Question 933062
You can try proving through contradiction.


{{{(x+b)(x+b)}}} is a perfect square and the equivalent trinomial can be found.
{{{(x+b)^2=x^2+2bx+b^2}}}.
You would assume that {{{2b=6}}} and {{{b^2=6}}}.  You would expect to find ONE unique number for b which satisfies both of these equations.  Let's try.


{{{2b=6}}} means {{{b=3}}}.


{{{b^2=6}}} means  {{{b=0+- sqrt(6)}}}.


These two results for b are not equal.