2

I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $x \in \mathbb R$ and $a \neq 0$. Throughout this derivation, I will use the definition: $\sqrt{x^2}=|x|$.

Here is my derivation, and I have placed a $\color{red}{\dagger}$ next to the part that I would like some clarificaiton about:

$ax^2+bx+c =0 \iff x^2+\frac{bx}{a}+\frac{c}{a}=0 \iff (x+\frac{b}{2a})^2+(\frac{c}{a}-\frac{b^2}{4a^2})=0 \iff (x+\frac{b}{2a})^2+(\frac{4ac-b^2}{4a^2})=0 $

Bringing the right summand over to the right side of the equation:

$(x+\frac{b}{2a})^2=(\frac{b^2-4ac}{4a^2}) \iff \left| x+\frac{b}{2a}\right|=\sqrt{b^2-4ac}\cdot\sqrt{\frac{1}{4a^2}} \iff \left| x+\frac{b}{2a}\right|=\left|\frac{1}{2a} \right| \cdot \sqrt{b^2-4ac} \quad \quad \color{red}{\dagger}$

My confusion stems from how the final expression above is equivalent to the syntax "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" .

For $\color{red}{\dagger}$, we have 4 total cases:

  1. $ a \lt 0$ and $x+\frac{b}{a} \lt 0$

  2. $ a \lt 0$ and $x+\frac{b}{a} \geq 0$

  3. $ a \gt 0$ and $x+\frac{b}{a} \lt 0$

  4. $ a \gt 0$ and $x+\frac{b}{a} \geq 0$

Case 1: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$

Case 2: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Case 3: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Case 4: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$

From the four scenarios, is the way we get to "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" simply by noting that for a fixed $a$, we have $x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ or } x=\frac{-b-\sqrt{b^2-4ac}}{2a}$? ...where the '$\text{ or }$' here is denoting the logical or.

At this point, we define "$x=\pm \alpha$" as meaning $x = \alpha \text { or } x=-\alpha$...therefore meaning that $x=\beta\pm \alpha$ is equivalent to $x=\beta+\alpha \text{ or } x=\beta - \alpha$.

Is that the proper understanding?

Frightened of Sinusoids
  • 133
  • A method which solves $ax^2+bx+c=0$ whether or not $a$ is negative is one which produces solutions for $2x^2-3x+1=0$ and for $-2x^2+3x-1=0$ - thankfully, it gives the same solution set for both problems! – aschepler Apr 12 '22 at 03:04
  • @aschepler I am not sure I understand the point you are trying to make. Could you please elaborate? – Frightened of Sinusoids Apr 12 '22 at 03:27

3 Answers3

1

Your analysis is spot on, up until the point where you concluded that

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. \tag1 $$

I didn't bother reading after that, because there is no point. All that is necessary from the point of (1) above is to consider that any equation of the form

$$[f(x)]^2 = C$$

may be dissected as follows:

  • If $C < 0$ then there can't be any real roots. So, without loss of generality (assuming that you are only concerned with real roots), assume that $C \geq 0.$

  • Then, you must have that $f(x) = \pm \sqrt{C}$.
    Bang, that makes it: game over right there.
    So, no further analysis is needed.

This begs the question: what happens if $C < 0$.

Then, if you let $D = \sqrt{-C}$, then the equation becomes

$$[f(x)] = \pm (iD) ~: ~i = \sqrt{-1}.$$

user2661923
  • 35,619
  • 3
  • 17
  • 39
  • But how have you concluded that $f(x)=\pm \sqrt{C}$? And what does $\pm$ mean? – Frightened of Sinusoids Apr 12 '22 at 03:48
  • @FrightenedofSinusoids Because the equation $[f(x)]^2 = C$ can only be solved if $f(x)$ is such that its square is equal to $C$. By definition, the only values $a$ in $\Bbb{R}$ such that $a^2 = C$ are $a = \sqrt{C}$ and $a = -\sqrt{C}$. That is, this is what the definition of $\sqrt{C}$ stands for. Namely, (when $C \geq 0$), the specific non-negative value, whose square is equal to $C$. – user2661923 Apr 12 '22 at 03:51
  • To me, the fact that $a^2=C \implies a = \pm \sqrt{C}$ is a consequence of the definition: $\sqrt{x^2}=|x|$. For then we have that $a^2=C \implies |a| = \sqrt{C} \implies \Big [(a \lt 0 \rightarrow a=-\sqrt{C}) \land (a \gt 0 \rightarrow a=\sqrt{C}) \ \Big]$. From our initial assumption, we have that $a \neq 0$, which means that $a \lt 0 \lor a \gt 0$ is a tautology. We then have that $\implies a= \sqrt{C} \lor a=-\sqrt{C}$. That is to say, if $x\in S={a \in \mathbb R | a^2=C}$, then $x \in T = {a\in \mathbb R|a=\sqrt{C} \lor a=-\sqrt{C}}$. – Frightened of Sinusoids Apr 12 '22 at 04:26
  • Further, if $x \in T$, then $x \in S$. So we say that $x \in S \iff x \in T$...which is equivalently saying: $f(x)^2=C \iff f(x)=\pm \sqrt {C}$. Would you agree with this? – Frightened of Sinusoids Apr 12 '22 at 04:26
  • 1
    @FrightenedofSinusoids To tell you the truth, I'm unsure which came first, the chicken or the egg. I would phrase it that for $C \geq 0$, $\sqrt{C}$ symbolizes the unique non-negative value whose square is $C$. From this, one can conclude that $\sqrt{x^2} = |x|$. You are going at it from the other direction. Actually, I am unsure which of us is right. I simply always thought of it as asking: what does $\sqrt{C}$ symbolize? – user2661923 Apr 12 '22 at 04:30
  • thank you for the input – Frightened of Sinusoids Apr 12 '22 at 04:31
  • 1
    @Frightened of Sinusoids: Maybe it would help to notice that $a^2 = C$ is equivalent to $(a - \sqrt C)(a + \sqrt C) = 0$ (basic difference of squares factorization), which in turn is equivalent to "$a - \sqrt C = 0$ or $a + \sqrt C = 0$", which in turn is equivalent to "$a = \sqrt C$ or $a = -\sqrt C$", which is usually abbreviated as $a = \pm \sqrt C.$ – Dave L. Renfro Apr 12 '22 at 08:19
  • @DaveL.Renfro +1: Nice save. – user2661923 Apr 12 '22 at 08:39
0

From your working step $\left(x+\frac{b}{2a}\right)^2=\left(\frac{b^2-4ac}{4a^2}\right),$

take square root (permissible because non-negative);

then apply your definition $\sqrt{x^2}=|x|,$ which implies that $\sqrt{x^2}=\pm x$ (what does $\pm x$ mean);

then simplify;

finally, verify (by substitution) that your two solutions indeed satisfy $ax^2+bx+c=0.$

This considers two instead of four separate cases.

ryang
  • 38,879
  • 14
  • 81
  • 179
0

An appealing way of understanding why $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be written as $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ is to take a set perspective.

The statement $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be thought of as saying:

$x \in S$ where $S:=\left\{x\in \mathbb R:\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \right\}$

The statement $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ can be though of as saying:

$ x \in T$ where $T:=\left\{x\in \mathbb R:x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}\right\}$

The objective is to show that $x \in S \rightarrow x \in T$ and $x \in T \rightarrow x \in S$.

There is a lemma that follows directly from the definition of $|\cdot|$, which reads as:

$|x|=C \iff x=C \text{ OR } x= -C \quad (*_1)$

To prove the $\rightarrow$ direction, simply exhaust all possible values of $x \in \mathbb R$ by splitting it into the two cases of $x \geq 0$ and $x \lt 0$. The $\leftarrow$ direction is trivial.

Prove:$\quad x \in S \rightarrow x \in T$

By assumption, we have that $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} $. Taking the square root of both sides, and applying the definition of $\sqrt{\cdot}$, we have:

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

Applying our lemma $(*_1)$, we have the logical statement:

$$x+\frac{b}{2a}=\sqrt{\frac{b^2 - 4ac}{4a^2}} \quad\text { OR }\quad x+\frac{b}{2a}=-\sqrt{\frac{b^2 - 4ac}{4a^2}} \quad (*_2)$$

The symbol $\pm$ is defined to capture the meaning of $(*_2)$ and is equivalently written as:

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

Subtraction gives us: $x=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$, which means that $x \in T$.

Prove:$\quad x \in T \rightarrow x \in S$

By assumption, we have that $x=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$. This means that $x= \sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}\text { OR } x=-\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$.

In the first case, we have that $\left(x+\frac{b}{2a}\right)^2=\left(\sqrt{\frac{b^2 - 4ac}{4a^2}} \right)^2=\frac{b^2 - 4ac}{4a^2}$, which means that $x \in S$.

In the second case, we have that $\left(x+\frac{b}{2a}\right)^2=\left(-\sqrt{\frac{b^2 - 4ac}{4a^2}} \right)^2=\frac{b^2 - 4ac}{4a^2}$, which means that $x \in S$. Therefore, in all cases, we have that $x \in T$.

In conclusion, we have: $x \in S \iff x \in T$, which means that the two statements:

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

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

are equivalent.

S.C.
  • 4,984