Do we include operators in front of the coefficient when applying algebraic formulas to equations

Question

I'm doing some simple refresher on college algebra after a long time, but I've come to find some inconsistencies in my understanding of how to apply formulas, do we include operators in front of the coefficient when applying algebraic formulas to equations?

I'm struggling in particular with these two formulas:

Quadratic formula, operators included here:

$2x²+14x-196 = 0$, when applied to the formula $\frac{-b\pm\sqrt{b²- 4ac}}{2a}$, gives $\frac{-14\pm\sqrt{(14)^2-4(2)(-196)}}{2(2)}$. The answer here is $x = -14$, $x = 7$ [verified].

Notice that the coefficient "$c$" has to take on the negative operator infront of it to give the answer.

However:

If we be consistent and say yes, operators are to be included in formulas:

When using the formula (what is this algebraic formula call btw?) $(a-b)^2 = a^2-2ab+b^2$ , when applied to $(2-3)^2$, gives: $2^2-2(2)(-3)+(-3)^2 = 4+12+9=25$. Which is clearly the wrong answer.

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1) I'm struggling with this inconsistency, can anyone please help?

2) What is the general rule, when applying formulas, do we bring along operators in front of the number? (Yes/no)?

Answer 1

You want to apply $(a-b)^2=a^2-2ab+b^2$ to $(2-3)^2$. Good, no ask yourself

1. What's $a$?
2. What's $b$?

Here $a=2$ and $b=3$.

You can't apply the formula with $a=2$ and $b=-3$, because $2-(-3)=2+3$.

Answer 2

It might help to not think of negative signs as an operator in of itself but rather the negative sign being an indicator that you want to add the inverse of say $b$.

Then we have for example $(a-b)^{2} = (a+(-b))^{2} = a^{2} -2ab + b^{2}$. So we can see that $(a-b)^{2}$ is really just a special case of $(a+b)^{2}$.

Edit:

For further clarification, your quadratic formula example is completely consistent with the example above, since $c = (-196)$. Where the whole formula is then:

$ax^{2} + bx + c$ where $a = 2$, $b = 14$ and $c = (-196)$. Hope this helped.