Question 100405
Consecutive even integers follow the form: {{{2x}}}, {{{2x+2}}}, {{{2x+4}}}, ...., etc.



So the product of two consecutive even integers is:



{{{2x(2x+2)=288}}}



{{{4x^2+4x=288}}} Distribute



{{{4x^2+4x-288=0}}} Subtract 288 from both sides



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{4*x^2+4*x-288=0}}} ( notice {{{a=4}}}, {{{b=4}}}, and {{{c=-288}}})





{{{x = (-4 +- sqrt( (4)^2-4*4*-288 ))/(2*4)}}} Plug in a=4, b=4, and c=-288




{{{x = (-4 +- sqrt( 16-4*4*-288 ))/(2*4)}}} Square 4 to get 16  




{{{x = (-4 +- sqrt( 16+4608 ))/(2*4)}}} Multiply {{{-4*-288*4}}} to get {{{4608}}}




{{{x = (-4 +- sqrt( 4624 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-4 +- 68)/(2*4)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-4 +- 68)/8}}} Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


{{{x = (-4 + 68)/8}}} or {{{x = (-4 - 68)/8}}}


Lets look at the first part:


{{{x=(-4 + 68)/8}}}


{{{x=64/8}}} Add the terms in the numerator

{{{x=8}}} Divide


So one answer is

{{{x=8}}}




Now lets look at the second part:


{{{x=(-4 - 68)/8}}}


{{{x=-72/8}}} Subtract the terms in the numerator

{{{x=-9}}} Divide


So another answer is

{{{x=-9}}}


So our solutions are:

{{{x=8}}} or {{{x=-9}}}




Now plug in {{{x=8}}} to find the two numbers


{{{2(8)=16}}}

{{{2(8)+2=18}}}


So one pair of numbers is 16 and 18




Now plug in {{{x=-9}}} to find the two numbers


{{{2(-9)=-18}}}

{{{2(-9)+2=-16}}}


So another pair of numbers is -18 and -16