# 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

• There is the $\pm$ sign... – Botond May 18 at 20:08
• 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 $x=\pm\sqrt{2}$. – Minus One-Twelfth May 18 at 20:08
• why isn't it represented as |x|=+-sqrt2 instead its written as x=+-sqrt2. Maybe I am just being picky or missing something – philalethesnew May 18 at 20:13
• would it be incorrect to write it as |x|=+-sqrt2 or does it need to be represented as x=+-sqrt2 – philalethesnew May 18 at 20:16
• 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 $x=\pm\sqrt{2}$. – Minus One-Twelfth May 18 at 20:21

## 4 Answers

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.

• :) very helpful – philalethesnew May 18 at 21:09
• 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? – philalethesnew May 19 at 1:04
• the third expression is missing the square root of b^2-4ac, not just b^2-4ac – philalethesnew May 19 at 1:20
• @philalethesnew Good catch, thanks. Edited. – 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?

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

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