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.


  • $\begingroup$ There is the $\pm$ sign... $\endgroup$ – Botond May 18 at 20:08
  • $\begingroup$ You would be equating it to a "plus or minus" square root though, right? If so, this is fine. For example, if $x^2=2$, then $x=\pm\sqrt{2}$. $\endgroup$ – Minus One-Twelfth May 18 at 20:08
  • $\begingroup$ why isn't it represented as |x|=+-sqrt2 instead its written as x=+-sqrt2. Maybe I am just being picky or missing something $\endgroup$ – philalethesnew May 18 at 20:13
  • $\begingroup$ would it be incorrect to write it as |x|=+-sqrt2 or does it need to be represented as x=+-sqrt2 $\endgroup$ – philalethesnew May 18 at 20:16
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    $\begingroup$ It would be wrong to write $|x|=\pm\sqrt{2}$. This is because $|x|$ is never negative. What is correct is that $|x|=\sqrt{2}$. From this, you can conclude that $x=\pm\sqrt{2}$. $\endgroup$ – Minus One-Twelfth May 18 at 20:21

You seem to be asking why after the step $\;\left(x+\cfrac b{2a}\right)^2=\cfrac{b^2-4ac}{4a^2}\;$ , they don't go to

$$\left|x+\cfrac b{2a}\right|=\sqrt{\frac{b^2-4ac}{4a^2}}$$

But we actually do! To write the above is exactly the same as to write

$$x+\cfrac b{2a}=\pm\sqrt\frac{b^2-4ac}{4a^2}$$

just as $\;x^2=a\implies |x|=a\;\; (\,a\ge 0)\;$ is exactly the same as $\;x^2=a\implies x=\pm a\;\;(a\ge0)\;$ , under the assumption, of course, that once we take the + sign and the second time we take the - sign.

  • $\begingroup$ :) very helpful $\endgroup$ – philalethesnew May 18 at 21:09
  • $\begingroup$ Hey Don. Thanks for help but i have to clarify. Out of your three main expressions, should the right side of the 1st expression be over 4a^2, not 2a? Same with your second expression, it should be over 4a^2. Only in your 3rd expression should the denominator become 2a and the numerator should be the square root of b^2-4ac? Am I correct? $\endgroup$ – philalethesnew May 19 at 1:04
  • $\begingroup$ the third expression is missing the square root of b^2-4ac, not just b^2-4ac $\endgroup$ – philalethesnew May 19 at 1:20
  • $\begingroup$ @philalethesnew Good catch, thanks. Edited. $\endgroup$ – DonAntonio May 19 at 8:18

Well actually it is done.

$$ax^2 + bx + c = 0$$

$$\implies x^2+\frac{bx}{a} +\frac{c}{a} = 0$$

Now completing the square:

$$\left(x+\frac{b}{2a}\right)^2= \frac{b^2-4ac}{4a^2}$$

Now since we know that,

$$x^2 = a, \quad a > 0$$

has two solutions, $x=\pm a$.

Now if $b^2 - 4ac > 0$, there are two solutions indeed!

$$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$.

Does that help?

  • $\begingroup$ After you write (x+(b/2a))^2=(b^2-4ac)/4a^2 $\endgroup$ – philalethesnew May 18 at 20:21
  • $\begingroup$ shouldn't the square root of the left side be written as |x+(b/2a)| not just x+(b/2a) $\endgroup$ – philalethesnew May 18 at 20:22
  • $\begingroup$ Yeah, what about it? $\endgroup$ – Vizag May 18 at 20:22
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    $\begingroup$ Read Minus One Twelfths comment below the question. $\endgroup$ – Vizag May 18 at 20:23

You can look at it differently: $$ax^2+bx+c=0 \iff \\ x^2+\frac bax+\frac ca=0\iff \\ \left(x+\frac b{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}=0 \iff \\ \left(x+\frac b{2a}\right)^2-\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2=0 \iff \\ \left[\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}\right]\cdot \left[\left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}\right]=0 \iff \\ \left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}=0 \ \ \text{or} \ \ \left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}=0 \iff \\ x=\frac{-b+\sqrt{b^2-4ac}}{2a} \ \ \text{or} \ \ x=\frac{-b-\sqrt{b^2-4ac}}{2a} \iff \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$


When they get to point of having $$\bigl{(}x+\frac{b}{2a}\bigr{)}^2 =-\frac{c}{a}+\bigl{(}\frac{b}{2a}\bigr{)}$$ the square root of both sides has no absolute value sign on the left because the right side can be $\pm$. $$\biggl{|}x+\frac{b}{2a}\biggr{|}\ne\sqrt{-\frac{c}{a}+\biggl{(}\frac{b}{2a}\biggr{)}}$$

One complete derivation explained simply is shown here.

  • $\begingroup$ You should clarify on the usage of $\pm$, the square root does not(!) include $\pm$ just because it is the square root. $\endgroup$ – Hirshy May 19 at 20:24

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