Question 483660
This problem is not correct
PLEASE, LOOK PERHAPS YOU MISS WORD EVEN??
Let x is the first number, then next even number is (x+2)
The produt of these numbers is {{{x(x+2)=x^2+2x}}}, three times their sum is {{{3(x+(x+2))=3(2x+2)=6x+6}}}
The equation
{{{x^2+2x-6=6x+6}}}
{{{x^2+2x-6-6x-6=0}}}
{{{x^2-4x-12=0}}}
{{{(x-6)(x+2)=0}}}
{{{x-6=0}}} or {{{x+2=0}}}
{{{x=6}}} or {{{x=-2}}}
If {{{x=6}}}, then the second number is {{{6+2=8}}}
If {{{x=-2}}}, then the second number is {{{-2+2=0}}}
Answer 6 and 8, or -2 and 0