<PRE><B>Question 127004</B>
First integer: x
Second integer: x + 1
Third integer: x + 2


3 times the square of the first:  {{{3x^2}}}

The product of the other two:  {{{(x+1)(x+2)=x^2+3x+2}}}


3 times the square of the first exceeds the product of the other two by 25:{{{3x^2=(x^2+3x+2)+25}}}


{{{3x^2-x^2-3x-27=0}}}


{{{2x^2-3x-27}}}


{{{(2x-9)(x+3)=0}}}


So {{{x=9/2}}} or {{{x=-3}}}.  We can exclude the first root because it is not an integer.  So the first integer is -3, the second would be -2, and the third, -1.


Check:
{{{3(-3)^2=27}}}
{{{(-1)(-2)+25=27}}}, answer checks</PRE>