# Why is it you can change the value of an equation when completing the square by having $a=1$ in equation form $ax^2+bx+c=0$?

When completing the square for a given polynomial in the form

$$Ax^2+Bx+C=0$$

the first step is to ensure that $A =1$. If $A$ is not $1$ then you would divide each term by $A$ to get $A=1$.

I was given the equation $3x^2+6x+5=0$ and I decided to use the quadratic formula.

Now, for this equation

$a=3\quad b=6\quad c=5$

However, I divided the whole equation by $3$ to get $a=1 \quad b=2 \quad c=5/3$

I proceeded to solve using the quadratic formula. This led me to an incorrect answer on my given test. Why? I, feeling idiotic, concluded that I had changed the original equations value by dividing by $3$. Then why is it that you can do such an operation when completing the square? Aren't you changing the equation as well? These kinds of mistakes really demotivate me because I want to master algebra but I feel inadequate every time stuff like this happens.

This is what I did and was marked incorrect for:

This is what my teacher did:

• If the other side is zero then you can indeed divide by $3$, and then apply whatever method you want after that. – Ian Nov 29 '17 at 2:15
• If you got an incorrect answer, you must have done something else wrong. The quadratic formula will give the same answer, whether you divide through by 3 or not. – Gerry Myerson Nov 29 '17 at 2:16
• Are you trying to factorise the left hand side or solve the equation? The former result will be changed when you reduce the coefficient of the lead term, the latter will not. (In any case, note that the particular equation you quoted has no real roots). – Deepak Nov 29 '17 at 2:19
• I added pictures for reference. I was not trying to factor, I just wanted to simplify the coefficient of x^2, yes I know I included a fraction 5/3 doing so but eh. – Sphygmomanometer Nov 29 '17 at 2:25
• $(12/3)-(20/3)\ne-10$. As I said, you did something else wrong. – Gerry Myerson Nov 29 '17 at 2:28