Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be  the square root of $4a^2$?
 A: $$ax^2+bx+c=0,~~~a\neq0$$
$$\begin{align}
&\Rightarrow x^2+\frac{b}{a}x+\frac{c}a=0\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)^2+\frac{c}{a}-\frac{b^2}{4a^2}=0\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{4a^2}\\
\\
&\Rightarrow\left(x+\frac{b}{2a} \right)=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\
\\
&\Rightarrow x+\frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{2|a|}\\
\end{align}$$
Case.(1) $a>0\Rightarrow |a|=a$
$$x+\frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{2a} \Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Case.(2) $a<0\Rightarrow |a|=-a$
$$x+\frac{b}{2a} =\pm\frac{\sqrt{b^2-4ac}}{-2a} \Rightarrow x=\frac{-b\mp\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Therefore, no matter what cases, you get the same formula.
A: Another way to approach it which does not rely on the study of the sign of $a$:
\begin{align*}
ax^{2} + bx + c = 0 & \Longleftrightarrow 4a^{2}x^{2} + 4abx + 4ac = 0\\\\
& \Longleftrightarrow (4a^{2}x^{2} + 4abx + b^{2}) = b^{2} - 4ac\\\\
& \Longleftrightarrow (2ax + b)^{2} = b^{2} - 4ac\\\\
& \Longleftrightarrow 2ax + b = \pm\sqrt{b^{2} - 4ac}\\\\
& \Longleftrightarrow x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
\end{align*}
Hopefully this helps!
A: I will make a guess that the proof you were looking at has the following equation in one of its steps:
$$ \left(x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2}. $$
This tells us that $x+\dfrac{b}{2a}$ is one of the square roots
of $\dfrac{b^2-4ac}{4a^2}.$
But it could be either the positive or negative square root.
The square roots of $\dfrac{b^2-4ac}{4a^2}$ are
$\dfrac{\sqrt{b^2 - 4ac}}{2a}$ and $-\dfrac{\sqrt{b^2 - 4ac}}{2a}.$
You can verify this by squaring each one:
\begin{align}
\left(\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}, \\
\left(-\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}. \\
\end{align}
Notice that $-\dfrac{\sqrt{b^2 - 4ac}}{2a}$ is just
$\dfrac{\sqrt{b^2 - 4ac}}{2a}$ with its sign flipped.
That is, one of these is a positive square root and one is a negative square root.
Which one is the positive root and which is the negative root depends on the sign of $a,$ but no matter which sign $a$ has we still have both roots, the positive and the negative.
Therefore we have found that the following statement is true:
$$
 \text{$x + \frac{b}{2a}$ is either
 $\frac{\sqrt{b^2 - 4ac}}{2a}$ or $-\frac{\sqrt{b^2 - 4ac}}{2a}$}.
$$
Another way to write the same statement is
$$ x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}. $$
Now just subtract $\dfrac{b}{2a}$ from both sides and you have the usual quadratic formula.
At no point in any of this did we ever take a square root of $4a^2.$
Instead, we looked at the expression $\dfrac{b^2-4ac}{4a^2}$ and found its square roots
— both of them — and included both roots in the answer.
