Question 1083261
Some (not all) quadratic equation can be solved by factoring.
Every quadratic equation can be solved by completing the square,
and every quadratic equation can be solved by using the quadratic formula.
People usually resort to the quadratic formula,
but professor Peter Alfeld of the university of Utah
believes we should complete squares.
I use what comes easier. When factoring works, it may be easier.
 
{{{3x^2+ax^2=9x+9a}}}
{{{(3+a)x^2-9x=9a}}}
Maybe dividing everything by {{{(3+a)}}} at this point would be easier.
Of course, we will have to add {{{a<>-3}}} as a requirement to any solution we find as a function of a.
{{{x^2-(9/(3+a))x=9a/(3+a)}}}
{{{x^2-(9/(3+a))x+(9/(2(3+a)))^2=9a/(3+a)+(9/(2(3+a)))^2}}}
{{{(x-9/(2a+6))^2=9a/(3+a)+(9/(2(3+a)))^2}}}
{{{(x-9/(2a+6))^2}}}{{{"="}}}{{{9a(4(a+3))/(2a+6)^2}}}{{{"+"}}}{{{9^2/(2a+6)^2}}}
{{{(x-9/(2a+6))^2}}}{{{"="}}}{{{(9(4a^2+12)+81)/(2a+6)^2}}}
{{{(x-9/(2a+6))^2}}}{{{"="}}}{{{9(4a^2+12a+9)/(2a+6)^2}}}
{{{(x-9/(2a+6))^2}}}{{{"="}}}{{{9(2a+3)^2/(2a+6)^2}}}
{{{x=9/(2a+6) +- 3(2a+3)/(2a+6)}}}
{{{x=(9 +- (6a+9))/(2a+6)}}}
So, the two solutions are
{{{x[1]=(6a+18)/(2a+6)=3}}} and {{{x[2]=6a/(2a+6)=3a/(a+3)}}} if {{{a<>-3}}}
 
WAIT!
Why did we not realize that we could solve by factoring?
{{{3x^2+ax^2=9x+9a}}}
{{{ax^2-9a+3x^2-9x=0}}}
{{{3x(x-3)=a(9-x^2)}}}
{{{a(x^2-9)+3x(x-3)=0}}}
{{{a(x+3)(x-3)+3x(x-3)=0}}}
{{{(a(x+3)+3x)(x-3)=0}}}
{{{(ax+3a+3x)(x-3)=0}}}
{{{((a+3)x-3a)(x-3)=0}}}
At this point, either {{{x-3=0}}} ---> {{{x=3}}} , or
{{{(a+3)x-3a=0}}} ---> {{{x=3a/(a+3)}}} as long as {{{a<>-3}}} .