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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: enter image description here

enter image description here This is what my teacher did: enter image description here

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  • $\begingroup$ If the other side is zero then you can indeed divide by $3$, and then apply whatever method you want after that. $\endgroup$
    – Ian
    Commented Nov 29, 2017 at 2:15
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    $\begingroup$ 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. $\endgroup$
    – Gerry Myerson
    Commented Nov 29, 2017 at 2:16
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    $\begingroup$ 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). $\endgroup$
    – Deepak
    Commented Nov 29, 2017 at 2:19
  • $\begingroup$ 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. $\endgroup$
    – Sphygmomanometer
    Commented Nov 29, 2017 at 2:25
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    $\begingroup$ $(12/3)-(20/3)\ne-10$. As I said, you did something else wrong. $\endgroup$
    – Gerry Myerson
    Commented Nov 29, 2017 at 2:28

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I am writing this answer so you can accept it and the site will show that the question has been answered

In the step:

$x = \frac{-2 \pm \sqrt{\frac{12}{3} - \frac{20}{3}}}{2} $

You proceeded to simplify $\frac{12}{3} - \frac{20}{3}$ as $-10$ which was the only mistake in your answer

These kinds of mistakes really demotivate me because I want to master algebra but I feel inadequate every time stuff like this happens.

Addressing this... To really master any field in maths, you need to understand the meaning behind what you are doing. One of the ways to do this is to know that everything except axioms are derived using logic and to understand that logic.

In case of the question you posted about finding the roots of a quadratic equation, you can do the following things to get better at them -

  • Understand that dividing a quadratic polynomial by any constant does not change its zeroes. This is because any quadratic polynomial can be expressed in terms of its zeroes as given below.

$p(x) = k(x-\alpha)(x-\beta)$

  • You can play around with these concepts on your own. For example, you can look at the graph of a quadratic polynomial in Desmos and see what happens when you change some values or multiply the polynomial by a constant.

  • Look at the derivation of the quadratic formula and understand what is happening or why things are working the way they are.

If something else comes to my mind, I will return to this answer and edit it.

Edit: Just saw that this question was 6 years and 1 month old. LOL

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