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I have come across a stage of the proof:
$$ \left(x+\frac b{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$ How does $\left(x+\frac b{2a}\right)^2$ not equal $\pm x\pm \frac b{2a}$ when taking the square root?

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    $\begingroup$ $\sqrt{x^2}=|x|$ (not $\sqrt{x^2}=x$), so you have to take the non negative one between $x+b/2a$ and $-x-b/2a$. $\endgroup$ – Milly Oct 17 '14 at 14:16
  • $\begingroup$ I mean when you are taking the square root you get positive or negative x plus or minus b/2a. It's a stage of the proof of the quadratic formula. $\endgroup$ – xp73r Oct 17 '14 at 14:24
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    $\begingroup$ Milly already provided you with a satisfying answer in the comment above, I will here only add another approach here that exploits the property $a^2-b^2=(a+b)(a-b)$ $$\eqalign{ \left(x+\dfrac{b}{2a}\right)^2a&=\dfrac{b^2-4ac}{4a^2} \\ & \iff \left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a^2}=0 \\ & \iff \left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{\sqrt{b^2-4ac}}{2a}\right)^2=0 \\ & \iff \left(x+\dfrac{b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}\right)\left(x+\dfrac{b}{2a}-{{}{}{}{}{}} \dfrac{\sqrt{b^2-4ac}}{2a}\right)=0. \\ &(b^2-4ac\geqslant0) }$$ $\endgroup$ – Hakim Oct 17 '14 at 14:53
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    $\begingroup$ You know that $b^2\geq 4ac$. Then $(x+b/2a)^2$ is the square of both $\pm (x+b/2a)$. When considered as square of $x+b/2a$ gives $x+b/2a = \sqrt{b^2-4ac}/2a$. When considered as square of $-(x+b/2a)$ gives $-x-b/2a=\sqrt{b^2-4ac}/2a$. That is why you have the $\pm$ before the square root in the formula. $\endgroup$ – Milly Oct 17 '14 at 14:56
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Because $\sqrt{\left(x+\frac b{2a}\right)^2} =\left(x+\frac b{2a}\right) \text{ or } -\left(x+\frac b{2a}\right) $. Both $x$ and $\frac{b}{2a}$ must have the same signs.

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