{x[1,2]=(-2 +- sqrt (1444 ))/2}}}

{{{x[1,2]=(-2 +- 38)/2}}}

{{{x[1]=(-2 + 38)/2}}}

{{{x[1]= 36/2}}}

{{{x[1]= 18}}}............one slution

{{{x[2]=(-2 - 38)/2}}}

{{{x[2]= -40/2}}}

{{{x[2]= -20}}}..........the other slution


check both solutions:
if {{{x = 18}}}, then second integer will be {{{18 + 2}}} which is {{{20}}}

let's check their product: 


{{{x(x+2) = 360}}}

{{{18*20 = 360}}}

{{{360= 360}}}...........



if {{{x = -20}}}, then second integer will be {{{-20 + 2}}} which is {{{-18}}}

let's check their product: 


{{{-20(-18) = 360}}}

{{{360 = 360}}}............in both cases we have correct answer

so, your integers are:

1. {{{18}}}and {{{20}}}

2. {{{-20}}} and {{{-18}}}