Hi;
is it possible to complete the square on a perfect square trinomial?
I don't understand what you are asking. Do you have an example?
This is the best example I could come up with:
If $a,b,f\ge 0$:
$$ax^2+by^2+cxy+dx+ey+f = (gx+hy+i)^2+jxy+kx+ly$$
Multiplying out gives:
$$ax^2+by^2+cxy+dx+ey+f = g^2x^2+h^2y^2+2ghxy+2gix+2hiy+i^2+jxy+kx+ly$$
Comparing like terms, we have:
$$g^2=a \Longrightarrow g=\sqrt{a}$$
$$h^2=b \Longrightarrow h=\sqrt{b}$$
$$2gh+j = c \Longrightarrow j=c-2\sqrt{ab}$$
$$i^2 = f \Longrightarrow i=\sqrt{f}$$
$$2gi+k = d \Longrightarrow k=d-2\sqrt{af}$$
$$2hi+l = e \Longrightarrow l=e-2\sqrt{bf}$$
Is that what you mean? Because this is one of many possible solutions.
In general, you have:
$$ax^2+by^2+cxy+dx+ey+f = (gx+hy+i)^2+jxy+kx+ly+m$$
But, this gives six coefficients on the LHS with seven variables on the RHS (in other words, you have to choose a value for one of the RHS coefficients). In my example above, I chose $m=0$. I could just as easily have chosen the value for $g, h, i, j, k,$ or $l$. That is what I mean when I say there are many ways this could be done.
Ooh! That is totally different from what I was describing.
$$4x^2+12x+9=(ax+b)^2=a^2x+2abx+b^2$$
By equating coefficients on the LHS and RHS, we have $a^2=4, 2ab=12, b^2=9$. This implies $a=\pm 2, b=\pm 3$. So there are two solutions: $4x^2+12x+9=(2x+3)^2=(-2x-3)^2$