<s>[INST] Is this valid when deriving quadratic equation? When deriving the quadratic formula, isn't the square root of $(x+\frac{b}{2a})^2$ the absolute value of $(x+\frac{b}{2a})$? It's usually just represented as $(x+\frac{b}{2a})$ without absolute value and then $\frac{b}{2a}$ is subtracted from left side and boom, theres the quadratic formula. I just don't understand why its not absolute value of $(x+\frac{b}{2a})$. For example the square root of $x^2$ is the absolute value of $x$, which is equal to $\pm{x}$. Sorry if this is confusing its more of a conceptual thing. Thank you in advance.
philalethesnew
 [/INST] You can look at it differently:
$$ax^2+bx+c=0 \iff \\
x^2+\frac bax+\frac ca=0\iff \\
\left(x+\frac b{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}=0 \iff \\
\left(x+\frac b{2a}\right)^2-\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2=0 \iff \\
\left[\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}\right]\cdot \left[\left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}\right]=0 \iff \\
\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}=0 \ \ \text{or} \ \ \left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}=0 \iff \\
x=\frac{-b+\sqrt{b^2-4ac}}{2a} \ \ \text{or} \ \ x=\frac{-b-\sqrt{b^2-4ac}}{2a} \iff \\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
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