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When I was in high school, they taught me to solve quadratic equations with this formula:

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

While reading Mathematics for the Nomathematician, Morris Kline suggests that a quadratic equation should be solved by making another equation with half of the coefficient of $b$:

$$y=x+\frac{b}{2}$$

Then:

$$x=y-\frac{b}{2}$$

and then proceed with the substitution on the quadratic equation, I guess I don't need to describe the rest as it may be a standard procedure.

Are there benefits on using the first formula? I also imagine that they could be related somehow, but I still can't see this relation.

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    $\begingroup$ Unless your quadratics are different from everyone else's, don't you want $-4ac+b^2$ under that square root? $\endgroup$ Aug 3, 2012 at 23:13
  • $\begingroup$ Assuming $a=1$, Kline is describing "completing the square". $\endgroup$ Aug 3, 2012 at 23:14
  • $\begingroup$ @Robert: Ah - this all suddenly makes sense to me. $\endgroup$
    – davidlowryduda
    Aug 3, 2012 at 23:14
  • $\begingroup$ @mixedmath Yes. I missed something. $\endgroup$
    – Red Banana
    Aug 3, 2012 at 23:16
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    $\begingroup$ About the $\sqrt{4ac+b^2}$: Conceivably they wrote the equation as $ax^2+bx=c$. $\endgroup$ Aug 3, 2012 at 23:55

3 Answers 3

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It's the same thing. What Kline is doing is what's in general known as a depression, which is a variable substitution done for a polynomial of degree $n$, such that the resulting polynomial has no term of degree $x^{n-1}$.

One way to see how depression works is to consider the Vieta formulae, which in the case of the quadratic $ax^2+bx+c=a(x-x_1)(x-x_2)$ looks like this:

$$\begin{align*} x_1 x_2 &= \frac{c}{a}\\ x_1 + x_2 &= -\frac{b}{a} \end{align*}$$

From this, if the sum of the roots is $-\dfrac{b}{a}$, then the mean of the roots is $-\dfrac{b}{2a}$. Thus, the depression substitution

$$y=x+\frac{b}{2a}$$

can be geometrically interpreted as shifting the parabola corresponding to your quadratic $ax^2+bx+c$ such that it is "centered" about the origin, and the two roots are laid out symmetrically.

If we depress our original quadratic, we get

$$\require{cancel}\begin{align*} ax^2+bx+c&=a\left(y-\frac{b}{2a}\right)^2+b\left(y-\frac{b}{2a}\right)+c\\ &=a\left(y^2-\frac{b}{a}y+\frac{b^2}{4a^2}\right)+b\left(y-\frac{b}{2a}\right)+c\\ &=ay^2\cancel{-by}+\frac{b^2}{4a}\cancel{+by}-\frac{b^2}{2a}+c\\ &=ay^2+\frac{b^2}{4a}-\frac{2b^2}{4a}+c\\ &=ay^2-\frac{b^2}{4a}+c=ay^2-\frac{b^2-4ac}{4a} \end{align*}$$

and I'm sure you know how easy it is to solve the equation

$$ay^2-\frac{b^2-4ac}{4a}=0$$

Having solved for $y$, you undo the depression you did, which means you have to add the term $-\dfrac{b}{2a}$ to get the actual roots you want. That's where that part of the quadratic formula comes from.

In short, I would not say that Kline's method is the most expedient, but it at least looks to me that this slower method allows for more cogitation on what the steps are supposed to "mean", as opposed to a lazy plug-and-chug.

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    $\begingroup$ This is ... great! $\endgroup$
    – Gigili
    Aug 4, 2012 at 3:05
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    $\begingroup$ I understand it (the process of solving a quadratic equation) mechanicaly, I can solve a quadratic equation but the procedure you explained is still a little obscure to me: It seems I can understand a little of this heuristic but no very deeply. Can you suggest some reading about it? $\endgroup$
    – Red Banana
    Aug 4, 2012 at 3:44
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    $\begingroup$ I've also googled a little about depresion(Using "depresion mathematics" as search subject), and the only thing I've found was about people getting depressed while studying/doing mathematics - The search couldn't be deeper because of my slow internet. $\endgroup$
    – Red Banana
    Aug 4, 2012 at 3:59
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    $\begingroup$ @Gustavo, the search term you should be using is polynomial depression; you'll see that the general idea is also used in deriving explicit solutions for the cubic and quartic equations. $\endgroup$ Aug 4, 2012 at 4:47
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    $\begingroup$ I'm also happy with something I just read: When doing mathematics, Stubbornness is more important than inteligence. It's cool because it's the only thing I have. Haha $\endgroup$
    – Red Banana
    Aug 6, 2012 at 4:41
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Let's try to follow Kline's advice (and all the time we assume the coefficients $\,a,b,c\,$ are real).

We have $\,ax^2+bx+c=0\,\,,\,a\neq 0\,$ and we make the substitution $$y:=x+\frac{b}{2}\Longleftrightarrow x = y-\frac{b}{2}$$ and from here, substituting in the equation

$$a\left(y-\frac{b}{2}\right)^2+b\left(y-\frac{b}{2}\right)+c=0$$ ...and now? But for messing up the equation's variable I can't see any real advance or simplification.

I think what he actually meant, or should have meant, is the following, which is only the rather well-known method of completing the square (CS): $$X^2\pm BX=\left(X \pm \frac{B}{2}\right)^2-\frac{B^2}{4}\,\,,\,\text{and from here}$$ $$0=ax^2+bx+c\,\Longrightarrow\,x^2+\frac{bx}{a}=-\frac{c}{a}\stackrel{CS}\Longrightarrow\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}=-\frac{c}{a}\,\Longrightarrow$$ $$\Longrightarrow\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\,\,\,\text{(see the numerator?!)}\,\,\stackrel{\text{sq.rt. in both sides}}\Longrightarrow$$ $$\Longrightarrow x_{1,2}+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\,\Longrightarrow\,x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ .

So, as you can see, the benefits is that you need that expression to solve quadratics..as simple as that!

Of course, the number $\,\Delta:=b^2-4ac\,$ , called the quadratic's discriminant serves to know beforehand about the above equation's possible solutions: $$\begin{align*}\Delta>0\Longrightarrow & \,\text{there exist two different real solutions to the equation}\\ \Delta=0\Longrightarrow & \,\text{there exists only one unique real solution to the equation}\\ \Delta<0\Longrightarrow & \,\text{the equation has no real solutions}\end{align*}$$

In the last case above there exist two conjugate complex non-real solutions (disregard this if you haven't yet studied complex numbers)

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  • $\begingroup$ What's the meaning of $:=$? It's the first time I see it. $\endgroup$
    – Red Banana
    Aug 4, 2012 at 16:16
  • $\begingroup$ := means the left side is defined by the right one $\endgroup$
    – DonAntonio
    Aug 11, 2012 at 1:16
  • $\begingroup$ "there exists only one unique real solution to the equation" - slightly strange wording, like, there are actually two solutions, but the other one isn't unique? :) $\endgroup$ Aug 11, 2012 at 10:23
  • $\begingroup$ @BenMillwood, I think that's a language fix. Where I live (Israel), even at 7th-8th grade in junion high school, children are taught to say "there are (there aren't) REAL solution(s) to this quadratic equation", apparently as preparation for complex numbers later in high school. The meaning, though, is correct, I believe, under the proper interpretation: one unique real solution means there's only one number that solves the quadratic eq. $\endgroup$
    – DonAntonio
    Aug 11, 2012 at 10:37
  • $\begingroup$ @DonAntonio: I see where you're coming from, but I think it's strange to use "unique" here - what are you saying that isn't said by "there exists only one real solution to the equation"? If you think the latter is untrue because there is a repeated solution, then you think there are two solutions, but then you say that there is one unique solution, which just leads me to wonder if the other one is unique, too... $\endgroup$ Aug 11, 2012 at 10:47
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The given answers are good, but let me show you the "typical" way of proving the original formula. The goal is to find the roots of the quadratic, so that:

$$Ax^2 + Bx + C = 0$$

The $C$ constant is a little messy, so let's move it to the other side:

$$Ax^2 + Bx = -C $$

Now, on the left-hand side, we can "complete the square", i.e., find out a value that will allow us to write the left-hand side as $(\_ + \_)^2$. This is easiest to do if the $x^2$ is by itself. So we'll divide both sides by $A$ before continuing.

$$x^2 + \frac{B}{A}x = -\frac{C}{A}$$

Now, when completing the square, the middle term is key. The middle term is twice the left-hand times the right-hand side. Here, since we only have $x^2$ on the left, the left-hand side of our square will obviously be $x$. The middle term is $\frac{B}{A}x$. The $x$ obviously comes from the left-hand side of our square. $\frac{B}{A}$ comes from the right side, but, remember, the middle term gets doubled in squares. Therefore, the right-hand side of the square should be $\frac{B}{2A}$, which means that the right-hand term will be the square of that, or $\frac{B^2}{4A^2}$. If we add this term to both sides we get:

$$ x^2 + \frac{B}{A}x + \frac{B^2}{4A^2} = -\frac{C}{A} + \frac{B^2}{4A^2} $$

The left-hand side is now a square. So we can simplify it to:

$$ \left(x + \frac{B}{2A}\right)^2 = -\frac{C}{A} + \frac{B^2}{4A^2}$$

Square-rooting both sides gets:

$$ x + \frac{B}{2A} = \pm \sqrt{-\frac{C}{A} + \frac{B^2}{4A^2}} $$

Subtracting $\frac{B}{2A}$ from both sides gets:

$$ x = \pm \sqrt{-\frac{C}{A} + \frac{B^2}{4A^2}} - \frac{B}{2A} $$

This is a quadratic formula, but not in the form we are used to seeing. However, it is just a matter of rearranging to get the quadratic formula.

First, let's simplify the interior of the square root. We will start by putting everything over the same denominator ($4A^2$).

$$ x = \pm \sqrt{\frac{-4AC + B^2}{4A^2}} - \frac{B}{2A} $$

The bottom part of the fraction is actually a square, so it can be pulled out of the square root.

$$ x = \frac{1}{2A} \pm \sqrt{-4AC + B^2} - \frac{B}{2A} \\ x = \frac{\pm \sqrt{-4AC + B^2}}{2A} - \frac{B}{2A} \\ x = \frac{\pm \sqrt{-4AC + B^2} - B}{2A} $$

This is the formula, just in the wrong order. Changing the order (both inside and outside the square root) yields:

$$ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $$

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